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\begin{document}
\title{Math 953: Algebraic Geometry}
\author{Brian Harbourne}
\address{
Department of Mathematics\\
University of Nebraska\\
Lincoln, NE 68588-0130 USA}
\email{bharbour@math.unl.edu}
\thanks{}
\date{May 7, 2011}
\begin{abstract}
These are notes for an introductory course on Algebraic Geometry
taught at the University of Nebraska-Lincoln, Spring 2011.\end{abstract}
\maketitle
\mysection{January 12, 2011}\label{lect1}
Fundamentally, Algebraic Geometry is the study of the solution sets of systems of polynomial equations.
So it seems appropriate to start at the beginning, as it is now known.
\subsection{In the beginning (of recorded history)}
The British Museum has a cuneiform table
[see \url{http://www.malhatlantica.pt/mathis/Babilonia/BM13901.htm} for a photo],
BM 13901 (ca. 1700 BC) which has a problem and solution which reads:
\begin{prob}
I totalled the area and (the side of) my square: it is 0;45. You put down 1, the unit. You break in half 1.
You multiply 0;30 and 0;30. You add 0;15 to 0;45. The square root of 1 is 1. Subtract 0;30 that you
multiplied (with itself) from 1 and 0;30 is (the side of) the square.
[see \url{http://www.maa.org/reviews/lsahoyrup.html}]
\end{prob}
In the sexagesimal (i.e., base 60) number system used by the Babylonians, 0;45 means 45/60.
Thus, in more familiar terms, the problem is to find $x$, given $x^2+1x=3/4$. To find $x$, the
tablet says to take half the coefficient of $x$, i.e., half of 1 which is $1/2=0;30$, square it and
add to both sides to get $x^2+x+(1/2)^2=3/4+1/4$, or $(x+1/2)^2=1$.
We have thus completed the square; taking square roots gives $x+1/2=1$, and subtracting
$1/2=0;30$ from both sides gives $x=1/2$; i.e., $x=0;30$. Never mind about the other solution
$x=-3/2$; since it's negative, it was regarded as being fictitious to the Babylonians.
In any case, 4700 years ago, the Babylonians knew how to solve a quadratic equation by completing the square.
In modern terms, if
$$x^2+ax=b,$$
then
$(x+a/2)^2=b+(a/2)^2$ so
$$x=-a/2\pm\sqrt{b+(a/2)^2}.$$
\subsection{Some time later in Medieval Italy}
In Italy in the Middle Ages, academicians put on public scholarly competitions.
(This lives on today in the system of concorsi in modern Italy;
to earn a university position in Italy, one must win a concorso,
a competition among the various candidates, except today
the problems are posed by a panel of professors, not by the candidates to
each other. However, recent reforms may result in this system finally being abandoned.
These competitions typically involved
two people who each would propose problems for the other to solve. The one who could better solve the other's
problems won the competition. You could make your reputation and establish your livelihood
by doing well in these competitions. Thus if you made a discovery, such as how to solve certain
equations, you had a strong incentive to keep it a secret, so you could pose problems
to opponents that perhaps only you knew how to do.
An important such competition occurred in 1535. A certain Antonio Maria Fior challenged
a certain self-taught Niccolo Fontana (who, due to a saber wound to his jaw suffered as a child
when French soldiers attacked his town,
was nicknamed Tartaglia, pronounced Tartalya and meaning Stutterer)
[see p. 99 of {\em The Ellipse: A Historical and Mathematical Journey}, by Arthur Mazer, Wiley, 2010].
Fior posed thirty problems to Fontana, each of which involved solving a
cubic equation of the form $ax^3+cx+d=0$. One of the problems, for example, was the following:
\begin{prob}
Two men together gain 1000 ducats. The gain of the first is the cube root of the gain of the second. What is
the gain of each?
\end{prob}
I.e., if $x$ is the gain of the first, then $x^3$ is the gain of the second so $x^3+x=1000$ and we must solve for $x$.
The source of the solution of the cubic is uncertain. Somehow Fior's teacher, Scipione del Ferro,
a professor in Bologna, came to know how to solve certain cases of cubic equations, and he
shared his method with Fior. Fior's challenge motivated Fontana to figure out those cases
and possibly others also, with the result that Fontana prevailed in the competition.
This attracted the attention of one Gerolamo Cardano, who pestered Fontana to reveal the secret.
Eventually, under an oath of secrecy, Fontana shared his solution with Cardano,
who (possibly after tracking down del Ferro's solution, thereby no longer being bound by
the oath) then published it anyway in his 1545 work {\em Ars Magnus},
but giving credit to del Ferro, Fior and Fontana. (Cardano seems to have been at
the forefront of the tradition of making one's reputation by publishing what one knows, rather
than keeping it secret.)
Note that one can always reduce a cubic equation $ax^3+bx^2+cx+d=0$ to the form $u^3+eu+f=0$: first divide
by $a$, then substitute $u-b/(3a)$ in for $x$ to eliminate the term of degree $3-1=2$ (this is analogous to
completing the square, which eliminates the term of degree $2-1=1$).
Now use the identity $(s+t)^3=s^3+3s^2t+3st^2+t^3$, rewritten as:
$$(s+t)^3-3st(s+t)=s^3+t^3.$$
This means $u=s+t$ is a solution of $u^3+eu+f=0$
if we choose $s$ and $t$ to solve the following system of equations:
$$-3st=e$$
and
$$s^3+t^3=-f.$$
But this is easy: substitute $s=-e/(3t)$ into the second equation and clear denominators to get
$$(t^3)^2+ft^3-e^3/27=0.$$
The Babylonians showed us how to solve this for $t^3$; then take cube roots to get $t$.
Knowing $t$ gives us $s$, since $s=-e/(3t)$, and from this we get $u=s+t$ and finally $x=u-b/(3a)$.
Cardano's student and former servant, Lodovico Ferrari, found the solution to a quartic equation.
As usual, it is enough to solve $x^4+cx^2+dx+e=0$. His solution is to rewrite
it as $x^4+2cx^2+c^2=cx^2-dx+c^2-e$, or
$$(x^2+c)^2 = cx^2-dx+c^2-e.$$
Now insert a supplementary variable $y$:
$$(x^2+c+y)^2 = (c+2y)x^2-dx+(c^2+2cy+y^2-e).$$
Note that the RHS is a quadratic polynomial in $x$.
It would be nice if it were of the form $(\alpha x+\beta)^2$, a perfect square.
The idea is to choose a value of $y$ to make this happen.
So choose $y$ so that the RHS is a perfect square in $x$; i.e.,
choose $y$ so that $(c+2y)x^2-dx+(c^2+2cy+y^2-e)$ has
discriminant $d^2-4(c+2y)(c^2-e+2cy+y^2)$
equal to 0. This involves solving a cubic, which thanks to del Ferro/Fontana/Cardano
we know how to do. Thus we get an equation of the form
$$(x^2+c+y)^2=(\alpha x+\beta)^2$$
hence $x=\pm\sqrt{-(c+y)\pm(\alpha x+\beta)}$.
\newpage
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex1.1}
Find an exact solution for $x^3+x=1000$.
\end{exer}
\begin{proof}[Solution by Ashley Weatherwax (also presented by her in class)]
Using the identity $(r+s)^3 - 3rs(r+s) = r^3 + s^3$, we get that $x = r+s$ is a solution to $x^3 + x = 1000$
whenever $$1 = -3rs \text{ and } r^3 + s^3 = 1000$$
Solving for $r$ in the first equation, we get $r = \frac{-1}{3s}$. Plugging this into the second equation, we get
$$(\frac{-1}{3s})^3 + s^3 = 1000$$
Multiplying through by $s^3$ and then substituting $v = s^3$, we get the above equation is quadratic in $v$:
$$\frac{-1}{27} + v^2 = 1000v \Rightarrow v^2 - 1000v - \frac{1}{27}=0$$
Using the quadratic equation to solve for $v$, we get
$$v = \frac{1000 \pm \sqrt{(1000)^2 - 4(1)(\frac{-1}{27})}}{2} = \frac{4500 \pm \sqrt{20250003}}{9}$$
But recall that $v = s^3$, so in fact we have
$$s = \sqrt[3]{\frac{4500 \pm \sqrt{20250003}}{9}}$$
Finally, recall that $r = \frac{-1}{3s}$, and so
$$r+s = \frac{-1}{3\sqrt[3]{\frac{4500 \pm \sqrt{20250003}}{9}}} + \sqrt[3]{\frac{4500 \pm \sqrt{20250003}}{9}}$$
As a final note, both solutions (one with both $+$ and one with both $-$) are equal, and are approximately 9.96666679053.
\end{proof}
\begin{exer}\label{ex1.2}
Find an exact solution for $x^3+3x^2-3x-11=0$.
\end{exer}
\begin{proof}[Solution by Nora Youngs]
To use the method given in class:
First we must remove the quadratic term, so we perform a variable change: Let $x=u-1$.
\begin{eqnarray*}
x^3+3x^2-3x-11&=&(u-1)^3+3(u-1)^2-3(u-1)-11\\
\,&=&u^3-3u^2+3u-1+3u^2-6u+3-3u+3-11\\
\,&=&u^3-6u-6\\
\end{eqnarray*}
The new equation is of the form $u^3+cx+d=0$ where $c=-6$, $d=-6$.
Thus, following again from the method $u=r+s$ is a solution if $s$ is a solution to $(s^3)^2-6x-\frac{(-6)^3}{27}=0$ and $r=\frac{6}{3(s)}$.
Solving the quadratic:
\begin{align*}
(s^3)^2-6x+8 &=0\\
(s^3)^2-6x+9 &= 1\\
(s^3-3)^2 &=1\\
s^3-3 &= 1\\
s^3 &= 4\\
s &=\sqrt[3]{4}\\
\end{align*}
Then $r=\frac{6}{3\sqrt[3]{4}}=\frac{6\sqrt[3]{16}}{12}=\frac{\sqrt[3]{16}}{2}=\sqrt[3]{2}$.
So $u=\sqrt[3]{4}+\sqrt[3]{2}$ is a solution to $u^3-6u-6=0$.\\
And therefore, $x=u-1=\sqrt[3]{4}+\sqrt[3]{2}-1$ is a solution to the original equation. \\
Note also that if we take the other option, $\sqrt 1$ to be $-1$ instead of $1$, we have $s=\sqrt[3]{2}$, $r=\sqrt[3]{4}$,
and the solution is $x=\sqrt[3]{2}+\sqrt[3]{4}-1$, the same as before.\\
\end{proof}
\begin{exer}\label{ex1.3}
Find an exact solution for $x^4+4x^3+10x^2-76x-104=0$.
\end{exer}
\begin{proof}[Solution presented in class by Zheng Yang but typed up by BH]
Making the substitution $u=x+1$ converts $x^4+4x^3+10x^2-76x-104=0$ to
$u^4+4u^2-88u-21=0$. Rewriting gives
$u^4+8u^2 +16 = 4u^2+88u+21+16$ or $(u^2+4)^2=4u^2+88u+37$.
Now introduce a new variable $y$ as follows:
$$(u^2+4+y)^2=(4+2y)u^2+88u+37+8y+y^2$$
We pick $y$ so that $(4+2y)u^2+88u+(37+8y+y^2)$ is a perfect square; i.e., we need
the discriminant $88^2-4(37+8y+y^2)(4+2y)$ to vanish. This is a cubic
(known as the resolvent cubic), which we now know how to solve.
One solution is $y=6$. Using this value of $y$ the equation
$(u^2+4+y)^2=(4+2y)u^2+88u+37+8y+y^2$ becomes
$(u^2+10)^2=16u^2+88u+121$ or
$$(u^2+10)^2=(4u+11)^2.$$
Thus $u^2+10=\pm(4u+11)$. Taking the plus sign gives
$u^2-4u-1=0$ which has root $u=2\pm\sqrt{5}$, and thus $x=1\pm\sqrt{5}$
is a root of the original polynomial.
\end{proof}
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\mysection{January 14, 2011}\label{lect2}
\noindent{\bf The scene expands to northern Europe.}
In 1824, Niels Henrik Abel and Paolo Ruffini independently prove that there is no
general formula for solutions of quintic equations just in terms of the usual field operations
(addition, subtraction, multiplication and division)
and taking radicals. However, early in the 1800s it was proved that any non-constant polynomial
with complex coefficients has a root (this is usually attributed to C.~F.~Gauss,
but it is not clear that he actually had the first correct proof):
\begin{thm}[Fundamental Theorem of Algebra] If $f\in\CC[x]$ is not a constant, then it has a root
(i.e., the field of complex numbers $\CC$ is algebraically closed).
\end{thm}
\begin{cor}
If $f\in\CC[x]$ has degree $\deg(f)=d>0$, then $f(x)=a(x-c_1)\cdots(x-c_d)$
for constants $a,c_1,\ldots,c_d\in\CC$; i.e., $f$ has $d$ roots, counted with multiplicity.
\end{cor}
\begin{proof} Let $f(c)=0$, so $c$ is a root. Using polynomial division, we have $f(x)=q(x)(x-c)+r(x)$
where $r$ is a constant (since the remainder term can always be taken to have degree less than the divisor,
$x-c$). Since $f(c)=0$, we have $r=0$, so $f(x)=q(x)(x-c)$, but $\deg(q(x))=\deg(f(x))-1$.
The result follows by induction.
\end{proof}
Thus a polynomial of degree $d$ determines a choice (unique up to reindexing)
of $d$ constants $c_1,\ldots,c_d$
(repeats allowed), and any such $d$ choices of constants determines a unique monic polynomial
$(x-c_1)\cdots(x-c_d)$ of degree $d$.
The following definition formalizes this interplay between algebra and geometry:
\begin{defn}
Given a subset $S\subseteq\CC[x]$, let $Z(S)$ be the set of simultaneous solutions to $f(x)=0$
for all $f\in S$; i.e., $Z(S)=\{c\in\CC: f(c)=0\ {\rm for\ all }\ f\in S\}\subseteq\CC$ is the {\em zero set} of $S$.
And given any subset $V\subset \CC$, let $I(V)\subseteq \CC[x]$ be the set of all $f\in\CC[x]$
such that $f(c)=0$ for all $c\in V$.
\end{defn}
Note that if $S\subseteq \CC[x]$, there is a unique smallest ideal, denoted $I(S)$, that contains $S$, called
the ideal generated by $S$. The intersection of any collection of ideals is itself an ideal;
$I(S)$ is the intersection of all ideals that contain $S$. Alternatively, $I(S)$ is the set of all
finite sums of the form $\sum_i g_if_i$ where $g_i\in\CC[x]$ and $f_i\in S$.
When $n>1$, note that $\CC[x_1,\ldots,x_n]$ is a UFD but not a PID. Nevertheless, we can define
$Z(S)$ and $I(S)$ for $S\subseteq \CC[x_1,\ldots,x_n]$ as before, and likewise also
$I(V)$ for $V\subseteq \CC^n$, and we have $Z(S)=Z(I(S))$ exactly as before.
\begin{defn}
We say a subset $V\subseteq\CC^n$ is an {\em algebraic subset} of $\CC^n$
if $V=Z(S)$ for some subset $S\subseteq\CC[x_1,\ldots,x_n]$.
\end{defn}
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex2.1} Let $V\subseteq \CC^n$.
Show that $I(V)$ is an ideal.
\end{exer}
\begin{proof}[Solution by Philip Gipson]
Certainly $0\in I(V)$ and if $f\in I(V)$ then $-f(v)=-0=0$ for all $v\in V$ and so $-f\in I(V)$.
If $f,g\in I(V)$ then for all $v\in V$ we have $f(v)+g(v)=0+0=0$ and so $f+g\in I(V)$.
Thus $I(V)$ is a group under addition.
If $f\in I(V)$ and $h\in \CC[x_1,\ldots,x_n]$, then for all $v\in V$
we have that $h(v)f(v)=0\cdot0=0$ and so $hf\in I(V)$.
Therefore $I(V)$ is an ideal.
\end{proof}
\begin{exer}\label{ex2.2}
If $S\subseteq \CC[x]$, show that $Z(S)=Z(I(S))$.
\end{exer}
\begin{proof}[Solution by Jason Hardin]
Suppose that $\alpha\in Z(S)$, so that $f(\alpha)=0$ for all $f\in S$.
Since we know that $I(S)=\left\{\displaystyle\sum_{i=1}^ng_if_i\ |\ g_i\in\mathbb{C}[x],f_i\in S, n\in\mathbb{N}\right\}$,
given any $\displaystyle\sum_{i=1}^ng_if_i\in I(S)$, we have
$\left(\displaystyle\sum_{i=1}^ng_if_i\right)(\alpha)=\displaystyle\sum_{i=1}^ng_i(\alpha)f_i(\alpha)=0$,
as $f_i\in S$ and thus $f_i(\alpha)=0$ for $i=1,\ldots, n$. So $\alpha\in Z(I(S))$ and $Z(S)\subseteq Z(I(S))$.
Conversely, if $f(\alpha)=0$ for all $f\in I(S)$, then of course
$f(\alpha)=0$ for all $f\in S$, as $S\subseteq I(S)$.
So $Z(I(S))\subseteq Z(S)$, and equality follows.
\end{proof}
\begin{exer}\label{ex2.3}
Show that the algebraic subsets of $\CC$ are precisely the
finite subsets together with $\varnothing$ and $\CC$; in particular,
no infinite proper subset of $\CC$ is an algebraic subset of $\CC$.
\end{exer}
\begin{proof}[Solution by Kat Shultis]
It is clear that
$\varnothing=Z(1)$ and $\CC = Z (0)$. Let $A \subseteq \CC$ be a finite set and write
$A = \{c_1, \ldots, c_n\}$. Let $f(x) = (x - c_1) \cdots (x - c_n)$ so that
$Z(f ) = A$. Thus we know that
$\varnothing$, $\CC$, and any finite subset of $\CC$ are algebraic subsets of
$\CC$. Now, let $A=Z(I)$ for
some ideal $I$ of $\CC[x]$ and assume that $A$ is an infinite set.
Choose any $f \in I$. Then for every $p \in A$, we have that
$f(p) = 0$. However, any non-zero polynomial $f$, has exactly the same
number of zeros as its degree, and since the degree of a non-zero
polynomial is finite, we cannot
have that $f$ vanishes at infinitely many points. Thus, if $A$ is an infinite
but proper subset of $\CC$, we
cannot have $A$ as an algebraic subset of $\CC$.
\end{proof}
\begin{exer}\label{ex2.4}
Give an example of an infinite proper subset of $\CC^2$ which is an algebraic subset.
\end{exer}
\begin{proof}[Solution by Doug Heltibridle]
Let $S=\{x+y\}$, then $Z(S)$ has infinitely many elements, but it is a proper subset of $\mathbb C^2$ as $(1,1)$
and $(-1,-1)$ are not solutions of $x+y$. Thus $V=Z(S)$ is an infinite proper subset of $\mathbb C^2$,
which is algebraic by construction.
\end{proof}
\begin{exer}\label{ex2.5}
Show that $(x_1,x_2)\subset\CC[x_1,x_2]$ is not a principal ideal.
\end{exer}
\begin{proof}[Solution]
Say $(x_1,x_2)=(f)$ for some $f\in \CC[x_1,x_2]$.
Clearly $f$ is not 0, and
$f$ divides $x_1$. Thus $\deg(f)\leq \deg(x_1)=1$. I.e., $f=ax+b$ for constants $a,b\in\CC$,
so either $f=ax_1$ and $a$ is non-zero, or $f=b$ and $b$ is non-zero.
In the latter case $(f)=\CC[x_1,x_2]$ hence $(x_1,x_2)\subsetneq (f)$, while in the
former case $(f)=(x_1)$, but $x_1$ does not divide $x_2$, so $x_2\not\in (f)$,
hence again $(f)\neq (x_1,x_2)$.
\end{proof}
\begin{exer}\label{ex2.6}
Let $I\subseteq J\subseteq\CC[x_1,\ldots,x_n]$ be ideals and let $V\subseteq W\subseteq \CC^n$.
Show that $Z(J)\subseteq Z(I)$ and that $I(W)\subseteq I(V)\subseteq \CC[x_1,\ldots,x_n]$.
\end{exer}
\begin{proof}[Solution]
If $p\in Z(J)$, then $f(p)=0$ for all $f\in J$, and since $I\subseteq J$ it follows that
$f(p)=0$ for all $f\in I$, hence $p\in Z(I)$, so $Z(J)\subseteq Z(I)$.
If $f\in I(W)$, then $f(p)=0$ for all $p\in W$, and since $V\subseteq W$ it follows that
$f(p)=0$ for all $p\in V$, hence $f\in I(V)$, so $I(W)\subseteq I(V)$.
\end{proof}
\begin{exer}\label{ex2.7}
Let $I_1,\ldots,I_r\subseteq\CC[x_1,\ldots,x_n]$ be ideals.
Define $\Pi_jI_j=I_1\cdots I_r$ to be the ideal generated by all elements of the form
$f_1\cdots f_r$, where $f_j\in I_j$ for $1\leq j\leq r$.
Show that $Z(\cap_jI_j)=Z(\Pi_jI_j)=\cup_jZ(I_j)$.
\end{exer}
\begin{proof}[Solution by Becky Egg]
Let $p \in Z( \cap_{j=1}^r I_j)$, but suppose that $p \notin \cup_{j=1}^r Z(I_j)$. So in particular,
$p \notin Z(I_j)$ for $1 \leq j \leq r$. So for each $j$, $1 \leq j \leq r$, there exists $f_j \in I_j$ such
that $f_j(p) \neq 0$. Note that $f_1 \cdots f_r \in \cap_{j=1}^r I(J)$, and so we have
\[(f_1 \cdots f_r)(p)=f_1(p)\cdots f_r(p)=0,\]
a contradiction, as each $f_i(p) \neq 0$. Thus $p \in \cup_{j=1}^r Z(I_j)$, and we have
$Z( \cap_{j=1}^r I_j) \subseteq \cup_{i=1}^n Z(I_r)$.
Now let $p \in \cup_{j=1}^r Z(I_j)$, and suppose that $p \in Z(I_k)$ for some $k$ with $1 \leq k \leq r$.
Choose $f_j \in I_j$ for $1 \leq j \leq r$, and note that
\[(f_1\cdots f_k \cdots f_r)(p)=f_1(p) \cdots f_k(p) \cdots f_r(p)=0,\]
as $f_k(p)=0$. Since $\prod_{j=1}^r I_j$ is generated by elements of the form $f_1 \cdots f_r$,
we have that $f(p)=0$ for all $f \in \prod_{j=1}^r I_j$, and hence $p \in Z(\prod_{j=1}^r I_j)$, so
$\cup_{j=1}^r Z(I_j)\subseteq Z(\prod_{j=1}^r I_j)$.
Finally, let $p \in Z( \prod_{j=1}^r I_j)$, and let $f \in \cap_{j=1}^r I_j$. Then $f^r \in \prod_{j=1}^r I_j$,
and so $f^r(p)=0$, i.e., $(f(p))^r=0$. So $f(p)=0$, and hence $p \in Z(\cap_{j=1}^r I_j)$.
Therefore we have
$$Z( \cap_{j=1}^r I_j) \subseteq \cup_{i=1}^n Z(I_r)\subseteq Z(\prod_{j=1}^r I_j)
\subseteq Z(\cap_{j=1}^r I_j)$$
and thus
\[Z(\cap_jI_j)=Z(\Pi_jI_j)=\cup_jZ(I_j).\]
\end{proof}
\begin{exer}\label{ex2.8}
If $I_j\subseteq\CC[x_1,\ldots,x_n]$ is a family of ideals, show that $\cap_j Z(I_j)=Z(\cup_jI_j)$
and that $\cup_j Z(I_j)\subseteq Z(\cap_jI_j)$, with equality if the family is a finite family.
Conclude that set ${\mathcal T}$ of all algebraic subsets of $\CC^n$ comprise the closed sets of a topology
on $\CC^n$; i.e., conclude that $\varnothing,\CC^n\in{\mathcal T}$, that ${\mathcal T}$ is closed under
arbitrary intersections and that ${\mathcal T}$ is closed under finite unions. This topology is
called the {\em Zariski} topology on $\CC^n$.
Given any algebraic subset $V\subseteq\CC^n$, we thus have the subspace topology on $V$
(in which a closed subset of $V$ is a set of the form $V\cap C$, where $C$ is a closed subset
of $\CC^n$), called the Zariski topology on $V$.
[Note: this is named after Oscar Zariski. His son,
the late Raphael Zariski, was a long-time professor of political science here at UNL.]
\end{exer}
\begin{proof}[Solution]
By Exercise \ref{ex2.2}, we have $Z(\cup_jI_j)=Z(I(\cup_jI_j))$.
Since $I_j\subseteq \cup_jI_j\subseteq I(\cup_jI_j)$, it follows from Exercise \ref{ex2.6} that
$Z(\cup_jI_j)=Z(I(\cup_jI_j))\subseteq Z(I_j)$ for every $j$, and hence that
$Z(\cup_jI_j)\subseteq \cap_jZ(I_j)$. If $p\in \cap_jZ(I_j)$, then $f(p)=0$
for all $f\in I_j$ for every $j$; i.e., $f(p)=0$ for all $f\in \cup_j I_j$, hence
$p\in Z(\cup_jI_j)$, so $\cap_jZ(I_j)\subseteq Z(\cup_jI_j)$ and thus
$\cap_j Z(I_j)=Z(\cup_jI_j)$.
If $p\in \cup_j Z(I_j)$, then $p\in Z(I_j)$ for some $j$, but $\cap_jI_j\subseteq I_j$
so by Exercise \ref{ex2.2} we have $p\in Z(I_j)\subseteq Z(\cap_jI_j)$ and hence
$\cup_j Z(I_j)\subseteq Z(\cap_jI_j)$. Exercise \ref{ex2.7} shows equality holds
when $j$ runs over a finite index set.
Clearly, $\varnothing,\CC^n\in{\mathcal T}$ since $\varnothing=Z(1)$ and
$\CC^n=Z(0)$. And if $C_j$ is a family of closed subsets, then
for each $j$ there is an ideal $I_j$ such that $Z(I_j)=C_j$, so
$\cap_jC_j=\cap_jZ(I_j)=Z(I(\cup_jI_j))\in{\mathcal T}$, while
$\cup_jC_j=\cup_jZ(I_j)=Z(\cap_jI_j)\in{\mathcal T}$ if $j$ runs over a finite index set.
Thus ${\mathcal T}$ satisfies the axioms for a topology.
\end{proof}
\begin{exer}\label{ex2.9}
Given any subset $V\subseteq\CC^n$, show that
$Z(I(V))$ is the Zariski closure $\overline{V}$ of $V$ (i.e., that $Z(I(V))$ is the intersection
of all algebraic subsets that contain $V$).
\end{exer}
\begin{proof}[Solution by Katie Morrison]
It is clear by definition that $V \subseteq Z(I(V))$ and $Z(I(V))$ is closed in the Zariski topology;
thus, $\overline{V} \subseteq Z(I(V))$. Thus it suffices to show for any ideal $J$ such that
$V \subseteq Z(J)$, that $Z(I(V)) \subseteq Z(J)$ since then $Z(I(V))$ will be in the intersection
of all such sets. Let $J$ be such an ideal, then $V \subseteq Z(J)$ implies that for all $f \in J$,
$f(v)=0$ for all $v \in V$. Each such $f$ lies in $I(V)$ by definition of $I(V)$. Thus, $J \subseteq I(V)$,
and so by Exercise 2.6, $Z(I(V)) \subseteq Z(J)$. Thus $Z(I(V))\subseteq \overline{V}$, and so $Z(I(V)) = \overline{V}$.
\end{proof}
\setcounter{thm}{0}
\mysection{January 19, 2011}\label{lect3}
\noindent{\bf Hilbert's Nullstellensatz and the Basis Theorem.}
We start by restating the Fundamental Theorem of Algebra:
\begin{thm}[Fundamental Theorem of Algebra] Let $I\subsetneq\CC[x]$ be an ideal.
Then $Z(I)\neq\varnothing $.
\end{thm}
\begin{proof}
Since $\CC[x]$ is a PID we know $I=(f)$ for some $f\in \CC[x]$.
Since $I\subsetneq\CC[x]$, we know $f$ is not a nonzero constant.
By Exercise \ref{ex2.2}, $Z(I)=Z(f)$. If $f=0$, then $Z(I)=Z(f)=\CC\neq\varnothing$.
If $f\neq0$, then $f$ is not a constant, so $\deg(f)>0$,
so $f$ has a root by the FTA, so $Z(I)=Z(f)\neq\varnothing$.
\end{proof}
There are various equivalent versions of the Nullstellensatz.
The FTA can be thought of as a special case of one of them.
\begin{thm}[Hilbert's Nullstellensatz, version 1] Let $I\subsetneq\CC[x_1,\ldots,x_n]$ be an ideal.
Then $Z(I)\neq\varnothing$.
\end{thm}
\begin{example}\label{example3.1.3}
Given a subset $V\subseteq \CC^n$,
Exercise \ref{ex2.9} shows that $Z(I(V))=\overline{V}$.
This raises the question of what happens when we start with an ideal
$J\subseteq\CC[x_1,\ldots,x_n]$ and consider $I(Z(J))$.
In the special case that $J=I(V)$, this is now easy.
Clearly $V\subseteq Z(I(V))$, hence $I(Z(I(V)))\subseteq I(V)$.
To show $I(V)\subseteq I(Z(I(V)))$, let $f\in I(V)$. Then $Z(I(V))\subseteq Z(f)$,
so $f\in I(Z(f))\subseteq I(Z(I(V)))$, hence $I(V)\subseteq I(Z(I(V)))$.
\end{example}
Example \ref{example3.1.3} answers what $I(Z(J))$ is when $J$ is an ideal of the form $J=I(V)$.
Another version of the Nullstellensatz
answers the question of how $I(Z(J))$ is related to $J$ in general.
\vskip\baselineskip
\begin{quotation}
{\bf Aside}: The Nullstellensatz (or zero points theorem) was only one of the results Hilbert proved around 1890
related to his work on invariant theory.
Another possibly more amazing result in this string is the
Hilbert Basis Theorem, which says that ideals of
$\CC[x_1,\ldots,x_n]$ are finitely generated.
\begin{thm}[Hilbert's Basis Theorem] Let $I\subseteq\CC[x_1,\ldots,x_n]$
be an ideal. Then $I=I(S)$ for some finite set
$S\subset \CC[x_1,\ldots,x_n]$.
\end{thm}
Note: this result holds if we replace $\CC$ by any field.
\vskip\baselineskip
The problem Hilbert was working on was to show that certain
rings of invariants were finitely generated. For example, consider
the multiplicative group $G=\{-1,1\}$. This acts on $\CC[x_1,\cdots,x_n]=R$ via
$(c*f)(x_1,\ldots,x_n) = f(cx_1,\ldots,cx_n)$, where $c\in G$ and $f\in R$.
The ring of invariants is the subset $R^G\subseteq R$ of all $f\in R$ such that
$c*f=f$ for all $c\in G$. It's easy to see that $\CC\subseteq R^G$ and that $x_ix_j\in R^G$.
It's not too hard to see that in fact every element of $R^G$ is a polynomial in the
expressions $x_ix_j$ with coefficients in $\CC$; i.e., $\{x_ix_j\}$ generate $R^G$ over $\CC$,
so $R^G$ is finitely generated.
Hilbert was mainly interested in an action where $G={\rm SL}_n(\CC)$,
the group of $n\times n$ matrices of determinant 1 with entries in $\CC$.
For example, in 1868 Paul Gordan (Emmy Noether was later to become Gordan's student)
found a finite set of generators for $R^G$ when $n=2$ and $G={\rm SL}_2(\CC)$.
Hilbert used his Basis Theorem to prove finite generation for all $n$, but without finding
an actual generating set. Gordan's reaction to Hilbert's proof is said to have been
``Das ist nicht Mathematik, das ist Theologie'' (although there is some question as to whether Gordan actually
did say this). In any case, Hilbert recognized that it would be desirable to provide
a constructive proof, and this led him to a partial solution, based on his Nullstellensatz.
At this point Gordan (according to Constance Reid's biography on Hilbert) responded
``I have convinced myself that theology also has its merits.''
\end{quotation}
\vskip\baselineskip
We need to recall some facts from commutative algebra in order to talk about additional
versions of the Nullstellensatz.
\begin{defn}
An ideal $I$ in a ring $R$ (we'll always assume rings are commutative with $1\neq0$)
is a {\em maximal} ideal if $I\subsetneq R$ and if $J$ is an ideal with $I\subseteq J\subsetneq R$, then
$I=J$.
\end{defn}
The facts we need are: (1) every proper ideal is (by Zorn's Lemma) contained in a maximal ideal; and
(2) an ideal $I$ in a ring $R$ is maximal if and only if $I$ is maximal.
Last semester
Tom Marley proved the following statement, whose proof is due to Artin and Tate
(see Atiyah-MacDonald, Corollary 5.24, for a proof):
\begin{thm}\label{Artin-Tate}
Let $k$ be a field and let $E$ be a finitely generated $k$-algebra which is a field; i.e.,
$E = k[x_1,\ldots, x_n]/I$ for some maximal ideal $I\subsetneq k[x_1,\ldots, x_n]$.
Then $E$ is a finite algebraic extension of $k$.
\end{thm}
This is often regarded as another version of the Nullstellensatz, but one in which the field
is arbitrary. When $k=\CC$ this is equivalent to version 1 above, as we shall see.
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex3.1}
Show that the Nullstellensatz is false if we replace $\CC$
by $\mathbb R$.
\end{exer}
\begin{proof}[Solution by Melissa DeVries]
Consider the ideal $I=(x^2+1)$ in ${\mathbb R}[x]$.
Note ${\mathbb R}[x]/I\cong\CC$ by sending ${\mathbb R} \to {\mathbb R}$
identically and $x\mapsto i$, so $I$ is a maximal ideal of ${\mathbb R}[x]$.
By Exercise \ref{ex2.2} (the proof is the same for $\mathbb R$ in place of $\CC$),
$Z(I)=Z(x^2+1)$. As $x^2+1$ has no real roots, $Z(I)=Z(x^2+1)=\varnothing$.
As ${\mathbb R}[x]$ has a proper ideal $I$ with an empty zero set, we see
the Nullstellensatz fails with $\mathbb R$ in place of $\CC$.
\end{proof}
\setcounter{thm}{0}
\mysection{January 21, 2011}\label{lect4}
Class started with Ashley and Zheng presenting solutions
to homework problems. We then looked at another version of the Nullstellensatz:
\begin{thm}[Hilbert's Nullstellensatz, version 2]\label{vers2Null}
Let $M\subseteq\CC[x_1,\ldots,x_n]$ be a maximal ideal.
Then there are constants $c_1,\ldots,c_n\in\CC$ such that
$M=(x_1-c_1, \ldots,x_n-c_n)$.
\end{thm}
\begin{proof}
By the Nullstellensatz, version 1, there is a point $(c_1, \ldots,c_n)\in Z(M)$.
Thus $x_i-c_i\in M$ for all $i$, hence $(x_1-c_1, \ldots,x_n-c_n)\subseteq M$.
But $(x_1-c_1, \ldots,x_n-c_n)$ is a maximal ideal
(since $\CC[x_1,\ldots,x_n]/(x_1-c_1, \ldots,x_n-c_n)\cong\CC$),
hence $(x_1-c_1, \ldots,x_n-c_n)=M$.
\end{proof}
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex4.1}
Show that versions 1 and 2 of the Nullstellensatz are equivalent.
\end{exer}
\begin{proof}[Solution by Anisah Nu'Man]
$(1 \Rightarrow 2:)$ Proof in notes page 9. $(2 \Rightarrow 1:)$ Let
$I \subsetneq \CC[x_1, \dots, x_n]$. Since every proper ideal is
contained in a maximal ideal, there exists an ideal $M\subseteq\CC[x_1,\dots x_n]$
such that $I \subseteq M$. By assumption we have
$I \subseteq M= (x_1-c_1, \dots,x_n-c_n)$ for some $c_i \in \CC$.
Also since $I \subseteq M$ we have $Z(M) \subseteq Z(I)$, from exercise 2.6.
Since $(c_1, \dots, c_n) \in Z(M) \subseteq Z(I)$ we have $Z(I) \not=\varnothing$.
\end{proof}
\begin{exer}\label{ex4.2}
Show that version 2 of the Nullstellensatz is equivalent to the following:
If $M\subset \CC[x_1,\ldots,x_n]$ is a maximal ideal, then
there is an isomorphism $h:\CC[x_1,\ldots,x_n]/M\to\CC$
such that $\CC\subset \CC[x_1,\ldots,x_n]\to\CC[x_1,\ldots,x_n]/M\to\CC$
is the identity on $\CC$.
[Aside: Note that this is essentially just Theorem \ref{Artin-Tate} in case $k=\CC$.]
\end{exer}
\begin{proof}[Solution]
Assume $M\subset \CC[x_1,\ldots,x_n]$ is a maximal ideal, such that
there is an isomorphism $h:\CC[x_1,\ldots,x_n]/M\to\CC$
inducing the identity on $\CC$. Then for each $i$ we have
$h(x_i)=c_i\in\CC$, so $x_i-c_i\in M$ for all $i$.
Thus $(x_1-c_1,\ldots,x_n-c_n)\subseteq M$. Since $(x_1-c_1,\ldots,x_n-c_n)$
is maximal, we have $(x_1-c_1,\ldots,x_n-c_n)=M$.
Conversely, assume $M=(x_1-c_1,\ldots,x_n-c_n)$ for some constants $c_i\in \CC$.
Define a ring homomorphism $H:\CC[x_1,\ldots,x_n]\to\CC$ by setting
$H|_{\CC}={\rm id}_{\CC}$ and
$H(x_i)=x_i$ for all $i$. Then $\ker(H)=M$, so $H$ induces an isomorphism
$h:\CC[x_1,\ldots,x_n]/M\to\CC$ which is the identity on $\CC$.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{subsection}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{January 24, 2011}\label{lect5}
\subsection{More on the Nullstellensatz}
\begin{defn}
Let $I\subseteq\CC[x_1,\ldots,x_n]$ be an ideal.
The {\em radical} $\sqrt{I}$ of $I$ is the ideal generated by all $f\in\CC[x_1,\ldots,x_n]$ such that
$f^r\in I$ for some $r\geq1$. If $I=\sqrt{I}$ we say $I$ is a radical ideal.
\end{defn}
\begin{thm}[Hilbert's Nullstellensatz, version 3] Let $J\subseteq\CC[x_1,\ldots,x_n]$ be an ideal
and let $f\in \CC[x_1,\ldots,x_n]$. If $Z(J)\subseteq Z(f)$, then
$f\in \sqrt{J}$.
\end{thm}
\begin{proof}
We have an inclusion $\CC[x_1,\ldots,x_n]\subset\CC[x_0,x_1,\ldots,x_n]$.
Thus we can regard $J$ and $f$ as being in $\CC[x_0,x_1,\ldots,x_n]$.
Given $Z(J)\subseteq Z(f)$, we thus have $\varnothing=Z(J\cup\{x_0f-1\})\subset\CC^{n+1}$.
By the Nullstellensatz, version 1, this means that $J$ together with $x_0f-1$ generates
the unit ideal; i.e., $1\in I(J\cup \{x_0f-1\})$, so there exist $a,h\in\CC[x_0,\ldots,x_n]$ and
$g\in J$ such that $1=ag+(x_0f-1)h$.
Substitute $1/y$ for $x_0$ in $1=ag+(x_0f-1)h$
and multiply by a large enough power $y^N$ to clear the denominator.
We get $a'g+(f-y)h'=y^N$, where
$a'=y^Na(1/y,x_1,\ldots,x_n),h'=y^{N-1}h(1/y,x_1,\ldots,x_n)\in \CC[x_1,\ldots,x_n,y]$.
Now substitute $f$ in for $y$ to get $a''g=f^N$. Since $g\in J$, this shows that $f\in\sqrt{J}$.
\end{proof}
\begin{thm}[Hilbert's Nullstellensatz, version 4]
Let $J\subseteq \CC[x_1,\ldots,x_n]$ be an ideal.
Then $I(Z(J))=\sqrt{J}$.
\end{thm}
We leave the proof as an exercise (see Exercise \ref{ex5.5}).
\subsection{The Basis Theorem}
We now look at some consequences of Hilbert's Basis Theorem.
\begin{defn}
A topological space $X$ is said to be {\em Noetherian} if it satisfies the Descending
Chain Condition on closed subsets; i.e., every chain ${\mathcal C}$
of closed sets of $X$ (i.e., every collection ${\mathcal C}$ totally ordered by inclusion),
has a minimal element (i.e., an element $C\in {\mathcal C}$ such that
$C\subseteq D$ for all $D\in{\mathcal C}$).
\end{defn}
\begin{example}
The reals $\mathbb R$ with the standard topology is not Noetherian. For example,
$\{[0,1+\frac{1}{n}]: n \geq 1\}$ is a chain of closed sets with no minimal element.
However, the reals with the finite complement topology (in which the closed sets,
other than the empty set and the whole space, are the finite subsets)
{\em is} Noetherian, since then clearly every chain of closed sets has a minimal element.
\end{example}
\begin{lem}
The Zariski topology on an algebraic set $V\subseteq\CC^n$ is Noetherian.
\end{lem}
\begin{proof}
Since closed subsets of closed subsets are closed, it's enough to
prove that the Zariski topology on $\CC^n$ is Noetherian.
Let $\{C_j\}$ be a chain of closed subsets of $\CC^n$.
Let $I_j=I(C_j)$. Since $\{C_j\}$ is a chain, so is $\{I_j\}$,
hence $\cup_jI_j$ is an ideal. By the Basis Theorem,
$\cup_jI_j$ is finitely generated, so $\cup_jI_j=I_t$ for some $t$.
By Exercise \ref{ex2.8},
$\cap_j Z(I_j)=Z(\cup_jI_j)=Z(I_t)=C_t$,
hence $C_t$ is a minimal element of $\{C_j\}$.
\end{proof}
\begin{defn}
A non-empty closed subset $C$ of a topological space $X$ is said to be {\em irreducible} if
$C$ is not the union $D_1\cup D_2$ of closed subsets $D_i\subsetneq C$.
\end{defn}
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex5.1}
Show that $(x-1)\subset\CC[x]$ is a radical ideal.
Show that $\sqrt{((x-1)^2)}=(x-1)$, so $((x-1)^2)$ is not a radical ideal.
\end{exer}
\begin{proof}[Solution presented in class by Nora Youngs]
Recall that $\CC[x]$ is a UFD.
Let $S=\{f\in \CC[x]: f^r\in (x-1){\rm for\ some\ }r\geq1\}$.
Suppose $f\in S$. By the Fundamental Theorem of Algebra, we can write
$f=(x-c_1)\cdots(x-c_n)$ for some complex numbers
$c_1,\ldots, c_n$. Then $f^r=(x-c_1)^r\cdots(x-c_n)^r$.
We know $f^r\in(x-1)$, so $f^r\in(x-1)g$ for some $g\in\CC[x]$.
By our previous factorization of $f$, we have $c_i=1$ for some $i$.
Thus $x-1$ divides $f$ so $f\in(x-1)$.
Hence $S\subseteq(x-1)$, so $\sqrt{(x-1)}=I(S)\subseteq(x-1)$
since $(x-1)$ is an ideal containing $S$.
However, if $f\in(x-1)$, then then $f^r\in(x-1)$ for $r=1$ so
$f\in S\subseteq I(S)=\sqrt{(x-1)}$. Thus $(x-1)\subseteq \sqrt{(x-1)}$.
By the double containment we have $\sqrt{(x-1)}=(x-1)$.
Now consider $((x-1)^2)$. Let
$S=\{f\in \CC[x]: f^r\in ((x-1)^2){\rm for\ some\ }r\geq1\}$.
Note that $x-1\in S$, as $(x-1)^2\in ((x-1)^2)$.
So $(x-1)=I((x-1))\subseteq I(S)=\sqrt{((x-1)^2)}$.
However, if $f^r\in ((x-1)^2)$ then again by
the Fundamental Theorem of Algebra, 1 is a zero of $f$ so $x-1\,|\, f$.
Thus $f\in (x-1)$, so $S\subseteq (x-1)$ and
$\sqrt{((x-1)^2)}=I(S)\subseteq (x-1)$. Therefore
$\sqrt{((x-1)^2)}=(x-1)\neq ((x-1)^2)$.
\end{proof}
\begin{exer}\label{ex5.2}
Let $I\subseteq \CC[x_1,\ldots,x_n]$ be an ideal.
Let $S=\{f\in\CC[x_1,\ldots,x_n]: f^r\in I\ {\rm for\ some}\ r\geq1\}$, hence $\sqrt{I}=I(S)$.
Show that $I(S)=S$; i.e., not only does $S$ generate $I$ but $S$ is itself an ideal.
\end{exer}
\begin{proof}[Solution presented in class by Douglas Heltibridle]
Note that if $I$ is a non-empty ideal, then $S\neq \emptyset $ as $I\subseteq S$. First to show that $S$
has the absorption property, let $g\in \mathbb C[x_1,...,x_n]$ and $f\in S$ such that $f^r\in I$.
Then $(gf)^r=g^rf^r\in I$ as $f^r\in I$, which is an ideal.
Next, we show that $S$ is closed under addition and contains inverses. Let $g\in \mathbb C[x_1,...,x_n]$
and let $f,h\in S$ with $f^r\in I$ and $h^k\in I$. Then $(g(f-h))^{2rk}=a_{2rk}f^{2rk}+a_{2rk-1}f^{2rk-1}h+...+a_1fh^{2rk-1}+a_0h^{2rk}$
and as each term has either $h^k$ or $f^r$ in it we know that it is in $I$ as $I$ is closed under addition. Thus $g(f-h)\in S$.
which means that $S$ is closed under addition and inverses. Thus $S$ is an ideal and it contains itself, which means that $I(S)=S$.
\end{proof}
\begin{exer}\label{ex5.3}
Let $I, J\subseteq \CC[x_1,\ldots,x_n]$ be ideals.
\begin{itemize}
\item[(a)] Show that $I\subseteq\sqrt{I}$.
\item[(b)] Show that $Z(\sqrt{I})=Z(I)$; conclude that $Z(I)=Z(J)$ if $\sqrt{I}=\sqrt{J}$.
\end{itemize}
\end{exer}
\begin{proof}[Solution by Kat Shultis]
(a) This is clear as if $f\in I$, then the first power of $f$ is in $I$, $f=f^1\in I$, so that $f\in\sqrt{I}$.
(b) We now have that $I\subseteq\sqrt{I}$, and so by Exercise 2.6, we know that $Z(\sqrt{I})\subseteq Z(I)$.
In order to show the other inclusion, let $p\in Z(I)$ and $f\in\sqrt{I}$ with $f^n\in I$. Then as $p\in Z(I)$ we have
that $f^n(p)=0$, which means that $f(p)=0$ because $\CC^n$ is an integral domain. Thus we have that for any
$p\in Z(I)$ and $f\in\sqrt{I}$, that $f(p)=0$, meaning that $Z(I)\subseteq Z(\sqrt{I})$, and hence $Z(I)=Z(\sqrt{I})$.
\end{proof}
\begin{exer}\label{ex5.4}
Show that versions 1 and 3 of the Nullstellensatz are equivalent.
\end{exer}
\begin{proof}[Solution]
The class notes show that version 1 implies version 3,
so assume version 3.
Let $I\subsetneq\CC[x_1,\ldots,x_n]$ be an ideal.
If $Z(I)=\varnothing$, then $Z(I)\subseteq Z(1)$,
hence $1\in\sqrt{I}$ so $1=1^r\in I$ for some $r\geq 1$,
contradicting $I\subsetneq\CC[x_1,\ldots,x_n]$.
\end{proof}
\begin{exer}\label{ex5.5}
Show that versions 3 and 4 of the Nullstellensatz are equivalent.
\end{exer}
\begin{proof}[Solution by Becky Egg]
First suppose that version 3 holds. Let
$J \subseteq \CC[x_1, \ldots, x_n]$ be an ideal, and
$g \in \sqrt{J}$. Then $g^k \in J$ for some $k$.
Given any $c \in Z(J)$, we have
\[0=(g^k)(c)=(g(c))^k.\]
So $g(c)=0$, and thus $g \in I(Z(J))$. Let
$h \in I(Z(J))$ and $c \in Z(J)$. Then $h(c)=0$
by definition, so $c \in Z((h))$. So $Z(J) \subseteq Z((h))$,
and hence by version 3 of the Nullstellensatz, $h \in \sqrt{J}$.
Thus $I(Z(J))=\sqrt{J}$.
Now suppose that version 4 holds. Let
$J \subseteq \CC[x_1, \ldots, x_n]$, and
$f \in \CC[x_1, \ldots, x_n]$ such that
$Z(J) \subseteq Z((f))$. By version 4, we
have $I(Z(J))=\sqrt{J}$. Since $I(Z((f)) \subseteq I(Z(J)))$,
we have $f \in \sqrt{J}$, and thus version 3 holds.
\end{proof}
\begin{exer}\label{ex5.6}
If $J$ is a radical ideal in $\CC[x_1,\ldots,x_n]$, show that
$J=I(V)$ for some algebraic subset $V\subseteq \CC^n$.
\end{exer}
\begin{proof}[Solution]
Since $I(Z(J))=\sqrt{J}$ by the Nullstellensatz, and since $J=\sqrt{J}$
by hypothesis, we have $I(V)=\sqrt{J}=J$ for $V=Z(J)$.
\end{proof}
\begin{exer}\label{ex5.7}
If $J$ is an ideal in $\CC[x_1,\ldots,x_n]$, show that
$\sqrt{J}=\cap_{M\in S}M$, where $S$ is the set of all maximal ideals containing $J$.
\end{exer}
\begin{proof}[Solution by Katie Morrison]
Let $M\subseteq \CC[x_1, \ldots, x_n]$ be a maximal ideal.
Then $\sqrt{M} \subsetneq \CC[x_1, \ldots, x_n]$ because
$1 \neq \sqrt{M}$ since $1^r= 1 \neq M$ for all $r\geq 0$.
Thus, $M \subseteq \sqrt{M} \subsetneq \CC[x_1, \ldots, x_n]$,
and so $M= \sqrt{M}$ since $M$ is maximal. Thus, every maximal
ideal is a radical ideal.
$(\subseteq)$ Observe that $J \subseteq \cap_{M \in S} M$, and
so $\sqrt{J} \subseteq \sqrt{\cap_{M \in S}M}$. We stated in
class (and will prove in Exercise 6.3) that $\sqrt{\cap_{M \in S}M}
= \cap_{M \in S} \sqrt{M}$. Then by the earlier observation,
$\cap_{M \in S} \sqrt{M} = \cap_{M \in S} M$. Thus,
$\sqrt{J} \subseteq \cap_{M \in S} M$.
$(\supseteq)$ For each ${\bf z} \in Z(\sqrt{J})$,
define $M_{{\bf z}}:= (x_1-z_1, \ldots, x_n-z_n)$ so that
$Z(M_{{\bf z}})={\bf z}$. Now $M_{{\bf z}}$ is maximal since
$\CC[x_1,\ldots, x_n]/M_{{\bf z}}\cong \CC$, which is a field.
Furthermore, since ${\bf z} \in Z(J)$ we have that
$J \subseteq I(Z(J)) \subseteq I({\bf z}) = M_{{\bf z}}$, and so
$J \subseteq M_{{\bf z}}$. Thus, $M_{{\bf z}} \in S$ for all ${\bf z} \in \sqrt{J}$.
Then $Z(\sqrt{J})=\cup_{{\bf z} \in Z(\sqrt{J})} Z(M_{{\bf z}}) \subseteq
\cup_{M \in S} Z(M)$. By Exercise 2.8,
$\cup_{M \in S} Z(M)\subseteq Z(\cap_{M \in S} M)$, and so $Z(\sqrt{J}) \subseteq Z(\cap_{M \in S} M)$. Then by Exercise 2.6,
$I(Z(\cap_{M \in S} M)) \subseteq I(Z(\sqrt{J}))$.
Finally since $\cap_{M \in S} M \subseteq I(Z(\cap_{M \in S} M))$
and $\sqrt{J} = I(Z(\sqrt{J}))$, we have that
$\cap_{M \in S} M \subseteq \sqrt{J}$, and so equality holds.
\end{proof}
\begin{exer}\label{ex5.8}
Consider the subring ${\mathbb Z}[\sqrt{-5}]$ of $\CC$ consisting of all
complex numbers of the form $a+b\sqrt{-5}$ where $a$ and $b$ are integers.
Show that 2 and 3 are irreducible elements of ${\mathbb Z}[\sqrt{-5}]$ but not prime.
[Hint: use the norm, and then factor 6 in two different ways.]
\end{exer}
\begin{proof}[Solution by Anisah Nu'Man]
For sake of a contradiction suppose $2$ is reducible. Then
there exists an
$x=a+b\sqrt{-5}$ and $y=c+d\sqrt{-5} \in {\mathbb Z}[\sqrt{-5}]$ such
that $x,y$ are not
units and $N(xy)=N(x)N(y)=N(2)$. Thus we have $N(x)N(y)=N(2)=4$
Since $x,y$ are not units we know the $N(x), N(y) \not = 1$, so
we have the $N(x)=N(y)=2$. Therefore we have
$N(x)=N(a+b\sqrt{-5})=a^{2}+5b^{2}=2$. This implies
that $b^{2}=0$ and $a^{2}=2$. But there is no integer
such that $a^{2}=2$. Thus the $N(x) \not = 2$, and so
we must have, without loss of generality, that $N(x)=1$ and
the $N(y)=4$. Thus $x$ is a unit and $2$ is irreducible.
Using an identical argument we can show that $3$ is also
irreducible. Last we have $(2)$ is not prime since
$6=(1+\sqrt{-5})(1-\sqrt{-5}) \in (2)$ but neither
$(1+\sqrt{-5})$ or $(1-\sqrt{-5})$ are in $(2)$.
Similarly, $(3)$ is not prime since $6\in (3)$, but
using the same factorization $6=(1+\sqrt{-5})(1-\sqrt{5})$
we have $(1+\sqrt{-5})$ and $(1-\sqrt{-5})$ are not in $(3)$.
\end{proof}
\begin{exer}\label{ex5.9}
In an integral domain (i.e., a commutative ring with $1\neq0$
and with no zero divisors) show
that every prime element is irreducible,
and in a UFD, show that every irreducible element is prime.
\end{exer}
\begin{proof}[Solution by Katie Morrison]
Let $D$ be an integral domain, and let $a \in D$ be a prime
element. Suppose $a=bc$ for some $b, c \in D$. Then
$a \vert bc$, and so $a \vert b$ or $a \vert c$ since
$a$ is prime. Without loss of generality, say $a \vert b$,
then there exists $k \in D$ such that $b = ak$. Observe that
$$b\cdot 1 = a k = (bc)k = b(c k).$$
Since $D$ is a domain, cancellation holds, and so we have that
$1 = c k$. Thus, $c$ is a unit, and so $a$ cannot be written
as the product of non-units.
Let $R$ be a UFD, and let $a$ be an irreducible. Suppose
$a \vert bc$ for some $b, c \in R$, i.e. $ak= bc$ for some
$k \in R$. Since $R$ is a UFD, there exist irreducibles
$p_1, \ldots, p_l, q_1, \ldots, q_m, r_1, \ldots, r_n \in R$ such
that $b$ can be uniquely written as $p_1\cdots p_l$, $c$ can be
uniquely written as $q_1, \ldots, q_m$, and $k$ can be uniquely
written as $r_1 \cdots r_n$. Then
$bc = p_1\cdots p_l\cdot q_1\cdots q_m$, $a k = a \cdot r_1\cdots r_n$,
and these decompositions are unique. Thus, the multiset
$\{p_1, \ldots, p_l, q_1, \ldots, q_m\}$ equals the multiset
$\{a, r_1, \ldots, r_n\}$, and so there exists some $p_i$ or
$q_j$ that equals $a$. Without loss of generality, say that
$a = p_i$ for some $1 \leq i \leq l$. Then we have that
$b = a\cdot \Pi_{j \neq i}p_j$, and so $a \vert b$. Thus,
$a$ is prime.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{January 26, 2011}\label{lect6}
Note: An algebraic set which is not irreducible is said to be {\em reducible}.
\begin{example}
Consider the algebraic set $V=Z(xy)\subset \CC^2$; here our polynomial ring is $\CC[x,y]$.
Then in the Zariski topology, $V=Z(x)\cup Z(y)$ is the union of two proper closed subsets, so $V$ is reducible.
\end{example}
\begin{example}
Now consider the algebraic set $V=Z(xy-1)\subset \CC^2$.
This time $V$ is irreducible. Every closed subset of $V$ is of the form
$V\cap C$ for some $C=Z(J)$, where $J\subseteq \CC[x,y]$ is an ideal.
But $\CC[x,y]$ is Noetherian, so $C$ is the intersection of a finitely
many closed subsets of the form $Z(f)$, where $f\in \CC[x,y]$. But consider
$V\cap Z(f)$, where $f$ is a non-zero element of $\CC[x,y]$.
Then any $p=(a,b)\in V\cap Z(f)$ satisfies $b=1/a$ and $0=f(a,b)=f(a,1/a)$.
For some $N\gg0$, $x^Nf(x,1/x)$ is a polynomial in $x$, and any non-zero root
of $f(x,1/x)$ is also a root of $h$ and vice versa. But $h$ has only finitely many roots,
hence the same is true of $f(x,1/x)$, and thus $V\cap Z(f)$ is finite.
Since any proper closed subset of $V$ is finite yet $V$ is infinite, we see that $V$ is
irreducible in the Zariski topology (but not, it is easy to see, in the standard topology).
\end{example}
\begin{lem}
Every non-empty closed subset $C$ of a Noetherian topological space $X$
is the union $C=C_1\cup\cdots\cup C_r$
of finitely many irreducible closed subsets $C_j$. Moreover,
the union can be chosen to be {\em irredundant} (i.e., so that none of the $C_i$
contains any of the others), in which case the decomposition
is unique up to order.
\end{lem}
\begin{proof} (The proof in class didn't use Zorn's lemma. For variety,
here's a different proof.)
Let ${\mathcal F}$ be the set of all non-empty closed subsets
of $X$ which are not finite unions of irreducible closed subsets.
If ${\mathcal F}$ is not empty, then by Zorn's lemma (using the fact
that $X$ is Noetherian, so that we know descending chains of closed sets
are bounded below), ${\mathcal F}$
has a minimal element $D$, and clearly $D$ cannot be irreducible.
Write $D=D_1\cup D_2$ for non-empty proper closed subsets $D_i$.
Since $D$ is minimal, neither $D_i$ is in ${\mathcal F}$, hence
each is the union of finitely many irreducible closed subsets, and thus so
is $D$, contradicting our assumption that $D\in{\mathcal F}$. Thus ${\mathcal F}$
is empty and hence every non-empty closed set is a finite union of
irreducible closed subsets.
Therefore, given any closed subset $C$, we can write $C=C_1\cup\cdots\cup C_r$
for some choice of irreducible closed subsets $C_i$. Whenever there is an $i$ and $j$
such that $C_i\subseteq C_j$, we can remove $C_i$ from the union.
Thus we obtain a union where we can assume that no $C_j$ contains any $C_i$.
Suppose we have two such unions, $C_1\cup\cdots\cup C_r=D_1\cup\cdots\cup D_s$.
Then $D_i=\cup_j(C_j\cap D_i)$ and since $D_i$ is irreducible,
$D_i=C_j\cap D_i$ for some $j$, hence $D_i\subseteq C_j$. Likewise,
$C_j\subseteq D_t$ for some $t$, hence $D_i\subseteq D_t$,
so $j=t$ and $D_i=C_j$. Thus each of the $D$'s equals one of the $C$'s and vice versa,
so $r=s$ and after reordering we can assume $D_l=C_l$ for $l=1,\ldots,r$.
Thus the decomposition is unique up to order.
\end{proof}
Note that the proof of the preceding lemma is similar to the proof of prime
factorization in the integers: given any positive integer $n$, we can write $n=p_1^{m_1}\cdots p_r^{m_r}$
for primes $p_i$ in an essentially unique way. If we rephrase this in terms of ideals, we have
for any $n$ that
$(n)=(p_1)^{m_1}\cap \cdots \cap (p_r)^{m_r}$; i.e., every ideal is the intersection of powers of prime
ideals in an essentially unique way (if we aren't silly, such as $(0)=(0)\cap (2)$).
This is an example of a primary decomposition. Nice ideals can be written as intersections
of prime ideals in a unique way (see Exercise \ref{ex6.4};
the preceding lemma is the geometric manifestation of this (i.e., the geometric version
of primary decompositions).
As we see already from the integers, ideals in general cannot be written as
intersections of prime ideals. We can however, in a Noetherian ring,
write every ideal as a finite intersection of primary ideals
(but we lose uniqueness in general). We pause to recall the definitions.
A {\em prime} ideal $P\subset R$ in a ring $R$ (commutative with $1\neq0$ as usual)
is a proper ideal such that if $fg\in P$, then either $f\in P$ or $g\in P$. A proper ideal $Q\subset R$
is said to be {\em primary} if $fg\in Q$, then either $f\in Q$ or $g\in \sqrt{Q}$. Note that prime ideals are
primary, but the reverse is not usually true. Writing an ideal $J$ as a finite intersection of primary ideals
is said to be a {\em primary decomposition} of $J$.
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex6.1}
Let $P\subset\CC[x_1,\ldots,x_n]$ be a prime ideal.
Show that $P=\sqrt{P}$.
\end{exer}
\begin{proof}[Solution by Anisah Nu'Man]
Clearly $P \subseteq \sqrt{P}$ since for $f\in P$ we have $f^1=f\in P$. Now let $g \in \sqrt{P}$,
thus $g^r \in P$ for some $r \geq 1$. Then $g \cdot g^{r-1}\in P$, and so either $g\in P$ or
$g^{r-1} \in P$. If $g\in P$ we are done. If not $g^{r-1}=g\cdot g^{r-1} \in P$. Therefore
either $g\in P$ or $g^{r-2} \in P$. If $g\in P$ we're done. If not, iteratively, this process must
stop and so we have $g\in P$. By double containment we have $P= \sqrt{P}$.
\end{proof}
\begin{exer}\label{ex6.2}
Let $C\subset\CC^n$ be an irreducible algebraic set
and let $P\subset \CC[x_1,\ldots,x_n]$ be a prime ideal.
Show that $I(C)$ is a prime ideal and show that
$Z(P)$ is irreducible. Conclude that a closed subset $D$
is irreducible if and only if $I(D)$ is prime.
\end{exer}
\begin{proof}[Solution by Kat Shultis]
First, we will show that $I(C)$ is prime by proving that if $I(C)$ is not prime, then
$C$ is reducible. So assume that $I(C)$ is not prime. This means that there exist
elements $f,g\in\CC[x_1,\ldots,x_n]\setminus I(C)$ such that $fg\in I(C)$. Consider
the two closed sets $Z(f)$ and $Z(g)$ in $\CC^n$. We know by Exercise 2.2 that
$Z(f)=Z((f))$ and $Z(g)=Z((g))$. We also know, by Exercise 2.7 that $Z(f)\cup Z(g)=Z(fg)$.
If $p\in C$, then $f(p)g(p)=0$ as $fg\in I(C)$, meaning that
$C\subseteq Z(fg)=Z(f)\cup Z(g)$. Notice that as neither $f$ nor $g$ is in
$I(C)$, that there exists some point $p\in C$ such that $f(p)\neq 0$, meaning that
$C\cap Z(f)\subsetneq C$, and similarly for $g$. Also, the intersection of finitely
many closed sets is closed, so that $C=(C\cap Z(f))\cup (C\cap Z(g))$ is a
decomposition of $C$, and $C$ is reducible.
Next, we show that $Z(P)$ is irreducible by contradiction. So, assume that
$Z(P)$ is reducible. In other words there exist finitely many Zariski closed and
irreducible sets $\{A_\alpha\}_{\alpha\in I}$ which are properly contained in
$Z(P)$ such that $Z(P)=\cup_{\alpha\in I} A_\alpha$. This means that for each
$\alpha\in I$ there exists an ideal $J_\alpha\subseteq C[x_1,\ldots,x_n]$, such
that $A_\alpha=Z(J_\alpha)$. By Exercise 2.7, this is equivalent to saying that
$$Z(P)=\bigcup_{\alpha\in I} Z(J_\alpha)=
Z\left(\bigcap_{\alpha\in I} J_\alpha\right)=Z\left(\prod_{\alpha\in I} J_\alpha\right).$$
As $P$ is prime, we know that $I(Z(P))=\sqrt{P}=P$ by exercise 6.1 and the
Nullstellensatz, version 4. Thus, by exercise 6.1 we have that
$$P=I(Z(P))=
I\left(Z\left(\prod_{\alpha\in I}J_\alpha\right)\right)\supseteq \prod_{\alpha\in I}J_\alpha.$$
By definition of a prime ideal, we must have that $J_\alpha\subseteq P$ for some
$\alpha$. Then, by exercise 2.6, we have that $Z(P)\subseteq Z(J_\alpha)$ for
some $\alpha\in I$ so that our original decomposition of $Z(P)$ was not actually
a decomposition, providing a contradiction, and showing that $Z(P)$ is irredicuble.
\end{proof}
\begin{exer}\label{ex6.3}
Let $J=\cap_{i=1}^rJ_i$ for ideals $J_i\subseteq\CC[x_1,\ldots,x_n]$.
Show that $\sqrt{J}=\cap_{i=1}^r\sqrt{J_i}$.
\end{exer}
\begin{proof}[Solution by Ashley Weatherwax]
We want to show that $\sqrt{\cap^r_{i=1}J_i}=\cap^r_{i=1}\sqrt{J_i}$.
Let $f\in \sqrt{\cap^r_{i=1}J_i}$. The for some $n$,
$f^n\in\cap^r_{i=1}J_i$. But then $f^n\in J_i$ for all $i$, and
so $f\sqrt{J_i}$ for all $i$. Therefore, $f\in \cap\sqrt{J_i}$.
Now let $f\in \cap\sqrt{J_i}$. Then $f\in\sqrt{J_i}$ for all $i$,
so there exists $n_i$ such that $f^{n_i}\in J_i$ for each $i$.
Let $m=n_1\cdots n_r$. As $J_i$ are ideals, $f^m\in J_i$ for all $i$,
hence $f^m\in \cap J_i$ and thus $f\in\sqrt{\cap J_i}$.
\end{proof}
\begin{exer}\label{ex6.4}
Let $J\subset\CC[x_1,\ldots,x_n]$ be an ideal.
Show that $\sqrt{J}$ can be written as the intersection
of finitely many prime ideals, none of which contains another.
[Aside: this is an analog of the fact that
a square-free non-zero integer is a product of distinct primes.]
Conclude that $\sqrt{J}$ is the intersection of all prime ideals that contain
$J$.
\end{exer}
\begin{proof}[Solution]
We have seen that $Z(J)=Z(\sqrt{J})$ is a finite union
$C_1\cup\cdots\cup C_r$ of Zariski closed irreducible subsets $C_i$.
Thus $I(C_i)$ is prime for each $i$, and
$I(C_1\cup\cdots\cup C_r)=\cap_iI(C_i)$, but
by the Nullstellensatz $I(Z(J))=\sqrt{J}$, so
$\sqrt{J}=\cap_i I(C_i)$, as claimed. Since
$\sqrt{J}$ is the intersection of some of the primes
which contain $J$, it is certainly the intersection of all primes
which contain $J$.
\end{proof}
\setcounter{thm}{0}
\mysection{January 28, 2011}\label{lect7}
\noindent{\bf More on primary decomposition.}
By Exercise \ref{ex6.4}, given any ideal $J\subseteq \CC[x_1,\ldots,x_n]$,
$\sqrt{J}$ is the intersection $P_1\cap\cdots\cap P_r$ for prime ideals $P_i$.
This is a primary decomposition of $\sqrt{J}$, which we may assume is {\em irredundant}
(i.e., we may assume $P_i\not\subseteq P_j$ if $i\neq j$).
In fact, the primes $P_i$ are the minimal primes that contain $J$, as we see from
the following lemma. The geometric version of this lemma states: if $C$ is an irreducible closed set contained
in a union $C_1\cup\cdots\cup C_r$ of closed sets $C_i$, then $C$ is already contained in $C_i$ for some $i$.
\begin{lem}\label{primelem}
Let $P, J_i\subseteq\CC[x_1,\ldots,x_n]$ be ideals with $P$ prime.
If $J_1\cap\cdots\cap J_r\subseteq P$, then $J_i\subseteq P$ for some $i$.
\end{lem}
\begin{proof}
If for each $i$ we have $J_i\not\subseteq P$, then for each $i$ we can pick $f_i\in J_i$ with $f_i\not\in P$.
Now $f=f_1\cdots f_r\in J_1\cap\cdots\cap J_r\subseteq P$, but because $P$ is prime
and $f_i\not\in P$ for all $i$, we must have $f\not\in P$. Thus by contradiction we must have
$J_i\subseteq P$ for some $i$.
\end{proof}
\begin{rem}\label{irredundantrem}
Let $P_1,\ldots,P_r\subset\CC[x_1,\ldots,x_n]$ be prime ideals.
This lemma shows that if $P_i\not\subseteq P_j$ holds whenever $i\neq j$,
then in fact we have the stronger property that $\cap_{i\neq j}P_i\not\subseteq P_j$.
However, for primary ideals $Q_1,\ldots,Q_r\subset\CC[x_1,\ldots,x_n]$,
even if $Q_i\not\subseteq Q_j$ holds whenever $i\neq j$,
there can be a $j$ such that $\cap_{i\neq j}Q_i\subseteq Q_j$.
For example, consider $Q_1=(x^4,y^2),Q_2=(x^3,xy,y^3),Q_3=(x^2,y^3)\subset\CC[x,y]$.
In Figure \ref{monomialidealgrid}, $Q_1$ is the ideal spanned by the monomials
$x^iy^j$ such that the integer lattice point $(i,j)$ is on or above
the blue lines; $Q_2$ corresponds to the lattice points on or above the heavy black lines,
$Q_3$ the red and $Q_1\cap Q_3$ the green.
\begin{figure}[h]
\setlength{\unitlength}{0.75cm}
\begin{picture}(5,4)(0,0)
\multiput(0,0)(0,1){4}{\line(1,0){5.5}}
\multiput(0,0)(1,0){6}{\line(0,1){3.5}}
\put(5.75,-.15){$x$}
\put(-.25,3.9){$y$}
{\color{blue}
\linethickness{3pt}
\put(0,2){\line(0,1){1.5}}
\put(0,2){\circle*{.3}}
\put(0,2){\line(1,0){4}}
\put(4,0){\line(0,1){2}}
\put(4,0){\circle*{.3}}
\put(4,-.1){\line(1,0){1.5}}
}
{\color{red}
\linethickness{2pt}
\put(-.2,3){\line(0,1){0.5}}
\put(-.2,3){\circle*{.3}}
\put(-.2,3){\line(1,0){2}}
\put(1.8,0){\line(0,1){3}}
\put(1.8,0){\circle*{.3}}
\put(1.8,0){\line(1,0){3.5}}
}
{\color{green}
\linethickness{2pt}
\put(-.3,3.1){\line(0,1){0.5}}
\put(-.3,3.1){\line(1,0){2}}
\put(1.7,2.1){\line(0,1){1}}
\put(1.7,2.1){\line(1,0){2}}
\put(3.7,0.1){\line(0,1){2}}
\put(3.7,0.1){\line(1,0){1.5}}
}
{%\color{gray}
\linethickness{2pt}
\put(-.51,2.8){\line(0,1){0.8}}
\put(-.51,2.8){\circle*{.3}}
\put(-.51,2.88){\line(1,0){1.15}}
\put(0.6,1){\line(0,1){1.88}}
\put(0.6,1){\circle*{.3}}
\put(0.6,1){\line(1,0){2}}
\put(2.6,-.2){\line(0,1){1.2}}
\put(2.6,-.2){\circle*{.3}}
\put(2.6,-.2){\line(1,0){2.5}}
}
\end{picture}
\caption{The monomial ideals $Q_1$ (blue), $Q_2$ (black), $Q_3$ (red) and $Q_1\cap Q_3$ (green)
from Remark \ref{irredundantrem}.}\label{monomialidealgrid}
\end{figure}
Then $Q_i\not\subseteq Q_j$ if $i\neq j$ holds but
$Q_1\cap Q_3=(x^4,x^2y^2,y^3)\subsetneq Q_2$.
In this case, $Q_1,Q_2$ and $Q_3$ all have the same associated prime,
$P=(x,y)$, but this is not essential.
Consider $Q'_1=(x^4,y^2,a),Q'_2=(x^3,xy,y^3,a,b),Q'_3=(x^2,y^3,b)\subset\CC[a,b,x,y]$.
Then the associated primes are $P'_1=(a,x,y)$, $P'_2=(a,b,x,y)$ and $P'_3=(b,x,y)$
respectively, and $Q'_i\not\subseteq Q'_j$ holds whenever $i\neq j$, but again
$Q'_1\cap Q'_3=(x^4,x^2y^2,y^3,ax^2,by^2,ab)\subsetneq Q'_2$.
\end{rem}
Let $Q\subset\CC[x_1,\ldots,x_n]$ be a primary ideal.
By Exercise \ref{ex7.1}, $\sqrt{Q}$ is prime; we say that
$\sqrt{Q}$ {\em belongs} to $Q$, or that $Q$ is $\sqrt{Q}$-primary.
Given an intersection $Q_1\cap\cdots\cap Q_r$ of primary ideals $Q_i$,
by Exercise \ref{ex7.2} we may assume that the primes belonging to each $Q_i$ are all different,
and it is clear that we can remove $Q_i's$ if need be from the intersection
$Q_1\cap\cdots\cap Q_r$ so that we end up with an intersection such that
for each $j$ we have
$\cap_{i\neq j}Q_i\not\subseteq Q_j$.
\begin{defn}
We say an intersection $Q_1\cap\cdots\cap Q_r$ of primary ideals $Q_i$
is {\em irredundant} if the primes belonging to each ideal $Q_i$ are different and if
for each $i$ we have $\cap_{j\neq i}Q_j\not\subseteq Q_i$.
\end{defn}
Given an irredundant primary decomposition $J=\cap Q_i$ of an ideal
$J\subseteq\CC[x_1,\ldots,x_n]$, we refer to the primes belonging to the ideals $Q_i$
as {\em belonging} to (or as being {\em associated} with) $J$.
The primes belonging to an ideal $J$ include the minimal primes containing $J$,
but there may be additional primes too (see Exercise \ref{ex7.4}).
\begin{defn}
Let $J\subseteq\CC[x_1,\ldots,x_n]$ be an ideal, and let $P_1,\ldots,P_r$ be the distinct associated primes
(i.e., the primes belonging to $J$). The minimal primes in the set $\{P_1,\ldots,P_r\}$
are called {\em isolated} primes. These correspond to the irreducible components of $Z(J)$.
The non-minimal primes in the set $\{P_1,\ldots,P_r\}$ are called {\em embedded} primes.
\end{defn}
Note that the isolated primes belonging to
an ideal $J$ are uniquely determined by $J$, since they are just the minimal primes containing
$J$. In fact, all of the primes belonging to $J$ are uniquely determined by
$J$ (see Atiyah-Macdonald, Theorem 4.5), as are the primary ideals
corresponding to the minimal primes (see Atiyah-Macdonald, Corollary 4.11).
The primary ideals corresponding to embedded primes are not in
general uniquely determined however (see Exercise \ref{ex7.4}).
One way that embedded primes arise naturally is when taking powers.
\begin{example}\label{embex}
Let $J=(x,y)\cap(x,z)\cap(y,z)=(xy,xz,yz)\subset\CC[x,y,z]$.
Then $J$ is radical; it is the ideal of the union of the coordinate axes in $\CC^3$.
It is easy to see that $J^2\subsetneq (x,y)^2\cap(x,z)^2\cap(y,z)^2$: $J^2$
is generated by products of pairs of the generators of $J$, and each such product is
in $(x,y)^2\cap(x,z)^2\cap(y,z)^2$ and has degree 4, hence $xyz\not\in J^2$
even though $xyz\in (x,y)^2\cap(x,z)^2\cap(y,z)^2$.
In fact, $J^2 = (x,y)^2\cap(x,z)^2\cap(y,z)^2\cap (x,y,z)^4$
is an irredundant primary decomposition of $J^2$, so
$(x,y,z)$ is an embedded prime.
\end{example}
It's good to become familiar working with symbolic algebra programs,
such as Macaulay 2.
Here is the preceding example worked out using Macaulay 2 (but working over a finite field
of characteristic 31991). This also demonstrates the use of the {\tt primaryDecomposition}
command.
\begin{verbatim}
> M2
Macaulay 2, version 0.9.2
--Copyright 1993-2001, D. R. Grayson and M. E. Stillman
--Singular-Factory 1.3c, copyright 1993-2001, G.-M. Greuel, et al.
--Singular-Libfac 0.3.2, copyright 1996-2001, M. Messollen
i1 : R=ZZ/31991[x,y,z]
o1 = R
o1 : PolynomialRing
i2 : X=ideal(y,z)
o2 = ideal (y, z)
o2 : Ideal of R
i3 : Y=ideal(x,z)
o3 = ideal (x, z)
o3 : Ideal of R
i4 : Z=ideal(x,y)
o4 = ideal (x, y)
o4 : Ideal of R
i5 : J=intersect(X,intersect(Y,Z))
o5 = ideal (y*z, x*z, x*y)
o5 : Ideal of R
i6 : M=ideal(x,y,z)
o6 = ideal (x, y, z)
o6 : Ideal of R
i7 : K=intersect(X^2,intersect(Y^2,intersect(Z^2,M^4)))
2 2 2 2 2 2 2 2 2
o7 = ideal (y z , x*y*z , x z , x*y z, x y*z, x y )
o7 : Ideal of R
i8 : J^2==K
o8 = true
i9 : toString primaryDecomposition(J^2)
o9 = {monomialIdeal matrix {{x, y^2}}, monomialIdeal matrix {{x^2, y}},
monomialIdeal matrix {{x, z^2}}, monomialIdeal matrix {{x^2, y^2, z^2}},
monomialIdeal matrix {{y, z^2}}, monomialIdeal matrix {{x^2, z}},
monomialIdeal matrix {{y^2, z}}}
\end{verbatim}
Note that {\tt primaryDecomposition} does not give a decomposition for which each primary
component has a different prime (which in principle it could do, by Exercise \ref{ex7.2}).
But we can see from the computer output that the associated primes of $J^2$ are $(x,y)$, $(x,z)$, $(y,z)$ and
$(x,y,z)$.
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex7.1}
Let $Q\subset\CC[x_1,\ldots,x_n]$ be a primary ideal.
Show that $\sqrt{Q}$ is prime.
\end{exer}
\begin{proof}[Solution by Philip Gipson]
Suppose $fg\in \sqrt{Q}$ but $f,g\not\in\sqrt{Q}$.
Then there exists an $n$ such that $f^ng^n=(fg)^n\in Q$.
Since $Q$ is primary, we know that either $f^n\in Q$ of $g^n\in \sqrt{Q}$.
If $f^n\in Q$, then by definition $f\in \sqrt{Q}$, contradicting
our hypothesis. If $g^n\in\sqrt{Q}$, then $g^{nk}=(g^n)^k\in Q$
for some $k$, so $g\in\sqrt{Q}$, again contradicting
our hypothesis. Therefore we conclude that either $f\in \sqrt{Q}$
of $g\in\sqrt{Q}$, and so $\sqrt{Q}$ is prime.
\end{proof}
\begin{exer}\label{ex7.2}
Let $Q_1,Q_2\subset\CC[x_1,\ldots,x_n]$ be primary ideals
belonging to the same prime $P$. Show that $Q_1\cap Q_2$
is also primary and belongs to $P$.
\end{exer}
\begin{proof}[Solution by Jason Hardin]
Observe that by Exercise \ref{ex6.3},
$\sqrt{Q_1\cap Q_2}=\sqrt{Q_1}\cap\sqrt{Q_2}=P\cap P=P$,
so if $Q_1\cap Q_2$ is primary, then it must be $P$-primary.
Let $ab\in Q_1\cap Q_2$. If $a\in Q_1\cap Q_2$, we're done.
So suppose $a\notin Q_1\cap Q_2$. Wlog, suppose $a\notin Q_1$.
Since $ab\in Q_1$ and $Q_1$ is primary, we must have
$b\in\sqrt{Q_1}=P=\sqrt{Q_1\cap Q_2}$. So $Q_1\cap Q_2$ is primary.
\end{proof}
\begin{exer}\label{ex7.3}
Let $J\subset \CC[x_1,\ldots,x_n]$ be an ideal such that $\sqrt{J}$
is a maximal ideal. Show that $J$ is primary. [Hint:
if $g\not\in\sqrt{J}$,
show that there are polynomials $a\in J$ and $b$ with $a+bg=1$,
and thus if $fg\in J$, then $(1-a)f=bfg\in J$
and hence $f\in J$. Or see Atiyah-Macdonald, Proposition 4.2.]
Conclude that primary ideals (unlike what happens in ${\mathbb Z}$)
are not always powers of prime ideals, by giving a simple example.
\end{exer}
\begin{proof}[Solution by Zheng Yang]
Suppose $fg\in J$ and $g\not\in\sqrt{J}$. Since $\sqrt{J}$ is a
maximal ideal (hence also prime), we have $(\sqrt{J},g)\supsetneqq
\sqrt{J}$ and so $(\sqrt{J},g)=(1)$. Then we have an equation
$a+bg=1$, for some $a\in\sqrt{J}$ (so $a^r\in J$ for some $r\geq1$)
and $b\in \CC[x_1,\ldots,x_n]$. On
the one hand, we have $(1-a)f=bfg\in J$ as $fg\in J$; on the other
hand, we have $1-a$ is not in $\sqrt{J}$ (otherwise,
$1=1-a+a\in\sqrt{J}$).
Thus $(1-a^r)f=(1+a+\cdots+a^{r-1})(1-a)f=bfg\in J$, hence
$f=(bg+a^r)f\in J$, since $a^r\in J$.
Thus, $f\in (1-a)^{-1}J\subseteq J$ (note
$J$ is an ideal).
The ideal $(x^2,y)\subset\CC[x,y]$ is an example of a primary ideal
that is not a power of a prime ideal; see Lecture 8. Here is another
example taken from Atiyah-Macdonald's book,
Exercise 4.4. In ${\mathbb Z}[t]$, $m=(2,t)$ is a maximal, as
${\mathbb Z}[t]/m\cong {\mathbb Z}/2{\mathbb Z}$. The ideal
$q=(4,t)$ is primary (because ${\mathbb Z}[t]/q\cong {\mathbb
Z}/4{\mathbb Z}$, of which the only zero-divisors are $\bar{2}$ and
it is nilpotent). But $m^2=(4,t^2,2t)\subsetneq q\subsetneq m$. So
$q$ is not a power of a power of prime ideal $m$.
\end{proof}
\begin{exer}\label{ex7.4}
Consider the ideal $J=(x^2,xy)\subset \CC[x,y]$.
Show $J=(x)\cap(x^2,y)$ and $J=(x)\cap(x^2,xy,y^n)$ for any $n\geq1$
are primary decompositions of $J$.
Conclude that the primes associated to $J$ are $(x)$ and $(x,y)$.
\end{exer}
\begin{proof}[Solution]
First, $(x)$, $(x^2,y)$ and $(x^2,xy,y^n)$ are primary,
since $(x)$ is prime and $\sqrt{(x^2,y)}=(x,y)=\sqrt{(x^2,xy,y^n)}$
are maximal. Thus $(x)\cap(x^2,y)$ and $(x)\cap(x^2,xy,y^n)$ are
irredundant primary decompositions. Also it is easy to see that
$J\subseteq (x)\cap(x^2,y)$ and $J\subseteq(x)\cap(x^2,xy,y^n)$.
Consider an element $f\in (x)\cap(x^2,y)$. Since $f\in (x)$, we see
$x|f$ so every term of $f$ is divisible by $x$; in particular,
$f$ has no constant term and no terms that are pure powers of $y$.
But $f\in (x^2,y)$, so every term of $f$ is divisible by either $x^2$
or $y$ (and hence also by $xy$ since $f$ has no terms that are
pure powers of $y$). Thus $f$ is in $(x^2,xy)$, so
$J=(x)\cap(x^2,y)$.
Now consider $f\in (x)\cap(x^2,xy,y^n)$ for any $n\geq 1$.
As before, $f$ has no terms that are pure powers of $y$, and since
$f\in (x^2,xy,y^n)$, every terms is divisible by either
$x^2$, $xy$ or $y^n$ but any term divisible by $y^n$ also is divisible by
$x$ and hence by $xy$, so $f\in(x^2,xy)$ and we have
$J=(x)\cap(x^2,xy,y^n)$.
This shows irredundant primary decompositions need note be unique.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{January 31, 2011}\label{lect8}
Let $f\in\CC[x_1,\ldots,x_n]$ and
$J=(f)\subseteq\CC[x_1,\ldots,x_n]$. Using the fact that $\CC[x_1,\ldots,x_n]$ is a UFD,
it is not hard to see that $J$ is primary if and only if $\sqrt{J}$ is prime,
and if $J$ is primary, then $J=(\sqrt{J})^r$ for some $r\geq1$.
(This is Exercise \ref{ex8.1}.) We also know by Exercise \ref{ex7.3} that
$M^r$ is primary for every $r\geq1$ if $M$ is a maximal idea,
and as we will see in a later lecture,
if $J$ is a prime ideal in a polynomial ring with $J$ generated by
monomials (i.e., if $J$ is a monomial ideal), then
$J^r$ is primary for every $r\geq1$.
But principal, monomial and maximal ideals are rather special in this regard,
since: it is not always true that an ideal is primary if
its radical is prime; it is not always true that every primary ideal is a power of a
prime ideal; and it is not always true that powers of prime ideals
are primary.
For example, $(x^2,y)\subset\CC[x,y]$ is primary by Exercise \ref{ex7.3},
since $\sqrt{(x^2,y)}=(x,y)$ is maximal, and thus the only prime that contains
$(x^2,y)$ is $(x,y)$, but $(x,y)^2\subsetneq (x^2,y)\subsetneq (x,y)$ so
$(x^2,y)$ is not $(x,y)^r$ for any $r$.
By Exercise \ref{ex7.1} we know the radical of a primary ideal is prime.
To show ideals with prime radical need not be primary,
consider $J=(x^2,xy)\subset\CC[x,y]$. Then $(x)\cap(x^2,y)$ is a primary decomposition,
so $(x^2,xy)$ is not primary (an alternative way to see this is
that although $xy\in J$, we have $x\not\in (x^2,xy)$ and
$y\not\in\sqrt{(x^2,xy)}=\sqrt{(x)\cap(x^2,y)}=\sqrt{(x)}\cap \sqrt{(x^2,y)}=(x)\cap (x,y)=(x)$),
even though $P=\sqrt{(x^2,xy)}=(x)$ is prime.
Note in this case that $J$ has an associated prime $P'=\sqrt{(x^2,y)}=(x,y)$ other than $P$,
and thus $P\subsetneq P'$; i.e., $P'$ is an embedded prime.
This must always be the case when a non-primary ideal $J$ has prime radical.
(For suppose $J$ is not primary, but has prime radical $P$.
Consider an irredundant primary decomposition
$J=Q_1\cap\cdots\cap Q_r$.
By assumption $P=\sqrt{J}=\sqrt{Q_1\cap\cdots\cap Q_r}=\cap_i\sqrt{Q_i}$,
so $P\subset \sqrt{Q_i}$ for all $i$ and by Lemma \ref{primelem} we know
$\sqrt{Q_j}\subseteq P$ for some $j$ and hence $P=\sqrt{Q_j}$;
we may as well assume $P=\sqrt{Q_1}$.
Since $J$ is not primary, we also must have $r>1$, so
$P\subsetneq \sqrt{Q_2}$. Thus $\sqrt{Q_2}$ is an embedded prime.)
We now would like to give an example of an ideal (in a polynomial ring, even)
which is a power of a prime ideal but
which is not primary. Such examples tend to be a bit complicated.
Example 3, p.~51 of Atiyah-Macdonald, gives an example of
an ideal $I$ which is power of a prime ideal yet
is not primary, but the ring is not a polynomial ring.
Example 8.18 of {\em Introduction to Algebraic Geometry} by Brendan Hassett
gives an example of an ideal $I$ which is a power of a prime ideal yet
is not primary, and in this case the ring is a polynomial ring, but in 6 variables.
It turns out that polynomial ring examples need at least 3 variables
(this will be an exercise in a later lecture). Our example will use 3 variables.
Thus we need a prime ideal $P\subset \CC[x,y,z]$ such that
$P^r$ is not primary and as noted above $P^r$
must have an embedded associated prime. We already
have an example of an ideal having a power with an embedded prime:
$J^2$ has an embedded associated prime
if $J$ is the ideal of Example \ref{embex}.
Unfortunately, $J$ is not prime, so we'd like to modify
$J$ to get a prime ideal without losing the square
having an embedded associated prime.
In order to make a good guess about what to try, it's helpful
to know what to pay attention to.
Note that the zero locus of the embedded prime of $J$ is a point
which is {\em singular} on the curve $Z(J)$. Being a singular point essentially
means that the curve crosses itself there.
(A point which is not singular is said to be {\em smooth}.
Smoothness is an important geometrical concept which
we'll discuss more later. The point here is that embedded primes are
related to smoothness. In fact, if $P\subset\CC[x_1,\ldots,x_n]$ is a prime ideal
and $Q$ is an embedded prime of $P^r$ for some $r>0$,
then $P\subsetneq Q$ so $Z(Q)\subsetneq Z(P)$, and
it turns out that the points of $Z(Q)$ must be singular points of $Z(P)$.)
Thus when we modify $J$ to get a prime ideal we want to preserve
the singularity at the origin.
One way to see that $J$ is not prime is to note that $Z(J)$ is not irreducible;
its irreducible components are the three coordinate axes.
The conceptual idea is to take the coordinate axes (as shown
at left in Figure \ref{coordaxesconnected}) and connect them together to
get (as shown at right in Figure \ref{coordaxesconnected}) an ``irreducible
curve'' $C\subset\CC^3$ that has three branches at the origin
(thus preserving the singularity), each branch tangent to a coordinate axis.
\setcounter{figure}{0}
\begin{figure}[h]
\caption{Coordinate axes and resulting irreducible heuristic curve $C$.}
\label{coordaxesconnected}
\vskip-1.8in\vbox to0in{\includegraphics[width=6in]{RoughSketch.pdf}}\vskip4.5in
\begin{picture}(0,0)
\put (130,50){$C$}
\end{picture}
\vskip-.25in
\end{figure}
To a very small observer located at the origin
the curve $C$ would appear to be the union of the coordinate axes.
If we can produce such a $C$ which is an algebraic set,
we might hope $I(C)$ and $J$ would behave
similarly in having embedded components at the origin.
This is in fact the case in the example which we now do more rigorously,
using computer calculations to guide us.
\begin{example}\label{primepower}
Let $C\subset\CC^3$ be the curve defined parametrically for all $t\in \CC$ by
$$t\mapsto (t(t-1)^2(t+1)^2,t^2(t-1)(t+1)^2,t^2(t-1)^2(t+1)).$$
Note that three values of $t$ map to $(0,0,0)$, these being $t=-1,0,1$,
and otherwise (it is easy to see) the map $\CC\to\CC^3$ is injective.
Let $P$ be the kernel of the homomorphism $h:\CC[x,y,z]\to\CC[t]$
which is the identity on $\CC$ and otherwise is defined by
\[
\begin{split}
h(x)=\ & t\ (t-1)^2(t+1)^2,\\
h(y)=\ & t^2(t-1)\ (t+1)^2,\\
h(z)=\ & t^2(t-1)^2(t+1).
\end{split}
\]
\end{example}
\vskip\baselineskip
We now discuss the ideal $P$.
Note that if $g\in P$, then for any point $p=(a,b,c)\in C$ we have
$(a,b,c)=(s(s-1)^2(s+1)^2,s^2(s-1)(s+1)^2,s^2(s-1)^2(s+1))$
for some value $s$ of $t$ and thus
$g(a,b,c)=0$, since
$$g(t(t-1)^2(t+1)^2,t^2(t-1)(t+1)^2,t^2(t-1)^2(t+1))=h(g)=0,$$
and so in particular
$$g(a,b,c)=g(s(s-1)^2(s+1)^2,s^2(s-1)(s+1)^2,s^2(s-1)^2(s+1))=0.$$
Thus $P\subseteq I(C)$, but for any $g\in I(C)$ we have
$g(a,b,c)=0$ for all $(a,b,c)\in C$ and hence
$g(t(t-1)^2(t+1)^2,t^2(t-1)(t+1)^2,t^2(t-1)^2(t+1))=0$
for all $t\in \CC$, so $h(g)=0$ and thus $g\in \ker{h}=P$.
In particular, $P=I(C)$.
Moreover, $P$ is a prime ideal (since $\CC[t]$ is a domain).
The curve $C$ passes through the origin when $t=0$, $t=-1$ and $t=1$,
and each of these three branches is tangent to a different coordinate axis,
giving the desired singularity at the origin.
We next will investigate the possible existence of embedded primes.
\vskip\baselineskip
Here we show some views of the curve $C$.
Figure \ref{Maplegraph} shows two plots of the curve $C$.
The one on the left was done using Maple (a very old version)
with the following commands (the color varies from blue to red
as the $z$-coordinate increases):
\vskip\baselineskip
\begin{verbatim}
> with(plots):
> spacecurve([t*(t-1)^2*(t+1)^2,t^2*(t-1)*(t+1)^2,t^2*(t-1)^2*(t+1),
numpoints=1000], t=-1.1..1.1,axes=NORMAL);
\end{verbatim}
\vskip\baselineskip
The plot on the right was drawn using
\LaTeX\ picture environment commands.
The \LaTeX\ picture environment
allows drawings to be done internally to \LaTeX. The graph was done by
plotting 3000 points on the curve. The coordinates of the points were
computed ahead of time using AWK (see the file
SpaceCurveData.tex for the very short AWK script).
This graph is drawn so that parts of the curve closer to
the observer are thicker, in an attempt to give perspective without color.
Finally, Figure \ref{stereoview} gives a stereoscopic view of the curve $C$
(created using 3D-Xplor-Math-J, available at \url{http://3d-xplormath.org/}).
\begin{figure}[h]
\caption{Two views of the curve $C$.}\label{Maplegraph}
\hskip-2in\includegraphics[width=4in]{spacecurve.pdf}
\vskip-2in
\begin{picture}(200,90)
\put (-35,60){$y$}
\put (25,100){$z$}
\put (120,50){$x$}
\put (-70,80){$C$}
\end{picture}
\vskip-1.5in
\hspace{3.5in}\hbox to0in{\input SpaceCurveData.tex \hss}
\end{figure}
\begin{figure}[h]
\vskip1in
\caption{Stereoscopic view of the curve $C$.}\label{stereoview}
\vbox to0in{\includegraphics[width=6in, height=3in]{StereoImageCurve.pdf}\vss}
\vskip3in
\end{figure}
\newpage
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex8.1}
Let $J=(f)\subset\CC[x_1,\ldots,x_n]$.
Show that $J$ is primary if and only if $\sqrt{J}$ is prime,
and if $J$ is primary, then $J=(\sqrt{J})^r$ for some $r\geq1$.
\end{exer}
\begin{proof}[Solution]
Since primary ideals have prime radicals, if $J$ is primary, then $\sqrt{J}$ is prime.
Before proving the converse, we note a fact about principal ideals.
Let $f=f_1^{m_1}\cdots f_s^{m_s}$ be a factorization of $f$ as a product
of irreducibles, where $m_i\geq1$ for all $i$ and where
$f_i$ and $f_j$ have no common irreducible factor
if $i\neq j$. Let $I=(\phi)$, where $\phi=f_1\cdots f_s$. Then
$\phi^m\in J$ for $m=\max(m_1,\ldots,m_s)$, so
$I\subseteq \sqrt{J}$. And
if $g\in \sqrt{J}$, then $g^s\in (f)$ so $f | g^s$, hence
$f_i|g^s$ for all $i$, so $f_i|g$ for all $i$, hence $\phi|g$
and we have $g\in I$. Thus $\sqrt(J)=I$.
Now assume $\sqrt(J)$ is prime. Then so is $\phi$,
hence $r=1$; i.e., the prime factorization of $f$ has only one prime factor.
Thus $J=(f_1^{m_1})=(\phi)^{m_1}=\sqrt{J}^{m_1}$, so if $ab\in J$ but $a\not\in J$,
then in the prime factorization of $ab$, $m_1$ factors of $f_1$ occur
but not all of them occur in the factorization of $a$, hence $f_1|b$, so
$f_1^{m_1}|b^{m_1}$ and thus $b^{m_1}\in J$, so $b\in\sqrt{J}$,
and so $J$ is primary.
Finally, if $J$ is primary, then $\sqrt{J}$ is prime, and as we saw above
$J=\sqrt{J}^{m_1}$.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{subsection}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{February 2, 2011}\label{lect9}
If $I=Q_1\cap\cdots\cap Q_r$ is an irredundant primary decomposition
we refer to each $Q_i$ as the $\sqrt{Q_i}$-primary component of
the decomposition. When $\sqrt{Q_i}$ is an isolated prime,
the $\sqrt{Q_i}$-primary component is uniquely determined by
$I$ (it is the same in any primary decomposition of $I$; see
Atiyah-Macdonald, Corollary 4.11).
When $I=Q^r$ for a prime ideal $Q$, then $Q$ is the unique
isolated prime, so the $Q$-primary component of $I$
is uniquely determined by $I$. It is the smallest $Q$-primary
ideal containing $I$, denoted
$Q^{(r)}$ and referred to as the $r$th {\em symbolic power} of $Q$.
Symbolic powers are very important geometrically.
The fact that powers and symbolic powers do not always agree is at the root
of various unsolved problems in algebraic geometry.
\subsection{A theorem of Zariski and Nagata and orders of vanishing}
Recall for any ideal $I\subseteq \CC[x_1,\ldots,x_n]$, that by the Nullstellensatz
$\sqrt{I}=\cap M$,
where the intersection is over all maximal ideals $M$ containing $I$.
A theorem of Zariski and Nagata (see Theorem 3.14, p.~106, of
Eisenbud's book {\em Commutative Algebra with a view toward Algebraic Geometry})
generalizes this:
\begin{thm}\label{ZarNagThm}
Let $I\subseteq \CC[x_1,\ldots,x_n]$ be a prime ideal.
Then $I^{(r)}=\cap\ M^r$, where
the intersection is over all maximal ideals $M$ containing $I$.
\end{thm}
\begin{defn}
Let $p\in\CC^n$ and let $M_p=I(p)\subset\CC[x_1,\ldots,x_n]$ be
the corresponding maximal ideal. Let $f$ be a non-zero element of
$\CC[x_1,\ldots,x_n]$.
If $f\in M^r$ but $f\not\in M^{r+1}$, we say the {\em order of vanishing} of $f$ at $p$
is $r$ and write $\ord_p(f)=r$. We regard the constant $0$ as having infinite
order of vanishing at all $p$.
\end{defn}
\begin{example}
If $f(p)\neq0$, then $f$ does not vanish at $p$ And we have
$\ord_p(f)=1$. If $p=(0,\ldots,0)$ and $f$ is not trivial,
then $\ord_p(f)$ is the degree of a term
of least degree. So for example $f=x^2y+y^5$ has order of vanishing
3 at the origin. If $p=(a_1,\ldots,a_n)$, then $\ord_p(f)$
is the degree of a term of $f(x_1+a_1,\ldots,x_n+a_n)$ of least degree;
i.e., find the degree of a term of least degree with respect to coordinates
centered at the point $p$.
If $I\subseteq\CC[x_1,\ldots,x_n]$ is an ideal,
the fact that $\sqrt{I}=\cap M$, where the intersection is over all maximal ideals
$M$ containing $I$, is just saying that the polynomials that vanish at all points
$p\in Z(I)$ are precisely the elements of the radical of $I$. Theorem \ref{ZarNagThm}
says that the polynomials that vanish to order at least $r$ at all points
of an irreducible closed set $V$ are exactly the elements of $Q^{(r)}$,
where $Q=I(V)$.
\end{example}
\subsection{Return to our example of the last lecture}
Let $J$ be the ideal of the union of the coordinate axes in $\CC^3$
and let $P$ be the ideal of the curve $C$ defined parametrically
in the previous lecture.
Recall that $J^2 = (x,y)^2\cap(x,z)^2\cap(y,z)^2\cap (x,y,z)^4$
is a primary decomposition of $J^2$. One way to see that
$J^2 \subsetneq (x,y)^2\cap(x,z)^2\cap(y,z)^2$ is to note that
$xyz\in (x,y)^2\cap(x,z)^2\cap(y,z)^2$ but (since $J=(xy,xz,yz)$
so $J^2$ has no elements of
degree less than 4) that $xyz\not\in J^2$.
Our goal is to find an element $f$ that plays the same role
for $P^2$ as $xyz$ does for $J^2$, in the sense that
$f$ is in the $P$-primary component of an irredundant primary decomposition
of $P^2$ but $f\not\in P^2$, thus showing that $P^2$ has an embedded prime
and hence $P^2$ is not primary.
By Theorem \ref{ZarNagThm}, it is enough to find an element
$f$ which is in $M^2$ for all maximal ideals $M$ containing $P$,
but such that $f\not\in P^2$. In order to do this we need to
find all such maximal ideals $M$, and we need to find our candidate element $f$.
It follows by Exercise \ref{ex9.1} that $C=\overline{C}$,
and hence that a maximal ideal $M_p$ contains $C$ if and only if $p\in C$.
We will verify $C=\overline{C}$ explicitly, with the help of Macaulay 2.
We will also use Macaulay 2 to find $f$ and verify that
$f\in M^2$ for all $p\in C$; it turns out that $f=xyz + {\rm higher\ order\ terms}$.
We will see that $f\not\in P^2$ follows from the fact that $Z(P)$
is singular at the origin.
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex9.1}
Let $f_1,\ldots,f_n\in\CC[t]$ and define the map
$h:\CC\to\CC^n$ given by $h(t)=(f_1(t),\ldots,f_n(t))$.
Show that the image $h(\CC)$ of $\CC$ under $h$
is a Zariski closed subset of $\CC^n$.
[Hint: Apply the going-up theorem, Atiyah-Macdonald, Theorem 5.10, p. 62.]
\end{exer}
\begin{proof}[Solution]
If $h$ maps $\CC$ to a point, then clearly $h(\CC)$ is closed.
Suppose $h$ is not constant. Then $f_i$ is a non-constant polynomial
for some $i$. Hence $\CC[f_i]\subseteq h^*(\CC[x_0,\ldots,x_n])\subseteq\CC[t]$.
Since $f_i$ is non-constant, $\CC[f_i]\subseteq\CC[t]$ is an integral extension
of rings, hence so is $h^*(\CC[x_0,\ldots,x_n])\subseteq\CC[t]$.
Let $p\in\overline{h(\CC)}$; then $I(p)$ is a maximal ideal, hence prime.
Since ${\rm ker}(h)=I(h(\CC))=I(\overline{h(\CC)})$, and since $p\in \overline{h(\CC)}$,
we have $I(\overline{h(\CC)})\subseteq I(p)$. Now by Theorem 5.10 of Atiyah-Macdonald,
for any prime ideal $Q\subseteq h^*(\CC[x_0,\ldots,x_n])$, there is a prime ideal
$P\subset \CC[t]$ such that $P\cap h^*(\CC[x_0,\ldots,x_n])=Q$.
In particular, there is a prime ideal $P\subset \CC[t]$ such that
$P\cap h^*(\CC[x_0,\ldots,x_n])=I(p)/{\rm ker}(h)$.
This means $(h^*)^{-1}(P)=I(p)$. Moreover, $P$ is maximal since if not
$P=(0)$ (these are the only choices in $\CC[t]$) but $P=(0)$ would mean
$I(p)={\rm ker}(h)$ which would in turn mean that $I(p)=I(h(\CC))$, which implies
$h(\CC)=p$. Since $P$ is maximal, $P=I(q)$ for some point $q\in\CC$,
and $(h^*)^{-1}(P)=I(p)$ means $h(q)=p$. I.e., $h(\CC)=\overline{h(\CC)}$,
so $h(\CC)$ is closed.
\end{proof}
\mysection{February 4, 2011}\label{lect10}
In order to show that $P^2$ is not primary when $P$ is the prime ideal
defined in Example \ref{primepower}, we want to show $C=\overline{C}$ and we need
to find a certain element $f$. Macaulay 2 will be helpful. We use it to gain some insight
about $P$ and $P^2$.
\begin{verbatim}
i1 : R=QQ[x,y,z];
i2 : S=QQ[t];
i3 : H=map(S,R,matrix{{t*(t-1)^2*(t+1)^2,t^2*(t-1)*(t+1)^2,t^2*(t-1)^2*(t+1)}});
o3 : RingMap S <--- R
i4 : P=ker H;
o4 : Ideal of R
i5 : toString P
o5 = ideal(x*y+x*z-2*y*z,y^3-8*x^2*z-7*y^2*z-8*x*z^2+23*y*z^2-z^3-8*y*z,
x^3-7/4*x^2*y+x*y^2-3/16*y^3-5/4*x^2*z+3/2*x*y*z-7/16*y^2*z+1/2*x*z^2
-5/16*y*z^2-1/16*z^3+1/2*x*y-1/2*x*z+1/2*y*z)
i6 : K=primaryDecomposition(P^2);
i7 : toString K
o6 = {ideal(x^2*y^2+2*x^2*y*z-4*x*y^2*z+x^2*z^2-4*x*y*z^2
+4*y^2*z^2,x*y^4-8*x^3*y*z-6*x*y^3*z-2*y^4*z-8*x^3*z^2+
8*x^2*y*z^2+16*x*y^2*z^2+14*y^3*z^2-8*x^2*z^3+38*x*y*z^3-46*y^2*z^3
-x*z^4+2*y*z^4-8*x*y^2*z-8*x*y*z^2+16*y^2*z^2,x^4*y+x^4*z-3*x^3*y*z
+1/4*x*y^3*z-x^3*z^2+4*x^2*y*z^2-7/4*x*y^2*z^2-1/2*y^3*z^2+2*x^2*z^3
-25/4*x*y*z^3+4*y^2*z^3-1/4*x*z^4+1/2*y*z^4-x^2*y*z-x^2*z^2+2*x*y*z^2,x^5
-1/32*y^5-5*x^3*y*z-5/4*x*y^3*z+15/32*y^4*z-5/2*x^2*y*z^2+25/2*x*y^2*z^2
-15/16*y^3*z^2-5/2*x^2*z^3+35/4*x*y*z^3-175/16*y^2*z^3+15/32*y*z^4
-1/32*z^5+x^3*y+1/8*x*y^3-x^3*z-x^2*y*z-7/8*x*y^2*z+1/4*y^3*z-x^2*z^2
+39/8*x*y*z^2-2*y^2*z^2-1/8*x*z^3-1/4*y*z^3-x*y*z,y^6-14*y^5*z+64*x^4*z^2
-80*x*y^3*z^2+95*y^4*z^2+128*x^3*z^3-640*x^2*y*z^3+624*x*y^2*z^3
-260*y^3*z^3-64*x^2*z^4+208*x*y*z^4-33*y^2*z^4+16*x*z^5-46*y*z^5
+z^6-16*y^4*z+128*x^2*y*z^2+112*y^3*z^2+128*x*y*z^3-368*y^2*z^3
+16*y*z^4+64*y^2*z^2), ideal(z,x^2*y^2,x*y^4,x^4*y,y^6,x^6)}
\end{verbatim}
\vskip\baselineskip
According to Macaulay 2, $P$ is generated (over $\mathbb Q$) by
\[
\begin{split}
g_1=&xy+xz-2yz,\\
g_2=&y^3-z^3-8x^2z-7y^2z-8xz^2+23yz^2-8yz,\\
g_3=&x^3-(3/16)y^3-(7/4)x^2y+xy^2-(5/4)x^2z+3/2xyz-(7/16)y^2z\\
&+(1/2)xz^2-(5/16)yz^2-(1/16)z^3+(1/2)xy-(1/2)xz+(1/2)yz.
\end{split}
\]
Certainly $g_1$, $g_2$ and $g_3$ are in $P$, because checking
this just amounts to plugging in the parametric equations and checking
to see that the expression simplifies to 0,
and this is independent of the ground field.
Let $p=(a,b,c)\in Z(P)$. We wish to show $p\in C$.
Since $g_1\in P$, we see that if any of the coordinates of $p$ are 0,
then one of the other two coordinates is also 0. For example, if $a=0$, then
since $g_1(a,b,c)=0$ we have $-2bc=0$ so either $b=0$ or $c=0$.
Now from $g_2\in P$, given that $a=0$, we see that $b=0$ if and only if $c=0$.
Thus $a=0$ implies $b=c=0$. Similarly, if $b=0$, then either $a=0$ or $c=0$,
and we just saw that $a=0$ implies $c=0$. If $b=c=0$, then $g_3\in P$ implies
$a=0$. Finally, if $c=0$, then as noted above either $a=0$ or $b=0$, but in either case
all three coordinates are 0. Thus if any coordinate of $p$ is 0, the point $p$ must be the origin,
and hence $p\in C$.
Now assume none of the coordinates of $p$ is 0.
If $a=b$, then $g_1$ gives
$a=c$. Likewise, if $a=c$, then also $a=b$, and if $b=c$, then $b=a$.
I.e., if any two of the coordinates are equal, all three are equal.
But if $a=b=c$, then $g_2$ gives that $a=0$.
We thus see that if none of the coordinates is 0, then no two of them are equal.
If $a=2b$, then $g_1(a,b,c)=0$ implies $c=0$, so now we also know that
$a\neq 2b$.
Now we wish to find a value of $t$ such that
$$t\mapsto (t(t-1)^2(t+1)^2,t^2(t-1)(t+1)^2,t^2(t-1)^2(t+1))=(a,b,c).\eqno{(\star)}$$
Let $t=\frac{b}{b-a}$, which is defined since $a\neq b$.
Let $(\alpha,\beta,\gamma)$ be the image of $t=\frac{b}{b-a}$ under the map
$(\star)$, so $\in C$. We also see that none of $\alpha$, $\beta$ or $\gamma$ is 0, since
this only happens if $t$ is $-1$, 0 or 1, but $t=-1$ is ruled out since
$2b\neq a$, $t=0$ is ruled out since $b\neq0$ and $t=1$ is ruled out since
$a\neq0$. It is easy to check that
$$\frac{\alpha}{\beta}=\frac{t-1}{t}=\frac{a}{b}$$
and hence that $(\alpha,\beta)=d(a,b)$ for some non-zero constant $d$, and thus
$$\frac{\alpha}{2\beta-\alpha}=\frac{a}{2b-a}.$$
Since $g_1(\alpha,\beta,\gamma)=0$ by definition of $P$,
we have $\frac{\gamma}{\beta}=\frac{\alpha}{2\beta-\alpha}$
and since $g_1(a,b,c)=0$ since $p\in Z(P)$, we have
$\frac{c}{b}=\frac{a}{2b-a}$. Thus $\frac{\gamma}{\beta}=\frac{c}{b}$
so $(\alpha,\beta,\gamma)=d(a,b,c)$.
If we show $d=1$, then $p=(a,b,c)=(\alpha,\beta,\gamma)\in C$,
and we conclude that $C=\overline{C}$.
Plug $(\alpha,\beta,\gamma)=d(a,b,c)$ into $g_2$. We get
$$0=(\beta^3-\gamma^3-8\alpha^2\gamma-7\beta^2\gamma-8\alpha\gamma^2+23\beta\gamma^2)-8\beta\gamma=
(b^3-c^3-8a^2c-7b^2c-8ac^2+23bc^2)d^3-8bcd^2,$$
and since $d\neq=0$, this means
$(b^3-c^3-8a^2c-7b^2c-8ac^2+23bc^2)-8bc/d=0$.
But $0=g_2(a,b,c)=(b^3-c^3-8a^2c-7b^2c-8ac^2+23bc^2)-8bc$,
so $d=1$.
Now consider the polynomial
$f=x^5-(1/32)y^5-5x^3yz-(5/4)xy^3z+(15/32)y^4z
-(5/2)x^2yz^2+(25/2)xy^2z^2-(15/16)y^3z^2-(5/2)x^2z^3+(35/4)xyz^3-(175/16)y^2z^3+
(15/32)yz^4-(1/32)z^5+x^3y+(1/8)xy^3-x^3z-x^2yz-(7/8)xy^2z+(1/4)y^3z-x^2z^2
+(39/8)xyz^2-2y^2z^2-(1/8)xz^3-(1/4)yz^3-xyz$. According to
the Macaulay 2 output above, $f\in P^{(2})$. We will give a proof of this in a moment,
and we will also show that $f\not\in P^2$, which proves that
$P^2$ is not primary.
To show that $f\in P^{(2})$, it suffices to show
$f\in M^2$ for all maximal ideals $M$ containing $P$.
To show $f\not\in P^2$, it suffices to show that $P\subseteq (x,y,z)^2$ and hence
$P^2\subseteq (x,y,z)^4$, but that $f\not\in (x,y,z)^4$.
But $f$ has a term of degree 3 (namely $xyz$),
so $f\not\in(x,y,z)^4$, and hence $f\not\in P^2$.
To show that $f\in M^2$ for all maximal ideals $M$ containing $P$,
we just need to show that $f\in M_p^2$ for all $p\in C$, since $C=\overline{C}$.
Since we have a parameterization of $C$,
we may assume every point $p=(a,b,c)\in C$ is of the form
$p=(t(t-1)^2(t+1)^2,t^2(t-1)(t+1)^2,t^2(t-1)^2(t+1))$ for some
value of $t$. Thus we just need to check that
$$f(x,y,z)\in (x-(t(t-1)^2(t+1)^2,y-t^2(t-1)(t+1)^2,z-t^2(t-1)^2(t+1))^2$$
holds for all $t$.
Alternatively, by doing a translation we just need to check that
$$f(x+(t(t-1)^2(t+1)^2,y+t^2(t-1)(t+1)^2,z+t^2(t-1)^2(t+1))\in (x,y,z)^2.$$
This is easy to do: just plug $x+(t(t-1)^2(t+1)^2$, $y+t^2(t-1)(t+1)^2$,
and $z+t^2(t-1)^2(t+1)$ into $f(x,y,z)$ for $x$, $y$ and $z$ respectively,
simplify and see if the resulting expression has any terms of total
degree in $x$, $y$ and $z$ less than 2.
Here's what we get when we do the substitution (with Macaulay 2 doing the actual algebra):
\begin{verbatim}
-10*x^2*t^15+10*x*y*t^15-5/2*y^2*t^15+10*x*z*t^15-5*y*z*t^15-5/2*z^2*t^15
+10*x^2*t^14-30*x*y*t^14+25/2*y^2*t^14+10*x*z*t^14+5*y*z*t^14-15/2*z^2*t^14
+25*x^2*t^13+5*x*y*t^13-75/4*y^2*t^13-35*x*z*t^13+45/2*y*z*t^13+5/4*z^2*t^13
-30*x^2*t^12+70*x*y*t^12-15/2*y^2*t^12-30*x*z*t^12-25*y*z*t^12+45/2*z^2*t^12
+5*x^3*t^10-35/2*x^2*y*t^10+35/4*x*y^2*t^10-5/8*y^3*t^10-55/2*x^2*z*t^10
+95/2*x*y*z*t^10-115/8*y^2*z*t^10+75/4*x*z^2*t^10-135/8*y*z^2*t^10-25/8*z^3*t^10
-85*x*y*t^11+55*y^2*t^11+35*x*z*t^11-30*y*z*t^11+15*z^2*t^11+20*x^2*y*t^9
-25*x*y^2*t^9+5/2*y^3*t^9-50*x*y*z*t^9+85/2*y^2*z*t^+35*x*z^2*t^9-25/2*y*z^2*t^9
-25/2*z^3*t^9+28*x^2*t^10-21/2*x*y*t^10-217/4*y^2*t^10+29/2*x*z*t^10
+49*y*z*t^10-67/4*z^2*t^10-25*x^3*t^8+95/2*x^2*y*t^8-5/4*x*y^2*t^8-5/8*y^3*t^8
+135/2*x^2*z*t^8-165/2*x*y*z*t^8-135/8*y^2*z*t^8-125/4*x*z^2*t^8
+425/8*y*z^2*t^8-85/8*z^3*t^8-52*x^2*t^9+118*x*y*t^9-89/4*y^2*t^9+6*x*z*t^9
+3/2*y*z*t^9-105/4*z^2*t^9-55*x^2*y*t^7+50*x*y^2*t^7-25/4*y^3*t^7+15*x^2*z*t^7
+90*x*y*z*t^7-225/4*y^2*z*t^7-80*x*z^2*t^7+105/4*y*z^2*t^7+65/4*z^3*t^7-6*x^2*t^8
-81*x*y*t^8+315/4*y^2*t^8+37*x*z*t^8-46*y*z*t^8-51/4*z^2*t^8+5*x^4*t^5
-5*x^3*y*t^5-5/4*x*y^3*t^5+5/16*y^4*t^5-5*x^3*z*t^5-20*x^2*y*z*t^5
+85/4*x*y^2*z*t^5-5/4*y^3*z*t^5-10*x^2*z^2*t^5+185/4*x*y*z^2*t^5
-185/8*y^2*z^2*t^5+15/4*x*z^3*t^5-45/4*y*z^3*t^5+5/16*z^4*t^5
+45*x^3*t^6-85/2*x^2*y*t^6-95/4*x*y^2*t^6+25/8*y^3*t^6-105/2*x^2*z*t^6
+15/2*x*y*z*t^6+615/8*y^2*z*t^6+25/4*x*z^2*t^6-465/8*y*z^2*t^6+225/8*z^3*t^6
+58*x^2*t^7-53*x*y*t^7-137/4*y^2*t^7-29*x*z*t^7+57/2*y*z*t^7+39/4*z^2*t^7
+5*x^3*y*t^4+5/4*x*y^3*t^4-5/8*y^4*t^4-5*x^3*z*t^4+5*x^2*y*z*t^4-115/4*x*y^2*z*t^4
+15/4*y^3*z*t^4+5*x^2*z^2*t^4-5/4*x*y*z^2*t^4+30*y^2*z^2*t^4+35/4*x*z^3*t^4
-95/4*y*z^3*t^4+5/8*z^4*t^4+52*x^2*y*t^5-51/2*x*y^2*t^5+43/8*y^3*t^5-36*x^2*z*t^5
-24*x*y*z*t^5-153/8*y^2*z*t^5+115/2*x*z^2*t^5-119/8*y*z^2*t^5+37/8*z^3*t^5
-2*x^2*t^6+133/2*x*y*t^6-109/4*y^2*t^6-101/2*x*z*t^6+19*y*z*t^6+89/4*z^2*t^6
-10*x^4*t^3+5*x^3*y*t^3+5/4*x*y^3*t^3-5/16*y^4*t^3+5*x^3*z*t^3+35*x^2*y*z*t^3
-85/4*x*y^2*z*t^3+5/2*y^3*z*t^3+10*x^2*z^2*t^3-165/4*x*y*z^2*t^3+85/8*y^2*z^2*t^3
+5/4*x*z^3*t^3+5/2*y*z^3*t^3-5/16*z^4*t^3-33*x^3*t^4+27/2*x^2*y*t^4+35/2*x*y^2*t^4
-17/8*y^3*t^4+27/2*x^2*z*t^4+31*x*y*z*t^4-327/8*y^2*z*t^4+23/2*x*z^2*t^4
+169/8*y*z^2*t^4-97/8*z^3*t^4-25*x^2*t^5+6*x*y*t^5+24*y^2*t^5+14*x*z*t^5
-25*y*z*t^5+8*z^2*t^5-5*x^3*y*t^2-5/4*x*y^3*t^2+5/8*y^4*t^2+5*x^3*z*t^2
-5*x^2*y*z*t^2+115/4*x*y^2*z*t^2-15/4*y^3*z*t^2-5*x^2*z^2*t^2+5/4*x*y*z^2*t^2
-30*y^2*z^2*t^2-35/4*x*z^3*t^2+95/4*y*z^3*t^2-5/8*z^4*t^2-20*x^2*y*t^3
+1/2*x*y^2*t^3-7/4*y^3*t^3+24*x^2*z*t^3-14*x*y*z*t^3+135/4*y^2*z*t^3
-21/2*x*z^2*t^3-15/4*y*z^2*t^3-33/4*z^3*t^3-17*x*y*t^4-9/4*y^2*t^4
+21*x*z*t^4-y*z*t^4-27/4*z^2*t^4+x^5-1/32*y^5-5*x^3*y*z-5/4*x*y^3*z
+15/32*y^4*z-5/2*x^2*y*z^2+25/2*x*y^2*z^2-15/16*y^3*z^2-5/2*x^2*z^3
+35/4*x*y*z^3-175/16*y^2*z^3+15/32*y*z^4-1/32*z^5+5*x^4*t-15*x^2*y*z*t
-5/4*y^3*z*t-5*x*y*z^2*t+25/2*y^2*z^2*t-5*x*z^3*t+35/4*y*z^3*t+8*x^3*t^2
-x^2*y*t^2-5/4*x*y^2*t^2+1/4*y^3*t^2-x^2*z*t^2-7/2*x*y*z*t^2-19/4*y^2*z*t^2
-21/4*x*z^2*t^2+3/4*y*z^2*t^2-9/4*z^3*t^2+4*x^2*t^3-x*y*t^3-5/4*y^2*t^3
-x*z*t^3+17/2*y*z*t^3-21/4*z^2*t^3+x^3*y+1/8*x*y^3-x^3*z-x^2*y*z-7/8*x*y^2*z
+1/4*y^3*z-x^2*z^2+39/8*x*y*z^2-2*y^2*z^2-1/8*x*z^3-1/4*y*z^3+3*x^2*y*t
+1/8*y^3*t-3*x^2*z*t-2*x*y*z*t-7/8*y^2*z*t-2*x*z^2*t+39/8*y*z^2*t-1/8*z^3*t
+2*x*y*t^2-2*x*z*t^2-y*z*t^2-z^2*t^2-x*y*z-y*z*t
\end{verbatim}
As can be seen by a careful inspection (most reliably done using
grep, for example), there are no constant terms and no terms which involve only $t$
(i.e., there are no terms of degree 0 in $x$, $y$ and $z$)
and there are no terms of degree 1 in $x$ and $y$ and $z$, so for every $t$ the polynomial $g$
is in $(x,y,z)^2$ and hence $f\in P^{(2)}$.
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex10.1}
If $f,g\in\CC[x_1,x_2]$ are non-constant polynomials with
no non-constant common factor and if $\deg(g)=1$, show
$|Z(f,g)|\leq \deg(f)$.
[Hint: apply the FTA.]
Give a simple example showing that equality can fail.
\end{exer}
\begin{proof}[Solution (with example by Philip Gipson)]
Since $\deg(g)=1$, we have $g=ax_1+bx_2+c$ and either $a\neq0$ or
$b\neq0$. Let's say $b\neq0$. Then $g=0$ implies $x_2=-ax_1/b-c$.
Plugging this in to $f$ gives a polynomial $h(x_1)=f(x_1,-ax_1/b-c)$
in one variable. If $h$ is identically 0, then $Z(g)\subseteq Z(f)$,
hence $f\in\sqrt{(g)}$, but $g$ is irreducible, so $g|f$,
contradicting $f$ and $g$ having no non-constant common factor.
Thus $h$ is a nonzero polynomial, of degree $r$ say, hence
has at most $r$ roots. I.e., there are at most $r$ points of $Z(g)$ at which
$f$ also vanishes, so $|Z(f,g)|\leq \deg(f)$.
For an example where $|Z(f,g)|< \deg(f)$, let $g(x_1,x_2)=x_1-x_2$
(hence $Z(g)$ is the diagonal line through the point $(1,1)$) and let
$f(x_1,x_2)=x_1x_2-x_1-x_2+1=(x_1-1)(x_2-1)$ (hence $Z(f)$ is the
union of the horizontal and vertical lines through the point $(1,1)$).
In this case, $Z(f,g)=Z(f)\cap Z(g)=\{(1,1)\}$ but
$\deg(f)=2$.
\end{proof}
\begin{exer}\label{ex10.2}
More generally, if $f,g\in\CC[x_1,x_2]$ are non-constant polynomials with
no non-constant common factor, show
there are non-zero polynomials $h_i\in\CC[x_i]$ such that
$Z(f,g)\subseteq Z(h_1)\cap Z(h_2)$. [Hint: to find
$h_i$, consider the gcd of $f$ and $g$ in $\CC(x_i)[x_j]$.]
Conclude that $Z(f,g)$ is a finite set. [Note: This is a weak version of
B\'ezout's Theorem. The classical version of
B\'ezout's Theorem says more precisely
that $|Z(f,g)|\leq \deg(f)\deg(g)$, with equality if you count
the points of $Z(f,g)$ with appropriate multiplicities, including {\em points at infinity}.]
\end{exer}
\begin{proof}[Solution]
We have $f,g\in\CC[x_1,x_2]\subseteq \CC(x_1)[x_2]$.
Let $(f,g)=(h)$ be the ideal generated
by $f$ and $g$ in $\CC(x_1)[x_2]$.
Since non-zero elements $\CC[x_1]$ are units in
$\CC(x_1)$, we may assume that $h\in\CC[x_1,x_2]$.
In fact, we claim that $h\in\CC[x_1]$.
To see this, note that $h|f$ and $h|g$ in $\CC(x_1)[x_2]$,
so there are elements $\phi,\gamma\in\CC[x_1]$ such that
$h|\phi f$ and $h|\gamma g$ in $\CC[x_1,x_2]$.
If $h\not\in\CC[x_1]$, then the prime factorization of $h$ in
$\CC[x_1,x_2]$ has an irreducible factor $e$ involving $x_2$,
and $e$ divides both $\phi f$ and $\gamma g$.
Since $x_2$ does not appear in $\phi$ nor in $\gamma$,
$e$ does not divide $\phi$ or $\gamma$, hence
$e|f$ and $e|g$, contrary to the assumption that $f$ and $g$ have
no non-constant common factor.
Thus $h\in\CC[x_1]$.
Continuing, since $(f,g)=(h)$, we have $h=af+bg$ for some
$a,b\in\CC(x_1)[x_2]$. Pick some polynomial $d\in\CC[x_1]$
such that $da,db\in\CC[x_1,x_2]$. (The coefficients of
$a$ and $b$ are elements of $\CC(x_1)$, so take $d$ to be a common denominator
for all of these coefficients.) Let $h_1=dh$, $A=da$ and $B=db$. Then $dh=daf+dbg$
is $h_1=Af+Bg$, with $A,B\in\CC[x_1,x_2]$ and $h_1\in \CC[x_1]$.
Since $f$ and $g$ are non-constant, $h$ cannot be identically 0, so neither can $h_1$.
Thus $h_1$, being a non-zero polynomial in $\CC[x_1]$, has finitely many roots.
But for any point $(p_1,p_2)=p\in Z(f,g)$ we have $h_1(p_1)=A(p)f(p)+B(p)g(p)=0$,
so $p_1$ is one of these finitely many roots.
Similarly, there is a non-zero polynomial $h_2\in\CC[x_2]$ such that
if $(p_1,p_2)=p\in Z(f,g)$, then $h_2(p_2)=0$. Thus $Z(f,g)\subseteq Z(h_1,h_2)
=Z(h_1)\cap Z(h_2)$,
but since $h_1$ is 0 for only finitely many values of $x_1$ and $h_2$ is 0 for only
finitely many values of $x_2$, we see $Z(h_1)\cap Z(h_2)$ is finite.
\end{proof}
\begin{exer}\label{ex10.3}
Show that every prime ideal $P\subset\CC[x_1,x_2]$
is either principal or maximal. [Hint: One way to do this is to
apply Exercise \ref{ex10.2}.]
\end{exer}
\begin{proof}[Solution 1, by Kat Shultis (with added details)]
We know from commutative algebra that
$$\dim(R[x_1,\ldots,x_n])=\dim(R)+n.$$
As $\CC$ is a field, it has dimension zero. Thus, in terms of Krull dimension,
$\dim(\CC[x_1,x_2])=2$, meaning that if $P\subset\CC[x_1,x_2]$ is a prime ideal
but not $(0)$ or maximal, then there is a maximal ideal $M$ such that
$(0)\subsetneq P\subsetneq M$ and no other prime ideals fit in this chain.
If $0\neq f\in P$ is an irreducible element
(which must exist since $P\neq (0)$ and $P$ is prime), then we have
$(0)\subsetneq (f)\subseteq P\subsetneq M$, and hence $(f)=P$ so $P$ is principal.
Thus all prime ideals of $\CC[x_1,x_2]$ are principal or maximal.
\end{proof}
\begin{proof}[Solution 2, by Douglas Heltibridle (with added details)]
First, assume $P\subset\CC[x_1,x_2]$ is a prime ideal that is not principal.
Pick generators $f_1,\ldots,f_r$ of $P$. Since $P$ is prime, each $f_i$ has an irreducible factor in $P$.
Let $g_i$ be such a factor for each $i$. Then $P=(f_1,\ldots,f_r)\subseteq (g_1,\ldots,g_r)\subseteq P$,
so $P=(g_1,\ldots,g_r)$. By choosing a subset of the $g_i$ we may assume that
$g_1,\ldots,g_r$ is a minimal generating set (i.e, that no proper subset of the $g_i$ generates $P$).
Thus if $i\neq j$ then $g_i$ does not divide $g_j$. Also, since $P$ is not principal, we know $r>1$.
Let $f=g_1$ and let $g=g_2$. Then both are irreducible and neither divides the other, so
in particular $f$ and $g$ have no non-constant common factor.
Thus we can apply Exercise 10.2 so that we have $h_1\in \mathbb C[x_1]$ and $h_2\in \mathbb C[x_2]$
such that $Z(f,g)\subseteq Z(h_1)\cap Z(h_2)$. We have $Z(h_1)\supseteq Z(h_1)\cap Z(h_2)$ and
$Z(h_2)\supseteq Z(h_1)\cap Z(h_2)$, and since $f,g \in P$ we have $Z(f,g)\supseteq Z(P)$.
Thus $Z(h_1)\supseteq Z(P)$ and $Z(h_2)\supseteq Z(P)$ and by version three of the Nullstellensatz
we know $h_1\in P$ and $h_2\in P$.
Now, if $h_1$ is not prime then we can factor it and one of its factors must be in $P$. Continuing to
factor we find an irreducible factor of $h_1$, which is also in $P$ and $\mathbb C[x_1]$. Similarly
we find an irreducible factor of $h_2$ in $P$ and $\mathbb C[x_2]$. These are of the form $x_1-z_1$
and $x_2-z_2$ for some $z_1,z_2\in \mathbb C$. So we know that $P\supseteq (x_1-z_1,x_2-z_2)$.
However, $\mathbb C[x_1,x_2]/(x_1-z_1,x_2-z_2)$ is a field, which means that $(x_1-z_1,x_2-z_2)$
is a maximal ideal. Thus $P=(x_1-z_1,x_2-z_2)$. Therefore we have that $P$ is maximal as desired.
\end{proof}
\begin{exer}\label{ex10.4}
Show that $P^m$ for a prime ideal $P\subset\CC[x_1,\ldots,x_n]$ is always $P$-primary if
$P$ is either maximal or principal.
Conclude that $P^m$ is $P$-primary for all primes
$P\subset\CC[x_1,\ldots,x_n]$ and all $m$ if and only if $n\leq2$.
\end{exer}
\begin{proof}[Solution]
If $P$ is maximal, then $P^m$ is $P$-primary by Exercise 7.3, since
$\sqrt{P^m}=P$. Say $P=(f)$. Since $P$ is prime, $f$ cannot be a non-zzero scalar.
If $f=0$, then $P^m=P$ for all $m\geq1$, hence $P^m$ is $P$-primary.
If $f$ is not a scalar, then $f$ is irreducible. Say $gh\in P^m=(f^m)\subseteq P=(f)$.
Thus $f^m$ divides $gh$. If $g\not\in P^m$, then the largest power of $f$ that divides
$g$ is $f^{m-1}$, hence (since $\CC[x_1,\ldots,x_n]$ is a UFD) $f$ must divide $h$,
so $h^m\in P^m$; i.e., $h\in P=\sqrt{P^m}$. Thus $P^m$ is $P$-primary.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{subsection}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{February 7, 2011}\label{lect11}
\subsection{The Example continued}
To finish the example from last lecture,
we must check that $P\subseteq (x,y,z)^2$.
This follows from the fact that no non-zero polynomial with terms
of degree less than 2 can vanish on $C$. For let $u\in P\subset\CC[x,y,z]$.
Intuitively, if $u$ had a non-zero constant term,
then $u$ does not vanish at the origin, but the origin is a point of $C$,
so $u\not\in I(C)=P$. If $u$ had
any non-zero linear terms, say $u=ax+by+cz + {\rm higher\ order\ terms}$ for
$a,b,c\in \CC$ not all 0, then the zero locus $Z(u)$ near $(0,0,0)$
would look like a plane (the plane in question would be the zero-locus $Z(ax+by+cz)$
of the non-zero
linear terms, which is the plane tangent to the surface defined by $u=0$).
But $C$ has branches with tangents
in three linearly independent directions, so $u=0$ could not encompass all of them.
We now show more rigorously that
$u$ has no terms of degree less than 2.
Since $(0,0,0)\in C$ and $u$ vanishes on $C$, we have $u(0,0,0)=0$
so $u$ has no constant term (or, if you prefer, the constant term is 0).
Now consider the image $h(u)=h(u(x,y,z))=u(h(x),h(y),h(z))$ of $u$ under the homomorphism
$h$ corresponding to our parameterization of $C$. In principle $h(u)$ is
a function of $t$, but since $u$ vanishes on $C$ (or because $u\in P=\ker h$),
in fact $u(h(x),h(y),h(z))$ is indentically zero, and hence $du(h(x),h(y),h(z))/dt=0$.
Thus
$$0=\frac{du(h(x),h(y),h(z))}{dt} =
\frac{\partial u}{\partial x}\frac{dh(x)}{dt}+
\frac{\partial u}{\partial y}\frac{dh(y)}{dt}+
\frac{\partial u}{\partial z}\frac{dh(z)}{dt}.$$
%(\partial u/\partial y)dh(y)/dt+(\partial u/\partial z)dh(z)/dt.$$
At $t=0$, explicit computation (using the fact that
$h(x)=t(t-1)^2(t+1)^2$, $h(y)=t^2(t-1)(t+1)^2$, and $h(z)=t^2(t-1)^2(t+1)$)
shows that $dh(y)/dt=0$ and $dh(z)/dt=0$, but
$dh(x)/dt=1$, thus we must have $\partial u/\partial x=0$ at $(0,0,0)$; i.e., there can be
no $x$ term. Similar calculations at $t=\pm1$ show that
$\partial u/\partial y$ and $\partial u/\partial z$ vanish at $(0,0,0)$, and hence there
are no $y$ or $z$ terms, and hence no terms of degree 1. Thus $u\in (x,y,z)^2$.
\begin{rem}
Let $f\in\CC[x_1,\ldots,x_n]$ and let $p=(a_1,\ldots,a_n)\in\CC^n$.
Then
$$f(x_1,\ldots,x_n)=\sum_{i_1.\ldots,i_n}
\frac{1}{i_1!\cdots i_n!}\frac{\partial^{i_1+\ldots+i_n}f}{\partial^{i_1}x_1\cdots\partial^{i_n}x_n}(p)
(x_1-a_1)^{i_1}\cdots(x_n-a_n)^{i_n}.$$
This is just a Taylor series expansion for $f$ near the point
$p$. Since $f$ is a polynomial, the expansion is itself a polynomial.
The terms of least degree in the $x_i-a_i$ in this expansion
give useful information about the behavior of $Z(f)$ near $p$.
For example, if $\frac{\partial f}{\partial x_i}(p)\neq0$ for some $i$, then
the zero locus of the linear terms define the hyperplane tangent to $Z(f)$
at the point $p$; i.e., the tangent hyperplane is
$$Z\Big(\sum_i\frac{\partial f}{\partial x_i}(p)(x_i-a_i)\Big).\eqno(\star)$$
For example, if $n=2$, $p=(0,0)$ and we take our variables
to be $x$ and $y$ with $f=y-x^3+x$, then from calculus
the tangent line to $y=x^3-x$ at $x=0$ is $y=-x$,
whereas $(\star)$ gives $Z(y+x)$, which is the same thing.
\end{rem}
\subsection{Algebraic maps}
\begin{defn}
We say a map $F:\CC^n\to\CC^m$ is an {\em algebraic map} if there exist
$g_1,\ldots,g_m\in\CC[x_1,\ldots,x_n]$ such that for every
$(a_1,\ldots,a_n)\in\CC^n$ we have
$$F((a_1,\ldots,a_n))=(g_1(a_1,\ldots,a_n),\ldots,g_m(a_1,\ldots,a_n)).$$
If $C\subseteq \CC^n$ and $D\subseteq \CC^m$ are algebraic subsets
(i.e., Zariski closed subsets), we say a map $f:C\to D$ is an {\em algebraic map}
if there exists an algebraic map $F:\CC^n\to\CC^m$ such that $f=F|_C$.
\end{defn}
\renewcommand{\thethm}{\arabic{section}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{February 9, 2011}\label{lect12}
\begin{example}
Simple examples of algebraic maps are identity maps ${\rm id}_C:C\to C$, where $C$
is an algebraic set, and the inclusion maps $i_C:C\to \CC^n$, where $C$ is a
Zariski closed subset of $\CC^n$. Moreover, if $f:C\to D$ and $h:D\to E$ are algebraic maps,
then so is $h\circ f$; i.e., compositions of algebraic maps are algebraic.
\end{example}
\begin{defn}
We say an algebraic map $f:C\to D$ is an {\em isomorphism} if there exists
an algebraic map $g:D\to C$ such that $f\circ g={\rm id}_D$ and
$g\circ f={\rm id}_C$, in which case we write $C\cong D$ and say
$C$ and $D$ are isomorphic algebraic sets.
\end{defn}
\begin{notation}
We refer to $\CC^n$ as {\em affine $n$-space}, and denote it as
$\AAA^n_\CC$, or just $\AAA^n$.
\end{notation}
\begin{example}
Let $C=Z(y-x^2)\subset\AAA^2$, and let $D=\AAA^1$.
Then $C\cong D$. Specifically, define $f:C\to D$ by $f((a,b))=a$.
Note that $(a,b)\in C$ means $b=a^2$.
Then $f$ is an algebraic map, and $g:D\to C$, where $g(a)=(a,a^2)$,
is an algebraic inverse.
\end{example}
\begin{example}
Let $C=\AAA^1$, and let $D=Z(y^3-x^2)\subset\AAA^2$.
Then $f:C\to D$ defined by $a\mapsto (a^3,a^2)$ is algebraic and a bijection,
but it turns out that $f$ is not an isomorphism.
If it were, there would be a polynomial $g\in\CC[x,y]$ such that
$a=g(a^3,a^2)$ for all $a\in \CC$. But this would mean that
$a-g(a^3,a^2)=0$ for all $a\in\CC$, hence $t-g(t^3,t^2)=0\in \CC[t]$,
so $g(t^3,t^2)=t$. But $g(x,y)$ is in the $\CC$-vector space span of all monomials
$x^iy^j$, hence $g(t^3,t^2)$ is in the span of the monomials
$t^{3i+2j}$; i.e., $g(t^3,t^2)$ is in the $\CC$-vector space span of
$\{1,t^2,t^3,t^4,\cdots\}$. But $\{1,t,t^2,t^3,t^4,\cdots\}$ is a vector space basis
for $\CC[t]$ hence $t$ is not in the span of $\{1,t^2,t^3,t^4,\cdots\}$.
Thus $g(t^3,t^2)=t$ is impossible and $f$ is not an isomorphism.
\end{example}
\begin{prop}\label{contProp}
Let $C\subseteq \CC^n$ and $D\subseteq \CC^m$ be algebraic sets.
An algebraic map $f:C\to D$ is continuous in both the standard and the Zariski topologies.
\end{prop}
\begin{proof} Let $F:\CC^n\to \CC^m$ be an algebraic map such that $F|_C=f$.
Since polynomials are continuous in the standard topology, an algebraic map
$F:\CC^n\to \CC^m$ is continuous in the standard topology, hence so is $f$,
since $f=F|_C$, and $C$ has the subspace topology.
Similarly, to show $f$ is continuous in the Zariski topology, it suffices to show
that $F$ is. Now, $F=(f_1,\ldots,f_m)$ for polynomials $f_i\in\CC[x_1,\ldots,x_n]$.
Each closed subset of $\CC^m$ is of the form $\cap_jZ(g_j)$ for some family
of polynomials $g_j\in\CC[y_1,\ldots,y_m]$, and
$F^{-1}(\cap_jZ(g_j))=\cap_jF^{-1}(Z(g_j))$,
so it's enough to show that $F^{-1}(Z(g_j))$ is closed.
But $F^{-1}(Z(g_j))=F^{-1}(g_j^{-1}(0))=(g_jF)^{-1}(0)=Z(g_j\circ F)$ is closed, since
$g_j\circ F=g_j(f_1,\ldots,f_m)\in\CC[x_1,\ldots,x_n]$.
\end{proof}
\begin{rem}
Since algebraic maps are continuous in both the standard and the Zariski topologies,
isomorphic algebraic sets are homeomorphic in both topologies.
\end{rem}
We now associate a ring, called the affine coordinate ring, to each algebraic set.
\begin{defn}
Let $V\subseteq\AAA^n$ be an algebraic set. The {\em affine coordinate ring}
(or simply the coordinate ring) of $\AAA^n$ is $\CC[x_1,\ldots,x_n]$, denoted
$\CC[\AAA^n]$. The affine coordinate ring $\CC[V]$ of $V$ is
$\CC[\AAA^n]/I(V)$. Note that $V$ comes with a canonical quotient homomorphism,
$\overline{\ \ \vbox to5pt{\vss}}\!:\CC[\AAA^n]\to\CC[V]$. This homomorphism is actually given by restriction;
i.e., $f\mapsto\overline{f}=f|_V$.
\end{defn}
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex12.1}
Let $C$ be $Z(xy-1)$ in $\AAA^2$.
Show that $C$ and $\AAA^1$ are homeomorphic
in the Zariski topology but not in the usual topology.
[Hint: use algebraic topology.]
Conclude that $C$ and $\AAA^1$ are not isomorphic
algebraic sets.
\end{exer}
\begin{proof}[Solution]
First note that $xy-1$ is irreducible. To see this, suppose $gh=xy-1$. Note that
$\deg(gh)=\deg(g)+\deg(h)$. Thus if neither $g$ nor $h$
is invertible, then $\deg(g)=\deg(h)=1$, which implies that $Z(xy-1)$ is either a line
or a union of two lines. But any line intersects either the horizontal or vertical
coordinate axis, while $xy-1=0$ has no solution where either $x=0$ or $y=0$.
Thus $xy-1$ must be irreducible. Since $xy-1$ is irreducible, $(xy-1)$ is prime
so $I(Z(xy-1))=(xy-1)$.
Now let $D$ be a non-empty proper closed subset of $C$. Then there is a
polynomial $f\in I(D)$ which does not vanish on all of $C$. Thus $f\not\in (xy-1)$,
so $f$ and $xy-1$ have no common factor. By Exercise 10.2, $Z(f,xy-1)$ is a finite
set, but $D\subseteq Z(f,xy-1)$ so $D$ is finite. Thus $C$ has the finite complement topology.
Since $C$ and $\AAA^1$ have the same cardinality there is a bijection, and since for both
the Zariski topology is the finite complement topology, any bijection is a homeomorphism
(and in fact it's not hard to write down an explicit bijection).
However, $C=Z(xy-1)$ is homeomorphic in the standard topology to $\AAA^1\setminus\{\rm origin\}$,
and thus $C$ is not simply connected. Since $\AAA^1$ is simply connected (contractible even),
$C$ and $\AAA^1$ are not homeomorphic. In particular, there do not exist
inverse algebraic maps $f$ and $g$ between $C$ and $\AAA^1$,
so $C$ and $\AAA^1$ are not isomorphic algebraic sets.
\end{proof}
\mysection{February 11, 2011}\label{lect13}
Given an algebraic subset $V\subseteq\AAA^n$,
the coordinate ring $\CC[V]$ on $V$ is the set of restrictions
of polynomials on $\AAA^n$. Among these are the constant functions,
and thus we have a canonical inclusion $i_V:\CC\subset \CC[V]$.
We say that $\CC[V]$ is a $\CC$-algebra.
If $W\subseteq\AAA^m$ is an algebraic subset, and if $h:\CC[W]\to\CC[V]$
is a ring homomorphism, we say $h$ is a $\CC$-homomorphism
if $h\circ i_W=i_V$. In particular, the canonical quotient homomorphism
$\overline{\ \ \vbox to5pt{\vss}}\!:\CC[\AAA^n]\to\CC[V]$ is a $\CC$-homomorphism, since
$\overline{\ \ \vbox to5pt{\vss}}(g)=g|_V$ is just restriction, and constants restrict to constants.
\noindent{\bf Construction I.}
Given algebraic subsets $C\subseteq\AAA^n$ and $D\subseteq\AAA^m$
and an algebraic map $f:C\to D$, we construct a $\CC$-homomorphism $f^*:\CC[D]\to \CC[C]$
and we show that if $\phi:C\to D$ is an algebraic map with $\phi^*=f^*$, then $\phi=f$.
First we define $f^*$ in case $C=\AAA^n$ and $D=\AAA^m$. In that case for each
$g\in\CC[\AAA^m]$ we define
$f^*(g)=g\circ f$. Since $f=(f_1,\ldots,f_m)$ for $f_i\in\CC[\AAA^n]$ and $g=g(y_1,\ldots,y_m)$,
where we assume that $\CC[\AAA^m]=\CC[y_1,\ldots,y_m]$ is a polynomial ring in the variables $y_i$,
we have $f^*(g)=g\circ f=g(f_1,\ldots,f_m)\in\CC[\AAA^n]$. Moreover, for any $g_i\in\CC[\AAA^m]$,
$f^*(g_1+g_2)=(g_1+g_2)\circ f= g_1\circ f+g_2\circ f = f^*(g_1)+f^*(g_2)$
and likewise $f^*(g_1g_2)=f^*(g_1)f^*(g_2)$, and clearly $f^*$ is the identity on constants,
so $f^*$ is a $\CC$-homomorphism.
Now consider the general case. Then $f=F|_C$ for some algebraic mapping
$F=(f_1,\ldots,f_m):\AAA^n\to\AAA^m$. Let $\gamma\in\CC[D]$ and pick $g\in\CC[\AAA^m]$
such that $\gamma=\overline{g}$.
We define $f^*(\overline{g})=\overline{F*^(g)}$. Note that $\overline{F*^(g)}=g\circ F|_C\in\CC[C]$.
Also note that $\overline{F*^(g)}=f^*(\overline{g})$: $\overline{F*^(g)}=g\circ F|_C$ is equal to
$g|_D\circ F|_C=\overline{g}\circ f=f^*(\overline{g})$
since $F(C)\subseteq D$. In particular, $f^*$ is well-defined, since $f^*(\gamma)=\overline{\gamma}\circ f$
is independent of the choice of $g$. As before $f^*$ is a $\CC$-homomorphism.
Finally, consider an algebraic map $\phi:C\to D$ with $\phi^*=f^*$, but suppose $\phi\neq f$.
Since $\phi\neq f$, there is a $c\in C$ such that $f(c)\neq \phi(c)$, and hence the
coordinates of the points $f(c)$ and $\phi(c)$ are not all the same. Say they differ in the $i$th coordinate.
Then $y_i(f(c))\neq y_i(\phi(c))$, and hence $f^*(\overline{y_i})\neq \phi^*(\overline{y_i})$,
so $\phi^*\neq f^*$, contradicting our hypothesis.
In the next lecture, given a $\CC$-homomorphism $h:\CC[D]\to\CC[C]$,
we will construct an algebraic map $f:C\to D$ such that $f^*=h$.
\setcounter{thm}{0}
\mysection{February 14, 2011}\label{lect14}
\noindent{\bf Construction II.}
Given algebraic subsets $C\subseteq\AAA^n$ and $D\subseteq\AAA^m$
and a $\CC$-homomorphism $h:\CC[D]\to \CC[C]$, we find an algebraic map $f:C\to D$
such that $h=f^*$.
We have the canonical quotients $\overline{\ \ \vbox to5pt{\vss}}\!:\CC[\AAA^n]\to\CC[C]$
and $\overline{\ \ \vbox to5pt{\vss}}\!:\CC[\AAA^m]\to\CC[D]$.
Suppose $\CC[\AAA^m]$ is a polynomial ring $\CC[y_1,\ldots,y_m]$ in variables $y$
and $\CC[\AAA^n]$ is a polynomial ring $\CC[x_1,\ldots,x_n]$ in variables $x$.
Let $\phi_i=h(\overline{y_i})$ and choose elements $f_i\in \CC[\AAA^n]$
such that $\overline{f_i}=\phi_i$ for $i=1,\ldots,m$. Let $H:\CC[\AAA^m]\to\CC[\AAA^n]$
be the unique $\CC$-homomorphism such that $H(y_i)=f_i$ for $i=1,\ldots,m$.
Then we have $\overline{H(g)}=h(\overline{g})$ for all $g\in\CC[\AAA^m]$.
Now $F=(f_1,\ldots,f_m)$ defines an algebraic map $F:\AAA^n\to\AAA^m$, and
clearly $H=F^*$.
First we claim that $F(C)\subseteq D$. To see this it's enough to check that
$g(F(c))=0$ for all $c\in C$ and all $g\in I(D)$. Note that $\overline{g}=0$
since $g\in I(D)$. Thus
$$g(F(c))=(F^*(g))(c)=(H(g))(c)
=(H(g))|_C(c) = \overline{H(g)}(c)=(h(\overline{g}))(c)=(h(0))(c)=0$$
for all $c\in C$ and all $g\in I(D)$.
Since as we now see $F(C)\subseteq D$, $F$ restricts to an algebraic map $f=F|_C:C\to D$.
Moreover, for any $\gamma\in\CC[D]$ there is a $g\in \CC[\AAA^m]$ with
$\overline{g}=\gamma$ and we have
$f^*(\overline{g})= g|_D\circ F|_C=g\circ F|_C=\overline{H(g)}=h(\overline{g})$
so $f^*=h$, as we wanted.
We thus have the following theorem:
\begin{thm}\label{mapsVShomoms}
The $*$-operation $f\mapsto f^*$ gives a bijective correspondence from algebraic maps
of algebraic sets to $\CC$-homomorphisms of their coordinate rings.
\end{thm}
\begin{proof} The mapping $f\mapsto f^*$ is injective by Construction I and it's surjective
by Construction II.
\end{proof}
\begin{cor}\label{isocor}
A mapping $f:C\to D$ of algebraic sets is an isomorphism if and only if
$f^*:\CC[D]\to\CC[C]$ is an isomorphism.
\end{cor}
\begin{proof}
Suppose $f$ is an isomorphism. Then there is a mapping $g:D\to C$ such that
$g\circ f={\rm id}_C$ and $f\circ g={\rm id}_D$. Thus
$f^*\circ g^*=(g\circ f)^*=({\rm id}_C)^*={\rm id}_{\CC[C]}$ and similarly
$g^*\circ f^*={\rm id}_{\CC[D]}$, so $f^*$ is an isomorphism with inverse $g^*$.
Conversely, suppose $f^*$ is an isomorphism and let $h$ be its inverse.
Then there is a mapping $g:D\to C$ with $h=g^*$, and we have
$(g\circ f)^*=f^*\circ g^*={\rm id}_{\CC[C]}=({\rm id}_C)^*$,
hence $g\circ f={\rm id}_C$ and similarly
$f\circ g={\rm id}_D$. Thus $f$ is an isomorphism with inverse $g$.
\end{proof}
\begin{example}
Exercise \ref{ex12.1} asks you to give a topological explanation for why
$C=Z(xy-1)$ in $\AAA^2$ is not isomorphic to $D=\AAA^1$.
We can use Corollary \ref{isocor} to give an algebraic reason.
If $f:D\to C$ were an isomorphism, then $f^*:\CC[C]\to\CC[D]$ would be an isomorphism too.
But $\CC[C]=\CC[x,y]/(xy-1)\cong \CC[x,\frac{1}{x}]$ and $\CC[D]=\CC[t]$.
Let $h: \CC[x,\frac{1}{x}]\to\CC[t]$ be any $\CC$-homomorphism.
Then since $h(x)$ has inverse $h(\frac{1}{x})$, but the only invertible elements
of $\CC[t]$ are non-zero constants, then $h(x)\in\CC$, hence ${\rm Im}(\CC[C])=\CC$,
so $h$ is not an isomorphism. Since $\CC[C]$ and $\CC[D]$ are not isomorphic,
neither are $C$ and $D$.
\end{example}
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex14.1}
Show that every algebraic mapping $f:\AAA^1\to\AAA^n$ is closed (i.e., that the image of
each Zariski-closed subset is closed).
\end{exer}
\begin{proof}[Solution by Jason Hardin]
Write $f=(f_1,\ldots,f_n)$. Let $C\subseteq\mathbb{A}^1$ be a closed subset. If $C=\emptyset$,
then $f(C)=\emptyset$ is closed. If $C=\mathbb{A}^1$, then Ex 9.1 shows that $f(C)$ is closed.
If $C\neq\emptyset,\mathbb{A}^1$, then by Ex 2.3 we know $C=\{c_1,\ldots,c_m\}$. For $i=1,\ldots,m$,
define ideals $I_i\subset\mathbb{C}[y_1,\ldots,y_n]$ by $I_i:=(y_1-f_1(c_i),\ldots,y_n-f_n(c_i))$.
Note that $Z(I_i)=(f_1(c_i),\ldots,f_n(c_i))=f(c_i)$. Thus,
$f(C)=\bigcup_{i=1}^mf(c_i)=\bigcup_{i=1}^mZ(I_i)=Z\left(\bigcap_{i=1}^mI_i\right)$ is closed.
\end{proof}
\begin{exer}\label{ex14.2}
For some $m$ and $n$, give an example of an algebraic mapping $f:\AAA^n\to\AAA^m$
which is surjective but not closed.
\end{exer}
\begin{proof}[Solution by Becky Egg]
Consider $f:\AAA^2 \rightarrow \AAA^1$ defined by $f(a_1,a_2)=a_1$. So $f$ is clearly surjective.
To see that $f$ is not closed, consider $V=Z(xy-1) \subseteq \AAA^2$. Note that $V$ is closed in
$\AAA^2$, with $V=\{(x,1/x)|x \neq 0\}$. So $f(V)=\AAA^1 \backslash \{0\}$. Note $f(V)=\AAA^1\backslash Z(1)$,
so $f(V)$ is open in $\AAA^1=\CC$. Since $\CC$ is connected, the only subsets of it which are both
open and closed are $\CC$ and $\emptyset$. So $f(V)$ is not closed, and hence $f$ is not a closed map.
\end{proof}
\begin{exer}\label{ex14.3}
Let $f:C\to D$ be an algebraic mapping of algebraic sets. If $C$ is irreducible, show that
the Zariski closure of $f(C)$ is irreducible.
\end{exer}
\begin{proof}[Solution by Zheng Yang (with added details)]
By Exercise 6.2, it suffices to show $I(Z(I(f(C)))$ is a prime ideal.
We justify and then use the fact for any subset $B\subset C$ that $I(f(B))=(f^*)^{-1}(I(B))$.
(Note $g\in I(f(B))$ if and only if $g$ vanishes on $f(B)$ if and only if $f^*(g)=g\circ f$ vanishes on $B$
if and only if $f^*(g)\in I(B)$ if and only if $g\in(f^*)^{-1}(I(B))$.)
But $I(C)$ is prime by Exercise 6.2, and contraction of a prime ideal is prime under $f^*$
so $I(f(C))=(f^*)^{-1}(I(C))$ is prime. But
$I(f(C))=I(Z(I(f(C)))$ by the Nullstellensatz (Theorem 5.1.3) and
$Z(I(f(C)))=\overline{f(C)}$ by Exercise 2.9. And by
Exercise 6.2 again $I(f(C))=I(\overline{f(C)})$ is prime.
\end{proof}
\begin{exer}\label{ex14.4}
For each $n\geq1$, show that an algebraic mapping $f:\AAA^n\to\AAA^1$ is either surjective or constant
(i.e, has image a single point).
\end{exer}
\begin{proof}[Solution by Nora Youngs]
Let $f:\AAA^n\rightarrow\AAA^1$ be an algebraic mapping. Then, by definition, there is some
$g_1\in \CC[x_1,...,x_n]$ such that for every $(a_1,...,a_n)\in \AAA^n$ we have $f(a_1,...,a_n)=g(a_1,...,a_n)$.
If $g_1$ is a constant polynomial, then $f$ is constant.
If $g_1$ is not a constant polynomial: Let $c\in \CC$ be given. To show that $f$ is surjective,
we need to show that $g_1(x_1,...,x_n)=c$ has a solution. Equivalently, we must show $g_1-c=0$ has a solution.
Let $I=(g_1-c)$. Note $g_1-c$ is not constant, so $I$ is a proper ideal of $\CC[x_1,...,x_n]$.
Thus, by Version 1 of the Nullstellensatz, $g_1-c$ has a zero.
Thus, there is a solution to $g_1-c=0$; hence, $f$ is surjective.
\end{proof}
\begin{exer}\label{ex14.5}
For some $m$ and $n$, give an example of an algebraic mapping $f:\AAA^n\to\AAA^m$
that is neither surjective nor constant.
\end{exer}
\begin{proof}[Solution by Philip Gipson]
Consider the embedding $f:\CC\to\CC^2$ via $f(x)=(x,0)$. This is certainly algebraic but neither
constant nor surjective.
\end{proof}
\begin{exer}\label{ex14.6}
Let $F:\AAA^{n+m}\to\AAA^m$ be the projection mapping
$(a_1,\ldots,a_n,b_1,\ldots,b_m)\mapsto(b_1,\ldots,b_m)$.
It is well known that $F$ is an open mapping in the standard topology (i.e.,
that the image of an open subset is open). Show that $F$ is
an open algebraic mapping for the Zariski topology.
\end{exer}
\begin{proof}[Solution by Katie Morrison]
It is that clear that $F$ is algebraic since it is given by the collection of polynomials
$f(x_1, \ldots, x_{n+m}) = (x_{n+1}, \ldots, x_{n+m})$. To see $F$ is open, let
$A \subseteq \AAA^{n+m}$ be an open set, then there exists some polynomials
$g_i \subseteq \CC[x_1, \ldots, x_{n+m}]$, $i \in I$, such that $\AAA^{n+m} \setminus A = \cap_{i \in I} Z(g_i)$
or equivalently, $A = \cup_{i\in I}( \AAA^{n+m}\setminus Z(g_i))$. Since
$F\left(\cup_{i\in I}( \AAA^{n+m}\setminus Z(g_i))\right) = \cup_{i\in I}F( \AAA^{n+m}\setminus Z(g_i))$
and the union of open sets is open, it suffices to show that each $F( \AAA^{n+m}\setminus Z(g_i))$ is
open, and so we restrict to $A = \AAA^{n+m}\setminus Z(g)$ for some $g \in \CC[x_1, \ldots, x_{n+m}]$.
Then $$A= \{(a_1, \ldots, a_{n+m})~:~g(a_1, \ldots, a_{n+m})\neq 0\},$$ and so
$$F(A)= \{(a_{n+1}, \ldots, a_{n+m})~|~ \exists~ a_1, \ldots, a_n \textrm{ s.t. }
g(a_1, \ldots, a_n, a_{n+1}, \ldots, a_{n+m})\neq 0\}.$$ Let
$i_{(b_1, \ldots,b_n)}:\AAA^m \to \AAA^{n+m}$ denote the inclusion map
where $(c_1, \ldots, c_m)\mapsto (b_1,\ldots, b_n, c_1,\ldots,c_m)$. Then we see
\begin{eqnarray*}
\AAA^m \setminus F(A) &=& \{(b_{n+1}, \ldots, b_{n+m})~:~ \forall~b_1, \ldots, b_n,~ g(b_1, \ldots, b_n, b_{n+1}, \ldots, b_{n+m})=0\}\\
&=& \cap_{(b_1, \ldots b_n) \in \CC^n} Z(g\circ i_{(b_1, \ldots b_n)}),
\end{eqnarray*}
which is a closed set since it is the intersection of closed sets. Thus we have
that $F(A)$ is open, and so $F$ is an open map.\qed
\end{proof}
\begin{exer}\label{ex14.7}
Consider the standard topology on the closed interval $X=[-1,1]$.
Give an example of continuous maps $f,g:X\to X$ such that
$f\circ g={\rm id}_X$ but such that $g\circ f\neq{\rm id}_X$.
Conclude that $g$ is not a homeomorphism.
\end{exer}
\begin{proof}[Solution by Nora Youngs]
Let $f=\sin(\pi x)$ and $g=\frac{\arcsin(x)}{\pi}$. [Note that $\arcsin$ has been scaled
to have image points only in the interval $[-1,1]$.
Then, for any $x\in [-1,1]$ $$(f\circ g)(x)=\sin\left(\pi\frac{\arcsin(x)}{\pi}\right)=\sin(\arcsin(x))=x$$
by definition of $\arcsin$.\\
However, $g\circ f$ is not the identity: Consider $x=1$.
Then $$(g\circ f)(1)=\frac{\arcsin(\sin(\pi\cdot1))}{\pi}=\frac{\arcsin(0)}{\pi}=0,$$ so $g\circ f$
is not the identity for $x=1$.
\end{proof}
\begin{exer}\label{ex14.8}
Let $f,g:\AAA^n\to\AAA^n$ be algebraic maps such that
$f\circ g={\rm id}_{\AAA^n}$.
Show that $g\circ f={\rm id}_{\AAA^n}$ and hence that
$g$ is an isomorphism. [Hint: Look up and apply Ax's Theorem,
sometimes also called the Ax-Grothendieck theorem.]
\end{exer}
\begin{proof}[Solution by Ashley Weatherwax]
We'll begin by showing that $g$ is bijective. Let $c, d \in \AAA^n$ such that $g(c) = g(d)$.
Then $f \circ g (c) = f \circ g(d)$. However, as we assumed that $f \circ g = id_{\AAA^n}$, we get immediately
that $c = d$, and thus $g$ is injective. Then, by the Ax-Grothendieck theorem, $g$ is, in fact, bijective.
Now, let $c \in \AAA^n$ and consider $g \circ f (c)$. As $g$ is bijective, there exists and $a \in \AAA^n$
such that $g (a) = c$. Then
$$g \circ f (c) = g \circ f (g(a)) = g(a) = c$$
as $f \circ g = id_{\AAA^n}$. Therefore $g\circ f = id_{\AAA^n}$ and $g$ is an isomorphism.
\end{proof}
\setcounter{thm}{0}
\mysection{February 16, 2011}\label{lect15}
\noindent{\bf MaxSpec.}
\begin{lem}\label{pteqvs}
Let $V\subset\AAA^n$ be an algebraic set. Then the following are equivalent:
\begin{itemize}
\item[(a)] specifying a point $v\in V$;
\item[(b)] specifying a $\CC$-homomorphism $\CC[V]\to \CC$; and
\item[(c)] specifying a maximal ideal $M\subset \CC[V]$.
\end{itemize}
\end{lem}
\begin{proof}
(a) implies (b): Having $v\in V$ is the same as $\{v\}\subseteq V$, which we know induces a $\CC$-homomorphism
$\CC[V]\to \CC[v]=\CC$.
(b) implies (c): Given a $\CC$-homomorphism $h:\CC[V]\to \CC$, let $M={\rm ker}(h)$.
Since $h$ is surjective and $\CC$ is a field, $M$ is a maximal ideal.
(c) implies (a): Given a maximal ideal $M\subset \CC[V]$, consider
$$\CC[\AAA^n] \xrightarrow{\ q_1\ }\CC[V] \xrightarrow{\ q_2\ }\CC[V]/M=\CC.$$
Since $q_2\circ q_1$ is surjective with image a field, $N={\rm ker}(q_2\circ q_1)$
is a maximal ideal of $\CC[\AAA^n]$, hence $N=I(v)$ for some point $v\in\AAA^n$, by
Theorem \ref{vers2Null}, version 2 of the Nullstellensatz.
Since clearly $I(V)\subseteq N=I(v)$, we see that $v\in V$.
\end{proof}
\begin{rem}
By Lemma \ref{pteqvs}, we see that the maximal ideals of $\CC[V]$
are exactly the ideal of the form $I(v)/I(V)$ for points $v\in V$.
We will denote $I(v)/I(V)$ by $I_V(v)$.
\end{rem}
\begin{cor}\label{invofmaxismax}
Let $V\subseteq\AAA^n$ and $W\subseteq\AAA^m$ be algebraic sets.
Let $h:\CC[W]\to\CC[V]$ be a $\CC$-homomorphism.
Then $h^{-1}(M)$ is a maximal ideal for every maximal ideal $M\subset\CC[V]$.
\end{cor}
\begin{proof}
Note that
$$\CC[W] \xrightarrow{h}\CC[V] \xrightarrow{q_2}\CC[V]/M=\CC$$
is a surjection to a field, so has kernel a maximal ideal $N=h^{-1}(M)$.
\end{proof}
\begin{example}
If $h:R\to S$ is a homomorphism of rings, it is not in general true that
the inverse image of a maximal ideal is maximal. For example,
let $h:{\mathbb Z}\to{\mathbb Q}$ be inclusion of the integers in the rationals.
Then $(0)\subset{\mathbb Q}$ is maximal, and $h^{-1}((0))=((0))$, but
$(0)\subset{\mathbb Z}$ is not maximal.
\end{example}
\begin{rem}
If $f:V\to W$ is the algebraic map corresponding to the homomorphism $h$
in Corollary \ref{invofmaxismax}, if $v\in V$ is the point with $I_V(v)=M$
and if $w\in W$ is the point with $I_W(w)=h^{-1}(M)$, then $w=f(v)$ and
$(f^*)^{-1}(I_V(v))=I_W(f(v))$. See Exercise \ref{ex15.1}.
\end{rem}
\begin{defn}\label{ZarTopMaxSpec}
Let $R$ be a commutative ring with $1\neq0$.
Then $\MaxSpec{(R)}$ is defined to be the topological space whose point set is
the set of maximal ideals of $R$, and where the closed sets are the subsets
$C\subseteq\MaxSpec{(R)}$ of the form $C=C_J$ for some ideal $J\subseteq R$, where
$C_J=\{M\in\MaxSpec{(R)}: J\subseteq M\}$. This topology is, as you might guess,
called the Zariski topology.
\end{defn}
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex15.1}
Let $f:V\to W$ be an algebraic map of algebraic sets.
Let $v\in V$. Show that $(f^*)^{-1}(I_V(v))=I_W(f(v))$.
\end{exer}
\begin{proof}[Solution by Becky Egg]
Let $V \subseteq \CC[x_1, \ldots, x_n]$ and $W \subseteq \CC[y_1, \ldots, y_m]$ be algebraic sets, and
$f:V \rightarrow W$ an algebraic map. Let $\overline{h}=h+I(W) \in I_W(f(v))$. So $h \in \CC[y_1, \ldots, y_m]$,
with $h(f(v))=0$. Then
\[[f^*(h)](v)=h(f(v))=0,\]
so $h\circ f=f^*(h) \in I(v)$, and hence $\overline{f^*h} \in I_V(v)$. So $\overline{h} \in (f^*)^{-1}(I_V(v))$. \\
Now let $\overline{g}=g+I(W) \in (f^*)^{-1}(I_V(v))$. So $g \in \CC[y_1, \ldots, y_m]$, and $f^*(g)=g \circ f \in I(v)$.
So $g(f(v))=0$ implies that $g \in I(f(v))$, and hence $\overline{g} \in I_W(f(v))$. Thus we have $(f*)^{-1}(I_V(v))=I_W(f(v))$.
\end{proof}
\begin{exer}\label{ex15.2}
Show that the sets of the form $C_J$ as given in Definition \ref{ZarTopMaxSpec}
do indeed comprise the closed sets of a topology.
\end{exer}
\begin{proof}[Solution 1, by Douglas Heltibridle]
Recall that in Definition 15.1.6 we have $C_J=\{M\in \MaxSpec(R):J\subseteq M\}$ where $J$ is
an ideal in $R$. As $(0)\subseteq M$ for all $M\in \MaxSpec(R)$ we have that $C_{(0)}=\MaxSpec(R)$,
and as $R\not\subseteq M$ for any $M\in \MaxSpec(R)$ we have $C_R=\emptyset$.
Now let $C_J$ and $C_I$ be closed sets. Since $J\cap I\subseteq J$ and $J\cap I\subseteq I$ we
have $I\subseteq M$ and $J\subseteq M$ both imply $J\cap I\subseteq M$. Thus $C_{I\cap J}\supseteq C_J\cup C_I$.
Now, let $M\in C_{J\cap I}$, and suppose that $I\not\subseteq M$ and $J\not\subseteq M$, then there
exists $i\in I$ such that $i\not\in M$ and $j\in J$ such that $j\not\in M$. However, as $I$ and $J$ are
ideals $ij\in I\cap J$. Since $M$ is prime, $ij\not\in M$ and thus $M\not\supseteq I\cap J$. Thus we
have a contradiction and so either $M\supseteq I$ or $M\supseteq J$, which means $M\in C_J\cup C_I$.
Therefore $C_{J\cap I}\subseteq C_J\cup C_I$ and we have equality. Thus the union of the arbitrary
closed sets $C_I$ and $C_J$ is itself a closed set.
Next, let $\{C_I\}_{I\in \mathcal{I}}$ be a family of closed sets with $\mathcal I$ the set of ideals in $R$.
First, $M\in \cap_{\mathcal{I}} C_I$ if and only if $M\in C_J$ for each $J\in \mathcal{I}$. Then, $M\in C_J$
for each $J\in \mathcal I$ if and only if $J\subseteq M$ for all $J\in \mathcal I$. Finally $J\subseteq M$
for all $J\in \mathcal{I}$ if and only if $\cup_{\mathcal{I}} J\subseteq M$. Therefore
$M\in C_{\cap_{\mathcal{I}}C_I}$ if and only if $M\in \cap_{\mathcal{I}}C_I$, which means
$\cap_{\mathcal{I}}C_I = C_{\cap_{\mathcal{I}}C_I}$. Thus arbitrary intersections of closed
sets are closed, and so we have that the closed sets of the form $C_J$ give a topology for $\MaxSpec(R)$.
\end{proof}
\begin{proof}[Solution 2, by Kat Shultis]
Let $C_J:=\{M\in \MaxSpec(R)|J\subseteq M\}$ be sets in $\MaxSpec(R)$ which are
defined for any ideal $J$ of $R$. Then $C_R=\emptyset$ and $C_{(0)}= \MaxSpec(R)$.
Thus we have the emptyset and the whole set in the collection of $C_J$. Let $I,J$ be
ideals in $R$. Then as $M\in \MaxSpec(R)$ are all prime, we know that $IJ\subseteq M$
if and only if $I\subseteq M$ or $J\subseteq M$. Hence $C_I\cup C_J= C_{IJ}$ so that the
collection of $C_J$ is closed under finite intersections. Next, let $\{I_\alpha\}_{\alpha\in A}$
be an arbitrary collection of ideals in $R$. Then, $J=\bigoplus_{\alpha\in A} I_\alpha$ is an
ideal of $R$. Also, $M\in C_J$ if and only if $\bigoplus_{\alpha\in A} I_\alpha=J\subseteq M$,
which is true if and only if $I_\alpha\subseteq M$ for all $\alpha\in A$. Hence
$C_J=\bigcap_{\alpha\in A} C_{I_\alpha}$, and our collection of $C_J$ is closed under
arbitrary intersections. Hence, the collection of $C_J$ define the closed sets of a topology on
$\MaxSpec(R)$.
\end{proof}
\begin{exer}\label{ex15.3}
Let $V\subseteq\AAA^n$ be an algebraic set. Define a map
$h_V:V\to\MaxSpec{(\CC[V])}$ by
$h_V:v\mapsto I_V(v)$. Show that $h_V$ is a homeomorphism in the Zariski topologies.
\end{exer}
\begin{proof}[Solution by Philip Gipson]
We already know from Lemma \ref{pteqvs} that this correspondence
is bijective, so all that remains to be shown is that it is Zariski-continuous. To that end consider
a closed set $C_J\subseteq \MaxSpec(\CC[V])$. We know that $C_J=\{M:J\subseteq M\}$ for
some ideal $J\subseteq \CC[V]$. Since each maximal ideal $M$ is uniquely expressible as $M=
I_V(v_M)$ for some $v_M\in V$ we have that
$$C_J=\{M:J\subseteq M\}=\{I_V(v_M):J\subseteq I_V(v_M)\}=\{I_V(v_M):v_M\in Z(J)\}$$
and therefore
$$h_V^{-1}(C_J)=\{h_V^{-1}(I_V(v_M)):v_M\in Z(J)\}=\{v_M:v_M\in Z(J)\}=Z(J).$$
Since $h_V$ is bijective, we also have $h_V(Z(J))=C_J$. Thus both $h_V$ and $h_V^{-1}$
are continuous and so $h_V$ is a homeomorphism.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{subsection}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{February 18, 2011}\label{lect16}
\subsection{More on MaxSpec}
Let $V$ and $W$ be algebraic sets, and let $f:\MaxSpec{(\CC[V])}\to\MaxSpec{(\CC[W])}$
be any map. Then $\widetilde{f}=h_W^{-1}\circ f\circ h_V$ (where $h_V$ is as defined in Exercise \ref{ex15.3})
is a map $\widetilde{f}: V\to W$ such that $f(I_V(v))=I_W(\widetilde{f}(v))$.
\begin{defn}
Let $V$ and $W$ be algebraic sets, and let $f:\MaxSpec{(\CC[V])}\to\MaxSpec{(\CC[W])}$
be a map. We say $f$ is {\em algebraic} if there is a $\CC$-homomorphism
$\phi:\CC[W]\to\CC[V]$ such that $f(I_V(v))=\phi^{-1}(I_V(v))$ for all $v\in V$.
\end{defn}
By Exercise \ref{ex16.2}, the algebraic maps $V\to W$ correspond bijectively via
$h_V$ and $h_W$ to the algebraic maps $\MaxSpec{(\CC[V])}\to\MaxSpec{(\CC[W])}$.
Thus we can in some sense regard an algebraic set $V$ as being the same thing as
$\MaxSpec(\CC[V])$. But the elements of $\CC[V]$ can be regarded as being functions
on $V$. We might ask in what sense do they give functions on $\MaxSpec(\CC[V])$.
If $f\in\CC[V]$, then $f:V\to \CC$ is algebraic, hence corresponds to the
$\CC$-homomorphism $f^*:\CC[t]=\CC[\CC]to\CC[V]$ defined by $t\mapsto f$.
Similarly, $f\in\CC[V]$ corresponds to the algebraic map
$\widetilde{f}:\MaxSpec(\CC[V])\to\MaxSpec(\CC[\CC])=\CC$.
There are several ways to think about $\widetilde{f}$. Given $v\in V$,
then $\widetilde{f}(I_V(v))=I_{\CC}(f(v))\subset\CC[\CC]$, or
$\widetilde{f}(I_V(v))=f+I_V(v)\in \CC[V]/I_V(v)$, or
$\widetilde{f}(I_V(v))=(f^*)^{-1}(I_V(v))=(t-f(v))\subset \CC[\CC]$.
The advantage of $\MaxSpec$ is that $\MaxSpec(R)$ makes sense for any commutative ring $R$
with $1\neq0$.
The problem with $\MaxSpec$ is that whereas for algebraic sets $V$ and $W$ the algebraic maps
$\MaxSpec{(\CC[V])}\to\MaxSpec{(\CC[W])}$ correspond to $\CC$-homomorphisms
$\CC[W]\to\CC[V]$, it's not clear how to define algebraic maps
$\MaxSpec{(R)}\to\MaxSpec{(S)}$ for arbitrary rings $R$ and $S$.
\subsection{France to the fore: Grothendieck and the Paris school}
The solution is Grothendieck's notion of {\em affine scheme}:
\begin{defn}
Let $R$ be a commutative ring with $1\neq0$. Then $\Spec(R)$ is a topological space
whose point set is the set of all prime ideals of $R$ and where the closed sets are the subsets
of the form $C_J=\{P\in\Spec{R}: J\subseteq P\}$ where $J\subseteq R$ is an ideal.
(Given ideals $J\subseteq I$, note that $C_I\subseteq C_J$.)
The topology is referred to as the Zariski topology on $\Spec(R)$, and we refer to
$\Spec(R)$ as an affine scheme.
\end{defn}
\begin{example}
If $R=\CC[V]$ for an algebraic set $V$, we can by Exercise \ref{ex16.3} regard $V$ as a
dense (but usually proper) subset of $\Spec(\CC[V])$. The points of $\Spec(\CC[V])$
are the prime ideals of $\CC[V]$, which correspond to the irreducible subsets of $V$;
i.e., for every irreducible subset of $V$ we get a point of $\Spec(\CC[V])$.
These points are not always closed. If $P\in\Spec(\CC[V])$, then
the Zariski closure $\overline{P}\subseteq\Spec(\CC[V])$ is
$$\overline{P}=\cap_{P\in C_J}C_J=\cap_{J\subseteq P}C_J=C_P=\{Q\in\Spec(\CC[V]):P\subseteq Q\}.$$
So, for example, if $V=\AAA^2$, where $\CC[\AAA^2]=\CC[x,y]$,
and if $P=(x)$, then $\overline{P}$ consists of all prime ideals that contain $x$.
In addition to $P$ itself, these are precisely the maximal ideals $(x,y-c)$ for $c\in \CC$; i.e., the ideals of points
on $Z(x)$.
\end{example}
Exercise \ref{ex15.3} suggests that it makes sense to identify
an algebraic set $V$ with $\MaxSpec(\CC[V])$. Under this identification
the points of $V$ are identified with the maximal ideals of the coordinate ring.
However, $\Spec$ has nicer formal properties than $\MaxSpec$.
One reason is that given any homomorphism $f:A\to B$ of commutative
rings (we'll always assume rings have $1\neq0$ and that a homomorphism
takes $1_A$ to $1_B$), if $P\subset B$ is a prime ideal, then
$f^{-1}(P)$ is a prime ideal of $A$, but it need not be true that
$f^{-1}(M)$ is a maximal ideal of $A$ just because $M$ is a maximal ideal
of $B$.
The first person to think of replacing the points of $V$ by the primes
of $\CC[V]$ may have been Emmy Noether in the 1920s (although
regarding the mathematics of her time---she died in 1935--Noether remarked
``it is all already in Dedekind''.)
Wolfgang Krull suggested thinking of prime ideals in arbitrary commutative rings
as points in a topological space
in some lectures in the 1930s, but his ideas were not taken seriously
at the time. The advantages of doing so became apparent to the Paris school
of the 1950s, especially in the work of Jean-Pierre Serre and Alexander Grothendieck.
Given a commutative ring $R$, Grothendieck referred to $\Spec(R)$
as an {\em affine scheme} and used it as the foundation for his general theory
of schemes (although Serre has said that no one invented schemes;
they were in the air in Paris in the 1950s.) [For a fun and interesting article
exploring the history of these ideas, and which I used as
a source for this paragraph, see:
\url{http://www.math.jussieu.fr/~leila/grothendieckcircle/mclarty1.pdf}.]
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex16.1}
Let $V$ and $W$ be algebraic sets, and let $f:\MaxSpec{(\CC[V])}\to\MaxSpec{(\CC[W])}$
be a map.
\begin{itemize}
\item[(a)] Show that $f$ is algebraic if and only if $\widetilde{f}: V\to W$ is algebraic.
\item[(b)] If $f$ is algebraic, show that $f(I_V(v))=(\widetilde{f}^*)^{-1}(I_V(v))=I_W(\widetilde{f}(v))$ for all $v\in V$.
\end{itemize}
\end{exer}
\begin{proof}[Solution by Katie Morrison (with minor changes)]
(a) $(\Leftarrow)$
If $\widetilde{f}$ is algebraic, we have the following diagram:
$$\begin{CD}
\MaxSpec(\CC[V]) @>f>> \MaxSpec(\CC[W])\\
@AA {h_V} A @AA h_W A\\
V @>\phantom{hit}\widetilde{f}\phantom{hit}>> W\\
@VV {I_V}V @VVI_W V\\
\MaxSpec(\CC[V]) @>>\phantom{hit}(\widetilde{f}^*)^{-1} \phantom{hit}>\MaxSpec(\CC[W])
\end{CD}$$
where $h_V, h_W$ are defined as in Exercise 15.3, and $\widetilde{f}^*:\CC[W]\to\CC[V]$,
defined by $g\mapsto g \circ \widetilde{f}$, is a $\CC$-homomorphism. Now the top
square commutes by definition of $\widetilde{f}$, so $f(I_V(v)) = I_W(\widetilde{f}(v))$, and the bottom
square commutes since $I_W(\widetilde{f}(v)) = (\widetilde{f}^*)^{-1}(I_V(v))$ by
Exercise 15.1. Thus, $f(I_V(v))=(\widetilde{f}^*)^{-1}(I_V(v))$ for the
$\CC$-homomorphism $\widetilde{f}^*$, so $f$ is algebraic.
$(\Rightarrow)$ If $f$ is algebraic, then there exists some $\CC$-homomorphism
$\phi^*:\CC[W]\to \CC[V]$ such that $f(I_V(v))=(\phi^*)^{-1}(I_V(v))$
(and hence we have the corresponding algebraic map $\phi:V\to W$)
and we obtain the following commutative diagram
$$\begin{CD}
\MaxSpec(\CC[V]) @>f>> \MaxSpec(\CC[W])\\
@AA {h_V} A@AA h_W A\\
V @>\phantom{hit}\phi\phantom{hit}>> W\\
@VV{I_V} V @VV I_W V\\
\MaxSpec(\CC[V]) @>>\phantom{hit}(\phi^*)^{-1} \phantom{hit}>\MaxSpec(\CC[W])
\end{CD}$$
Since $\widetilde{f}$ also makes the diagram commute and the maps
$h_V$ and $h_W$ are bijective by Lemma \ref{pteqvs}, we have $\widetilde{f}=\phi$,
so $\widetilde{f}$ is algebraic.
(b) Since $f$ is algebraic, $\widetilde{f}$ is algebraic as well by part (a). Then
$f(I_V(v)) = I_W(\widetilde{f}(v))$ since the diagram from part (a) commutes, and
$I_W(\widetilde{f}(v)) = (\widetilde{f}^*)^{-1}(I_V(v))$ by Exercise 15.1. Thus
$f(I_V(v))=(\widetilde{f}^*)^{-1}(I_V(v))=I_W(\widetilde{f}(v))$ for all $v\in V$.
\end{proof}
\begin{exer}\label{ex16.2}
Let $V$ and $W$ be algebraic sets. Show that $f\mapsto\widetilde{f}$
gives a bijection from the set of algebraic maps $\MaxSpec(\CC[V])\to\MaxSpec(\CC[W])$
to the set of algebraic maps $V\to W$.
\end{exer}
\begin{proof}[Solution]
Since $h_V$ and $h_W$ are bijective and $h_W\circ \widetilde{f}=f\circ h_V$,
we see that $f\mapsto\widetilde{f}$ is a bijection from the set of all maps
$\MaxSpec(\CC[V])\to\MaxSpec(\CC[W])$ to the set of all maps $V\to W$, and by Exercise \ref{ex16.1}
$\widetilde{f}$ is algebraic if and only if $f$ is. Hence $f\mapsto\widetilde{f}$
gives a bijection from the set of algebraic maps $\MaxSpec(\CC[V])\to\MaxSpec(\CC[W])$
to the set of algebraic maps $V\to W$.
\end{proof}
\begin{exer}\label{ex16.3}
Let $V$ be an algebraic set. Let $i_V : V \to \Spec(\CC[V])$ be the map $i_V : v\to I_V(v)$.
Show that $i_V$ is a continuous injective map such that the closure of $i_V(V)$ is $\Spec(\CC[V])$.
\end{exer}
\begin{proof}[Solution]
Note that $\MaxSpec(\CC[V])$ is a subset of $\Spec(\CC[V])$,
and that a subset $C\subseteq \MaxSpec(\CC[V])$ is closed
if and only if there is an ideal $J\subseteq \CC[V]$ such that
$C= \{M\in\MaxSpec(\CC[V]): J\subseteq M\}$. But a subset $D\subseteq \Spec(\CC[V])$ is closed
if and only if there is an ideal $J\subseteq \CC[V]$ such that
$D= \{P\in\Spec(\CC[V]): J\subseteq P\}$. Thus
the closed sets $C$ of $\MaxSpec(\CC[V])$ are precisely the sets of the form
$C=D\cap \MaxSpec(\CC[V])$ where $D$ is closed in $\Spec(\CC[V])$.
Thus $\MaxSpec(\CC[V])$ is a topological subspace of $\Spec(\CC[V])$,
hence the inclusion $\MaxSpec(\CC[V])\subseteq \Spec(\CC[V])$ is continuous.
However, $h_V$ is a homeomorphism and $i_V$ factors as
$$V\xrightarrow{\ h_V\ }\MaxSpec(\CC[V])\subseteq \Spec(\CC[V]).$$
Thus $i_V$ is a composition of injective continuous maps, hence is itself
injective and continuous. We now show that $i_V(V)$ is dense in
$\Spec(\CC[V])$. Since $V$ is an algebraic set, we have $V=Z(J)$ for some
ideal $J\subseteq\CC[x_1,\ldots,x_n]$ for some $n$. By Exercise \ref{ex5.7},
$\sqrt{J}=\cap M$, where the intersection is over all maximal ideals $M$ of
$\CC[x_1,\ldots,x_n]$ containing $J$. Since $\CC[V]=\CC[x_1,\ldots,x_n]/\sqrt{J}$,
we see that $\sqrt{(0)}=\cap \overline{M}$ where now the intersection is
over all maximal ideals $\overline{M}\subset\CC[V]$ containing $0$.
Thus if $D$ is a closed subset of $\Spec(\CC[V])$ containing
all maximal ideals $\overline{M}$ of $\CC[V]$, then
$D=C_J$ for some ideal $J\subseteq\CC[V]$ such that
$J\subseteq \overline{M}$ for all maximal ideals $\overline{M}$, hence
$J\subseteq\sqrt{(0)}$, so $D=C_J=C_{(0)}=\Spec(\CC[V])$,
so $i_V(V)$ is dense.
\end{proof}
\begin{exer}\label{ex16.4}
Let $R$ be a commutative ring.
Show that $P\in\Spec(R)$ is a closed point if and only if $P$ is a maximal ideal.
\end{exer}
\begin{proof}[Solution by Anisah Nu'Man]
Suppose $P$ is a maximal ideal. Clearly by definition we have $P\in C_P$.
Let $Q \in C_P$. Then $P\subseteq Q$ and $Q$
is a prime ideal of $R$. Since $Q$ is a proper ideal we have $Q=P$.
Therefore we have $C_P = \{P\}$, and so is a closed
point. Now Suppose $P\in\Spec(R)$ is a closed point, hence $C_P = \{P\}$.
Since every ideal is contained in
a maximal ideal there exists a maximal ideal $M$ such that
$P\subseteq M$. Thus $M\in C_P = \{P\}$ so we have $M=P$. Therefore $P$ is a maximal
ideal.
\end{proof}
\begin{exer}\label{ex16.5}
Let $R$ be a commutative ring. Since every maximal ideal is a prime ideal we have
the inclusion $\MaxSpec(R)\subseteq\Spec(R)$.
Show that this is continuous in the Zariski topology, but give an example to show that
$\MaxSpec(R)$ need not be dense in $\Spec(R)$. [Hint: for the example, do not pick
$R$ to be of the form $\CC[V]$ for some algebraic set $V$, since by
Exercises \ref{ex15.3} and \ref{ex16.3}, $\MaxSpec(\CC[V])$ is dense in $\Spec(\CC[V])$.]
\end{exer}
\begin{proof}[Solution]
The proof that the inclusion $\MaxSpec(R)\subseteq\Spec(R)$ is continuous
(i.e., that $\MaxSpec(R)$ is a topological subspace of
$\Spec(R)$) was given in the solution to Exercise \ref{ex16.3}.
For the example, let $R$ be any integral domain which is not a field but which has
a single maximal ideal $M$. Since $R$ is not a field, $(0)\subsetneq M$, and since
$R$ is an integral domain, $(0)$ is a prime ideal. But now $\MaxSpec(R)$ is the single
closed point of $\Spec(R)$, hence not dense.
Here is a specific example of such an $R$. Let $R$ be the set of rational numbers with odd
denominators. It's easy to check this is a subring of the rationals, and hence a domain.
Suppose $M\subseteq R$ is a maximal ideal. Let $a\in M$ be an element. The $a=f/g$
where $f$ and $g$ are integers with $g$ odd. Since $g$ is a unit, we have $(a)=(f)$.
Since $M\subsetneq R$, we cannot have $a$ being odd (since then $a$ would be
invertible). Since odds are invertible, we can assume $f$ is either 0 or a power of $2$.
Thus $a\in (2)$, and hence $M\subseteq (2)$. Since 2 is not invertible, we see
$(2)\subsetneq R$, hence $M=(2)$. I.e., $(2)$ is a maximal ideal of $R$ and it is
the only maximal ideal of $R$, but $2\neq 0$.
\end{proof}
\mysection{February 21, 2011}\label{lect17}
\subsection{Morphisms of affine schemes}
A morphism $\phi:\Spec(S)\to\Spec(R)$ of affine schemes
is a pair $\phi:(f,f^\#)$, where $f^\#:R\to S$ is a ring homomorphism
and $f:\Spec(S)\to\Spec(R)$ is the map $f(Q)=(f^\#)^{-1}(Q)$.
Clearly, $f^\#$ determines $f$. The utility of defining a morphism
as a pair is clearer for morphisms of schemes that are not affine.
As we will see, schemes are built by gluing together affine schemes,
in much the same way that an $n$-dimensional manifold is
built by gluing together open balls of ${\mathbb R}^n$.
I.e., a scheme in general is locally an affine scheme and morphisms of
schemes are defined by gluing together morphisms of affine schemes,
but if $\phi=(f,f^\#):X\to Y$ is a morphism of schemes $X$ and $Y$,
$f:X\to Y$ is a globally defined continuous map, but there is no globally defined
ring of functions and hence no globally defined homomorphism of rings $f^\#$.
Instead $f^\#$ is essentially a collection of locally defined homomorphisms,
whose relationships to each other are codified by a {\em sheaf} and depend on $f$.
One of the most important examples of non-affine schemes are projective
spaces. But instead of defining them scheme theoretically to start with, we
again take a historical approach first.
\begin{example}
Let $f^\#:\CC\to\CC$ be the identity and let $g^\#:\CC\to\CC$ be complex conjugation.
Define $\phi_i:\Spec(\CC)\to\Spec(\CC)$ as
$\phi_1=(f,f^\#)$ and $\phi_2=(g,g^\#)$, where
$f=g={\rm id}_{\CC}$. Then both $\phi_1$ and $\phi_2$ are morphisms,
but $\phi_1\neq\phi_2$.
\end{example}
\subsection{Projective Space}
Intuitively, projective space is a compactification of affine space obtained by
adding points at infinity. For example, $\pr1$ is the one point compactification
of $\AAA^1$; in the standard topology, $\pr{1}_\CC$ is the 2-sphere.
But $\pr2$ is not the one point compactification of $\AAA^2$. A basic result about
$\pr2$ is that any two lines intersect in a single point. We regard parallel lines
as being equivalent, and for each equivalence class of parallel lines we add a point
at infinity. For any two parallel lines we regard this point at infinity as where the two lines
meet. Of course, two lines are parallel exactly when they have the same slope
(which can be infinite if the lines are vertical). Thus the points at infinity can be regarded
as an $\CC=\AAA^1$ (representing the possible slopes) plus a point
representing infinity; i.e., the points at infinity that we add to $\AAA^2$ to get
$\pr2$ themselves comprise a $\pr1$, so $\pr2=\AAA^2\cup\pr1$. I.e., the points at infinity
for $\pr2$ form a {\em line} at infinity, and this line is a $\pr1$.
In general, $\pr{n}$ is defined in such a way that $\pr{n}=\AAA^n\cup\pr{n-1}$.
More rigorously, $\CC\setminus\{0\}$ acts on $\CC^{n+1}\{0\}$ by scalar multiplication.
Then $\pr{n}$ is the quotient space $(\CC^{n+1}\{0\})/(\CC\setminus\{0\})$; its points are
the orbits of $\CC\setminus\{0\}$ under the action (i.e., the lines through the origin).
It's conventional in this context to think of $\CC[\CC^{n+1}]$ as being $\CC[x_0,\ldots,x_n]$.
Each non-zero point $p=(a_0,\ldots,a_n)$ is contained in a unique orbit which we denote by
$[p]$, and each orbit is $[p]$ for some non-zero point $p\in\CC^{n+1}$.
One can think of $\AAA^n$ as a subset of $\pr{n}$ in the following way.
For each $0\leq i\leq n$, let $U_0=Z(x_0=1)$. Thus there is a bijection
$\AAA^n\to U_i$ given by inserting a 1 in the $0$th spot:
$(a_1,\ldots,a_n)\mapsto(1,a_1,\ldots,a_n)$. Every line through the origin,
except those lying in the plane $x_0=0$ intersect $U_0$ in a unique point.
The lines through the origin lying in the plane $x_0=0$ form a $\pr{n-1}$.
Thus $\AAA^n\cong U_0$ and $U_0\cup\pr{n-1}=\pr{n}$.
Figure \ref{p2} shows a figurative way to think of $\pr2$, with the points at infinity comprising a line
$z=0$. Also shown are two lines which in $\AAA^2$ are parallel, but which intersect at a point at infinity.
\setcounter{figure}{0}
\begin{figure}[h]
\setlength{\unitlength}{0.75cm}
\begin{picture}(5,4)(0,0)
\put(3.5,0){$y=0$}
\put(0,3.6){$x=0$}
\put(1,2){$z=0$}
\linethickness{1pt}
\put(1.85,.85){\circle*{.2}}
\put(0,0){\line(0,1){3.3}}
\put(.5,.175){\line(2,1){2.3}}
\put(.2,.3){\line(3,1){2.3}}
\put(-.3,3){\line(1,-1){3.2}}
\put(0,0){\line(1,0){3.3}}
\end{picture}
\caption{Figurative representation of $\pr2$.}\label{p2}
\end{figure}
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex17.1}
Let $\phi:(f,f^\#)$ be a morphism $\phi:\Spec(S)\to\Spec(R)$ of affine schemes.
Show that $f:\Spec(S)\to\Spec(R)$ is continuous in the Zariski topology.
\end{exer}
\begin{proof}[Solution by Katie Morrison]
Let $C \subseteq \Spec(R)$ be a Zariski-closed set. Then $C$ is the intersection of sets of
the form $C_J$ for ideals $J \subseteq R$. Since the pre-image of an intersection of sets is
the intersection of the pre-images of the sets and the intersection of closed sets is closed, it
suffices to show that the pre-image of $C=C_J$ is closed. Tracing through definitions of $C_J$ and $f$, we see
\begin{eqnarray*}
f^{-1}(C_J)&=&\{P\in \Spec(S)~:~f(P) \supseteq J\}\\
&=& \{P \in \Spec(S)~:~(f^\#)^{-1}(P) \supseteq J\}\\
&=&\{P \in \Spec(S)~:~P\supseteq f^\#(J)\}\\
&=&\{P \in \Spec(S)~:~P\supseteq (f^\#(J))\}\\
&=&C_{(f^\#(J))}.
\end{eqnarray*}
Thus, the pre-image of closed sets is closed, and so $f$ is continuous.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{February 23, 2011}\label{lect18}
\noindent{\bf Homogeneity.}
\begin{defn}
Let $F\in\CC[x_0,\ldots,x_n]$. We say $F$ is {\em homogeneous} if either $F=0$ or each
monomial term appearing in $F$ has the same total degree, this being $\deg(F)$.
And if $I\subseteq\CC[x_0,\ldots,x_n]$ is an ideal, we say $I$ is {\em homogeneous} if
$I=(F_1,\ldots,F_r)$ for some homogeneous polynomials $F_i$.
\end{defn}
\begin{example}
Consider $F=x_0^2-x_1\in\CC[x_0,x_1]$. Then $F$ is not homogeneous.
On the other hand $H=3x_0^2-x_0x_1\in\CC[x_0,x_1]$ is homogeneous, since each term
has the same total degree, 2.
\end{example}
\begin{example}
The ideal $I=(3x_0^2-x_0x_1, x_0^3)\subset\CC[x_0,x_1]$ is homogeneous, since each generator is
homogeneous. However the ideal $J=(x_0^2-x_1)\subset\CC[x_0,x_1]$ is not homogeneous
(see Exercise \ref{ex18.1}). On the other hand,
the ideal $L=(x_0^2-x_1,x_1)\subset\CC[x_0,x_1]$ is homogeneous since
$L=(x_0^2,x_1)$ has a set of homogeneous generators.
\end{example}
{\em Terminology}: Let $R=\CC[x_0,\ldots,x_n]$ and $F\in R$.
Let $F_i$ be the sum of the monomial terms of $F$ of total degree $i$.
If $F$ has ho terms of degree $i$, set $F_i=0$. Note that $F_i$ is homogeneous for all $i$, and
$F=\sum_i F_i$. We refer to $F_i$ as the {\em homogeneous component} of $F$ of degree $i$.
If we denote by $R_j$ the $\CC$-vector space span of the monomials in $R$ of total degree $j$,
note that $R_iR_j=R_{i+j}$. This gives a ring structure to $R_0\oplus R_1\oplus \cdots$, and
the map $h:R_0\oplus R_1\oplus \cdots\to R$ induced by sending each monomial to itself
is a ring isomorphism (see Exercise \ref{ex18.4}). We refer to $R_j$ as the {\em homogeneous component}
of $R$ of degree $j$. If $I\subseteq R$ is a homogeneous ideal, then
$h$ also induces an isomorphism $I_0\oplus I_1\oplus \cdots\to I$ where
$I_j=I\cap R_j$ (see Exercise \ref{ex18.5}). We refer to $I_j$ as the {\em homogeneous component}
of $I$ of degree $j$. Note that $R_iR_j=R_{i+j}$
\begin{example}
For $F=6x_1^4+x_0x_1+x_0^2-x_1$, we have $F_0=0$, $F_1=-x_1$, $F_2=x_0x_1+x_0^2$, $F_3=0$, $F_4=6x_1^4$
and $F_i=0$ for $i>4$.
\end{example}
\begin{rem}\label{ZeroLocusOfHomogF}
Let $F\in\CC[x_0,\ldots,x_n]$. One can check homogeneity $F$ by looking
at the zero-locus of $F$: by Exercise \ref{ex18.6}, $F$ is homogeneous if and only if
$Z(F)\subseteq \CC^{n+1}$ is either the origin or a union of lines through the origin.
For example, consider $F=x^2+y\in\CC[x,y]$. Then $F$ is not homogeneous
and $Z(F)$ is neither the origin nor a union of lines through the origin, but
$Z(G)$ for $G=x^2+xy$ is the union of two lines through the origin.
\end{rem}
\begin{notation}
For each point $p\in\CC^{n+1}\setminus\{{\rm origin}\}$,
let $[p]\in\pr{n}$ denote the orbit of $p$ under the action of $\CC^*$
on $\CC^{n+1}\setminus\{{\rm origin}\}$. Note that every point of $\pr{n}$ is of the form
$[p]$ for some $p\in\CC^{n+1}\setminus\{{\rm origin}\}$.
\end{notation}
If $F$ is a non-constant homogeneous polynomial $F\in\CC[x_0,\ldots,x_n]$,
then $F$ does not give a well-defined function on $\pr{n}$, since
$F(tp)=t^dF(p)$, where $t=\deg(F)$. Thus $F$ can take different values at
different representatives $tp\in[p]$. But the zero-locus of $F$ is well-defined, since
for any $t\neq0$ we have $F(p)=0$ if and only if $F(tp)=0$.
\begin{defn}
A projective algebraic subset $V$ of $\pr{n}$ is a subset of the form
$V=Z(F_1,\ldots,F_r)$ for homogeneous polynomials $F_i\in\CC[x_0,\ldots,x_n]$,
where $Z(F_1,\ldots,F_r)$ means $Z(F_1)\cap\cdots\cap Z(F_r)$.
More generally, if $I\subseteq \CC[x_0,\ldots,x_n]$
is a homogeneous ideal, we define $Z(I)$ to be $\cap Z(F)$,
where the intersection is over all homogeneous $F\in I$.
By Exercise \ref{ex18.7}, $Z(I)=Z(F_1,\ldots,F_r)$ for any set of homogeneous generators
$F_i$ of $I$.
\end{defn}
The projective algebraic subsets of $\pr{n}$ comprise the closed sets of a topology on $\pr{n}$, called the Zariski
topology. However, the interplay between closed sets and ideals is more complicated than for affine space:
\begin{example}
If $I\subseteq \CC[\AAA^n]$ has $Z(I)=\varnothing\subseteq \AAA^n$, then $I=(1)$.
However, if $I\subseteq \CC[x_0,\ldots,x_n]$ has $Z(I)=\varnothing\subseteq \pr{n}$, then
$I=(1)$ is only one of many possibilities. We will see the other possibilities next time.
\end{example}
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex18.1}
Show that the ideal $J=(x_0^2-x_1)\subset\CC[x_0,x_1]$ is not homogeneous.
\end{exer}
\begin{proof}[Solution]
By Exercise \ref{ex18.3}. if $J$ were homogeneous, then we would have
$x_0^2, x_1\in J$. But this would mean, in particular, that $x_0^2-x_1$ divides $x_1$,
which it does not. Thus $J$ is not homogeneous. [Note that it is not enough to
just show that $x_0^2-x_1$ is not homogeneous, since for example neither of the given
generators of $(x_0^2-x_1, x_0^2+x_1)$ is homogeneous, but
$(x_0^2-x_1, x_0^2+x_1)=(x_0^2,x_1)$ is homogeneous since we can find another set of generators
each element of which is homogeneous.]
\end{proof}
\begin{exer}\label{ex18.2}
Let $0\neq F\in\CC[x_0,\ldots,x_n]$ be homogeneous, $d=\deg(F)$.
\begin{itemize}
\item[(a)] Show that $F(tx_0,\ldots,tx_n)=t^dF(x_0,\ldots,x_n)$.
\item[(b)] (Euler's formula) Show that $\sum_ix_i\frac{\partial F}{\partial x_i}=dF$.
\end{itemize}
\end{exer}
\begin{proof}[Solution by Jason Hardin]
(a) Since $F(x_0,\ldots,x_n)$ is homogeneous of degree $d$, each term of $F(x_0,\ldots,x_n)$
has the form $cx_0^{e_0}\cdots x_n^{e_n}$, where $c\in\mathbb{C}$ and $e_i$ are non-negative
integers satisfying $e_0+\cdots+e_n=d$. The corresponding term in $F(tx_0,\ldots,tx_n)$ is
$c(tx_0)^{e_0}\cdots(tx_n)^{e_n}=t^{e_0+\cdots+e_n}cx_0^{e_0}\cdots x_n^{e_n}=t^dcx_0^{e_0}\cdots x_n^{e_n}$.
So each term of $F(tx_0,\ldots,tx_n)$ is $t^d$ times the original term of $F(x_0,\ldots,x_n)$. Factoring out the
common factor $t^d$, we obtain the result.
(b) By (a), $F$ satisfies the equation $F(tx_0,\ldots,tx_n)=t^dF(x_0,\ldots,x_n)$. Differentiating both sides
of this equation with respect to $t$ via the chain rule of partial differentiation, we obtain
$$\sum_{i=0}^n\frac{\partial F(tx_0,\ldots,tx_n)}{\partial tx_i}\cdot\frac{\partial tx_i}{\partial t}=dt^{d-1}F(x_0,\ldots,x_n).$$
Since $\dfrac{\partial tx_i}{\partial t}=x_i$, this yields
$$\sum_{i=0}^nx_i\frac{\partial F(tx_0,\ldots,tx_n)}{\partial tx_i}=dt^{d-1}F(x_0,\ldots,x_n).$$
Finally, setting $t=1$ we obtain the desired result:
$$\sum_{i=0}^nx_i\frac{\partial F(x_0,\ldots,x_n)}{\partial x_i}=dF(x_0,\ldots,x_n).$$
\end{proof}
\begin{exer}\label{ex18.3}
Let $I\subseteq \CC[x_0,\ldots,x_n]$ be an ideal. Show that $I$ is homogeneous if and only if
whenever $F\in I$, then $F_i\in I$ for each homogeneous component $F_i$ of $F$.
\end{exer}
\begin{proof}[Solution, presented in class by Jason Hardin]
Suppose $I$ is homogeneous. Write $I=(F^1,\ldots,F^m)$, where $F^1,\ldots,F^m\in\mathbb{C}[x_0,\ldots,x_n]$
are homogeneous polynomials. Let $F\in I$. Then there are polynomials
$G^1,\ldots,G^m\in\mathbb{C}[x_0,\ldots,x_n]$ such that $F=\sum_{j=1}^mG^jF^j$.
Since $F^j\in I$, we have $G^j_iF^j\in I$ for all $i$, and $G^j_iF^j$ is homogeneous as it is a
product of homogeneous polynomials. Thus, $(G^jF^j)_i=G^j_{i-\deg(F^j)}F^j\in I$ for all $i$
and $j=1,\ldots,m$. Hence, we have $F_i=\sum_{j=1}^m(G^jF^j)_i\in I$, as required.
Conversely, suppose that whenever $F\in I$ we have $F_i\in I$ for all $i$. Write $I=(F^1,\ldots,F^m)$
with $\deg(F^j)=d_j$. Let $J$ be the ideal generated by all the homogenous components of $F^1,\ldots,F^m$,
i.e., $J=(F_0^1,\ldots,F_{d_1}^1,F_0^2,\ldots,F^2_{d_2},\ldots, F^m_0,\ldots,F^m_{d_m})$. Observe that $J$ is
a homogeneous ideal as it's generated by homogeneous polynomials. We have $I\subseteq J$ since each
generator of $I$ is the sum of its homogeneous components, which are all in $J$. Also, $J\subseteq I$ since
$F^j_i\in I$ by the hypothesis. Thus, $I=J$ is homogeneous.
\end{proof}
\begin{exer}\label{ex18.4}
Show that the map $h:R_0\oplus R_1\oplus \cdots\to R$ induced by sending each monomial to itself
is a ring isomorphism.
\end{exer}
\begin{proof}[Solution by Nora Youngs (with minor additions)]
Since the monomials of degree $i$ give a basis of $R_i$,
$h$ restricted to $R_i$ is the identity, so $h$ is defined by $(r_0,r_1,...)\mapsto r_0+r_1+\cdots$.
It is now easy to check that $h$ preserves addition and multiplication,
so we will show that $h$ is injective and surjective.
Suppose $h(r_0,r_1,...)=h(s_0,s_1,...)$.
Then, $r_0+r_1+\cdots=s_0+s_1+\cdots$.
Since $r_i, s_i$ are the homogeneous components of degree $i$ and the
polynomials are equal, we must have $r_i=s_i$ for all $i$.
Thus, $(r_0,...)=(s_0,...)$ and so $h$ is injective.
To see that $h$ is surjective, let $f\in R$. If $f=0$, then $h(0)=f$, so
assume $f\neq0$. Write $f=f_0+\cdots+f_r$ where
$f_i$ is the homogeneous component of $f$ of degree $i$ and $r=\deg(f)$.
Then $h(f_0,\ldots,f_r)=f_0+\cdots+f_r=f$, so $h$ is also surjective.
\end{proof}
\begin{exer}\label{ex18.5}
Let $I\subseteq R=\CC[x_0,\ldots,x_n]$ be a homogeneous ideal.
Under the isomorphism $h:R_0\oplus R_1\oplus \cdots\to R$, show that
$h(I_0\oplus I_1\oplus \cdots)=I$, where $I_j=I\cap R_j$.
\end{exer}
\begin{proof}[Solution]
Since $h$ restricted to $R_i$ is the identity, $h$ restricted to $I_j$ is also the identity.
But $I=\sum_j I_j$, so $h(I_0\oplus I_1\oplus \cdots)=I$.
\end{proof}
\begin{exer}\label{ex18.6}
Let $F\in \CC[x_0,\ldots,x_n]$ and let $p\in \CC^{n+1}$. Show that $F$ is homogeneous
if and only if whenever $F(p)=0$, then $F(cp)=0$ for all $c\in \CC$.
\end{exer}
\begin{proof}[Solution]
First suppose $F$ is homogeneous. If $F=0$, then clearly $F(cp)=c^dF(p)$ for all $c\in\CC$.
So assume $F$ is not the zero polynomial and let $d=\deg(F)$.
Then $F(cp)=c^dF(p)$ by Exercise \ref{ex18.2}(a), so if $F(p)=0$, then also $F(cp)=0$
for all $c\in\CC$.
Conversely, suppose $F$ is not homogeneous. Write $F=F_m+\cdots+F_M$ as
the sum of its homogeneous components, where $m$ is the degree of the term of
least degree, and $M$ is the degree of the term of largest degree (and hence $mm\geq 0$.
Also, by assumption, $F_m\neq0$. If $m=0$, then $F$ has a non-zero constant term,
so $F(0)\neq0$. Pick any point $p\in Z(F)$. Then the line through the origin and $p$
is not contained in $Z(F)$; i.e., $F(cp)=0$ does not hold for all $c\in \CC$.
Suppose $m>0$. Note that $Z(F)\subsetneq\CC^{n+1}$.
(If we had $Z(F)=\CC^{n+1}$, then $F\in\sqrt{(0)}=(0)$ by the Nullstellensatz.)
Since $\CC^{n+1}$ is irreducible (since $I(\CC^{n+1})=(0)$ is prime),
$Z(F_m)\cup Z(F_M)\neq\CC^{n+1}$, so we can pick a point $p\not\in Z(F_m)\cup Z(F_M)$.
Thus $F(tp)$ is a polynomial in the single indeterminate $t$ whose homogeneous component
of least degree is $F_m(tp)$ and whose homogeneous component of maximum degree is $F_M(tp)$.
Thus $F(tp)$ is not the zero polynomial, so there are values of $t$ for which $F(tp)\neq0$.
Since $m>0$, we see $t|F(tp)$, so 0 is a root of $F(tp)$, but
$F(tp)$ is not a pure power of $t$, so $0$ cannot be its only root.
Thus $F$ vanishes at some point $q\neq0$ on the line $tq$, but not at every point of the line.
\end{proof}
\begin{exer}\label{ex18.7}
If $F_1,\ldots,F_r$ is any set of homogeneous generators for a homogeneous ideal
$I\subseteq \CC[x_0,\ldots,x_n]$, show that $Z(F_1,\ldots,F_r)=Z(I)$.
\end{exer}
\begin{proof}[Solution by Zheng Yang]
If $F_i$ vanishes on $\underline{c}$ for all $i$, then so does any
polynomial in $I$, as these $F_i$'s are generators for $I$, hence $Z(F_1,\ldots,F_r)\subseteq Z(I)$.
For the reverse inclusion, assume $F(\underline{c})=0$ for some
$\underline{c}$ and all $F$ in $I$. Then since each homogeneous component
$F_j$ of $F$ is also in $I$ by Exercise \ref{ex18.3}, we have that $F_j$ vanishes at
$\underline{c}$ too, so $Z(F_1,\ldots,F_r)=Z(F_1)\cap\cdots\cap
Z(F_r)\supseteq Z(I)$.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{subsection}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{February 25, 2011}\label{lect19}
\subsection{Homogeneity continued}
\begin{notation}\label{compProjAffzeroloci}
Let $J\subseteq\CC[x_0,\ldots,x_n]$ be a homogeneous ideal.
Then $Z(J)$ could denote a subset either of projective space $\pr n$ or affine space $\AAA^{n+1}$.
To distinguish which is meant, we will use $\ZA{n+1}(J)\subseteq\AAA^{n+1}$ or
$\ZP{n}(J)\subseteq\pr n$. Of course, $\ZA{n+1}(J)$ and $\ZP{n}(J)$ are related:
if $p\in\AAA^{n+1}\setminus\{\rm origin\}$, then
$p\in\ZA{n+1}(J)$ if and only if $[p]\in \ZP{n}(J)$.
\end{notation}
\begin{example}
As noted last lecture, $\ZA{n+1}((1))=\ZP{n}((1))=\varnothing$,
but if $M=(x_0,\ldots,x_n)\subset\CC[x_0,\ldots,x_n]$,
and $I$ is $M$-primary and homogeneous, then $\ZP{n}(I)=\varnothing$,
even though $\ZA{n+1}(I)=\{\rm origin\}$.
\end{example}
The previous example shows that different homogeneous ideals can define different
projective zero loci, even if the ideals do not have the same radical.
But as we will see from the projective Nullstellensatz,
this phenomenon is limited to the empty zero locus.
First a lemma characterizing the homogeneous ideals with empty projective zero locus.
Because of this lemma, the ideal $(x_0,\ldots,x_n)$ is sometimes referred to as the
{em irrelevant} ideal.
\begin{lem}
Let $J\subseteq\CC[x_0,\ldots,x_n]$ be a homogeneous ideal.
Then $\ZP{n}(J) =\varnothing$ if and only if $(x_0,\ldots,x_n)\subseteq \sqrt{J}$
\end{lem}
\begin{proof}
If $\ZP{n}(J)=\varnothing$, then $\ZA{n+1}(J)\subseteq\{\rm origin\}$ (by notational remark
\ref{compProjAffzeroloci} above), hence $(x_0,\ldots,x_n)=I(\{\rm origin\})\subseteq I(\ZA{n+1}(J))=\sqrt{J}$.
Conversely, $\ZA{n+1}(J)=\ZA{n+1}(\sqrt{J})\subseteq \ZA{n+1}((x_0,\ldots,x_n))=\{\rm origin\}$,
hence $\ZP{n}(J)=\varnothing$ (again notational remark \ref{compProjAffzeroloci}).
\end{proof}
\begin{defn}
Given any subset $S\subseteq\pr{n}$, let $I(S)$ be the ideal generated
by all homogeneous $F\in \CC[x_0,\ldots,x_n]$ such that
$\ZP{n}(F)$ contains $S$. (For emphasis, we may sometimes
write $\IP{n}(S)$ for $I(S)$, although in principle this is not necessary,
since given a set $A$, whether $I(A)$ means the ideal of all polynomials
vanishing on $A$ or the ideal generated by all homogeneous polynomials
vanishing on $A$ is determined by whether $A$ is a subset of affine space or projective space.)
We also write $\CC[\pr n]$ for $\CC[x_0,\ldots,x_n]$, and refer to it as the
{\em homogeneous coordinate ring} of $\pr n$.
\end{defn}
\begin{thm}[Projective Nullstellensatz]\label{ProjNullstellensatz}
Let $J\subseteq\CC[\pr n]$ be a homogeneous ideal.
If $\ZP{n}(J)$ is not empty, then $I(\ZP{n}(J))=\sqrt{J}$.
\end{thm}
\begin{proof}
To show $I(\ZP{n}(J))=\sqrt{J}$, since both ideals are homogeneous, it suffices to show
that any non-zero homogeneous element of one is an element of the other.
So let $F\in I(\ZP{n}(J))$ be non-zero and homogeneous. Since $\ZP{n}(J)\neq\varnothing$,
$F$ is not constant. Let $p\in\ZA{n+1}(J)$. If $p$ is the origin, then (since every homogeneous
polynomial vanishes at the origin), $F(p)=0$. If $p$ is not the origin, then again $F(p)=0$
since $p\in\ZA{n+1}(J)$ implies $[p]\in\ZP{n}(J)$. Thus $F\in\sqrt{J}$ by the Nullstellensatz,
hence $I(\ZP{n}(J))\subseteq \sqrt{J}$.
For the reverse containment, let $F\in \sqrt{J}$ be non-zero and homogeneous.
Then $F^r\in J\subseteq I(\ZP{n}(J))$ so $F\in I(\ZP{n}(J))$.
\end{proof}
\subsection{Comparing conics, over ${\mathbb R}$ and over $\CC$,
in $\AAA^2$ and in $\pr 2$.}\label{19:conicclasses}
There are eight different kinds of zero loci in ${\mathbb R}^2$
for degree 2 polynomials in ${\mathbb R}[x,y]$. Here are examples of each one.
\begin{itemize}
\item[(1)] ellipses: $x^2+y^2-1=0$;
\item[(2)] hyperbolas: $x^2-y^2-1=0$;
\item[(3)] parabolas: $y-x^2=0$;
\item[(4)] pairs of intersecting lines: $x^2-y^2=0$;
\item[(5)] pairs of parallel lines: $(x+y)(x+y+1)=0$;
\item[(6)] doubled lines: $(x+y)^2=0$;
\item[(7)] single points: $x^2+y^2=0$; and
\item[(8)] the empty set: $x^2+y^2+1=0$.
\end{itemize}
Over $\CC$ this reduces to only five different kinds. Note for example
that under the coordinate change in which we replace $y$ by $iy$,
$x^2+y^2-1=0$ becomes $x^2-y^2-1=0$.
Moreover, $\ZA{2}(f)$ is always an infinite set for a degree 2 polynomial $f(x,y)\in\CC[x,y]$
(see Exercise \ref{ex19.2}), so cases (7) and (8) above do not occur
over $\CC$.
\begin{itemize}
\item[(1)] ellipse-hyperbola category: $x^2+y^2-1=0$;
\item[(2)] parabolas: $y-x^2=0$;
\item[(3)] pairs of intersecting lines: $xy=0$;
\item[(4)] pairs of parallel lines: $(x+y)(x+y+1)=0$;
\item[(5)] doubled lines: $(x+y)^2=0$;
\end{itemize}
To compare this to what happens in projective space, we introduce
the notion of homogenization.
\begin{defn}
Let $0\neq g\in\CC[x_1,\ldots,x_n]$ and let $d=\deg(g)$.
Define the {\em homogenization} $g_H$ of $g$ to be the polynomial
$g_H(x_0,\ldots,x_n)=x_0^d\,g(\frac{x_1}{x_0},\ldots,\frac{x_n}{x_0})\in\CC[x_0,\ldots,x_n]$.
Note that $g_H$ is indeed a homogeneous polynomial.
We refer to $\ZP{n}(g_H)$ as the {\em projective closure}
of $\ZA{n}(g)$.
\end{defn}
\begin{example}
If $f=x_1^2+x_2^2-1$, then $f_H=x_1^2+x_2^2-x_0^2$.
If $f=x_2-x_1^2$, then $f_H=x_0x_2-x_1^2$.
\end{example}
Further simplification occurs when classifying zero loci of degree 2 homogeneous
polynomials in $\pr 2$. We can regard $\AAA^2$ as an open subset of $\pr2$; the points
$(a,b)\in\AAA^2$ can be identified with the points $[(a,b,1)]\in\pr2$. The points
of $\AAA^2$ at infinity can be identified with $[(a,b,0)]\in\pr2$.
If $0\neq f(x,y)\in\CC[x,y]$, suppose $[(a,b,0)]$ is a point at infinity which is a limit point
(in the standard topology)
of a sequence $[(a_i,b_i,1)]$ of points in the zero locus $\ZA{2}(f)$
in the finite affine plane $\AAA^2$. Then either
$a_i\stackrel{i\to\infty}{\xrightarrow{\hspace*{0.75cm}}}\infty$ or
$b_i\stackrel{i\to\infty}{\xrightarrow{\hspace*{0.75cm}}}\infty$; say the latter
and assume for simplicity that $b\neq0$.
Then $[(\frac{a_i}{b_i},1,\frac{1}{b_i})]$ is the same sequence of points
in $\pr2$, but the limit is $[(\frac{a}{b},1,0)]=[(a,b,0)]$.
Since $f(a_i,b_i)=0$ for all $i$, we have $f_H(a_i,b_i,1)=f(a_i,b_i)=0$,
hence we also have $b_i^df_H(\frac{a_i}{b_i},1,\frac{1}{b_i})=f_H(a_i,b_i,1)=0$
and hence $f_H(a,b,0)=\lim_if_H(\frac{a_i}{b_i},1,\frac{1}{b_i})=0$.
Thus $\ZP{2}(f_H)$ contains (and in fact is equal to) the union of $\ZA{2}(f)$
and the limit points at infinity, hence the name projective closure.
Thus after including the points at infinity in our categories above we have
the ellipse-hyperbola category, $\ZP{2}(x^2+y^2-z^2)$, the parabola
category, $\ZP{2}(yz-x^2)$, the two intersecting lines, $\ZP{2}(xy)$,
the parallel lines, $\ZP{2}((x+y)(x+y+z))$, and the doubled line,
$\ZP{2}((x+y)^2)$. But after a change of coordinates where we substitute
$y$ for $x+iy$, $z$ for $x-iy$ and $x$ for $z$,
$\ZP{2}(x^2+y^2-z^2)$ becomes $\ZP{2}(yz-x^2)$; i.e.,
the parabola is just another instance of the ellipse-hyperbola category.
And the two parallel lines become a pair of intersecting lines, so in $\pr2$
this suggests we have only 3 categories:
\begin{itemize}
\item[(1)] ellipse-hyperbola-parabola category (i.e., the irreducible conics): $\ZP{2}(yz-x^2)$;
\item[(2)] two intersecting lines (the reducible conics): $xy=0$;
\item[(3)] the doubled lines (the ``non-reduced'' conics): $(x+y)^2=0$;
\end{itemize}
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\multiput( 0.546446 , 4.42155 )( -0.00851327 , -0.00382585 ){ 20 }{\circle*{.05}}
\multiput( 0.37618 , 4.34504 )( -0.00704631 , -0.00548927 ){ 20 }{\circle*{.05}}
}}
\hbox to0in{\hskip-1in\hbox to0in{
% AWK script for points of hyperbola: echo 1 | awk '{for(i=0;i<50;i++) {t=i*3.75/50+.25; s=(i+1)*3.75/50+.25; n=40; a=t; b=1/t; c=s; d=1/s; cc=(c-a)/n; dd=(d-b)/n; print "\\multiput(",a,",",b,")(",cc,",",dd,")\{",n,"\}\{\\circle*\{.05\}\}"}}'
% other branch: echo 1 | awk '{for(i=0;i<50;i++) {t=-i*3.75/50-.25; s=-(i+1)*3.75/50-.25; n=40; a=t; b=1/t; c=s; d=1/s; cc=(c-a)/n; dd=(d-b)/n; print "\\multiput(",a,",",b,")(",cc,",",dd,")\{",n,"\}\{\\circle*\{.05\}\}"}}'
\put(-.4,4){$y$}
\put(0,-4){\line(0,1){8}}
\put(4.1,-.4){$x$}
\put(-4,0){\line(1,0){8}}
\put(1.2,1.2){$xy=1$}
\color{red}
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\multiput( 0.325 , 3.07692 )( 0.001875 , -0.0144231 ){ 40 }{\circle*{.05}}
\multiput( 0.4 , 2.5 )( 0.001875 , -0.00986842 ){ 40 }{\circle*{.05}}
\multiput( 0.475 , 2.10526 )( 0.001875 , -0.00717703 ){ 40 }{\circle*{.05}}
\multiput( 0.55 , 1.81818 )( 0.001875 , -0.00545455 ){ 40 }{\circle*{.05}}
\multiput( 0.625 , 1.6 )( 0.001875 , -0.00428571 ){ 40 }{\circle*{.05}}
\multiput( 0.7 , 1.42857 )( 0.001875 , -0.00345622 ){ 40 }{\circle*{.05}}
\multiput( 0.775 , 1.29032 )( 0.001875 , -0.0028463 ){ 40 }{\circle*{.05}}
\multiput( 0.85 , 1.17647 )( 0.001875 , -0.00238474 ){ 40 }{\circle*{.05}}
\multiput( 0.925 , 1.08108 )( 0.001875 , -0.00202703 ){ 40 }{\circle*{.05}}
\multiput( 1 , 1 )( 0.001875 , -0.00174419 ){ 40 }{\circle*{.05}}
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\multiput( 1.375 , 0.727273 )( 0.001875 , -0.000940439 ){ 40 }{\circle*{.05}}
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\multiput( 2.575 , 0.38835 )( 0.001875 , -0.000274776 ){ 40 }{\circle*{.05}}
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\multiput( 2.725 , 0.366972 )( 0.001875 , -0.00024574 ){ 40 }{\circle*{.05}}
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\multiput( 3.175 , 0.314961 )( 0.001875 , -0.000181708 ){ 40 }{\circle*{.05}}
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\multiput( 3.325 , 0.300752 )( 0.001875 , -0.000165856 ){ 40 }{\circle*{.05}}
\multiput( 3.4 , 0.294118 )( 0.001875 , -0.000158697 ){ 40 }{\circle*{.05}}
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\multiput( 3.55 , 0.28169 )( 0.001875 , -0.000145702 ){ 40 }{\circle*{.05}}
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\multiput( 3.925 , 0.254777 )( 0.001875 , -0.000119427 ){ 40 }{\circle*{.05}}
\color{blue}
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\multiput( -0.4 , -2.5 )( -0.001875 , 0.00986842 ){ 40 }{\circle*{.05}}
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\multiput( -0.625 , -1.6 )( -0.001875 , 0.00428571 ){ 40 }{\circle*{.05}}
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\multiput( -0.775 , -1.29032 )( -0.001875 , 0.0028463 ){ 40 }{\circle*{.05}}
\multiput( -0.85 , -1.17647 )( -0.001875 , 0.00238474 ){ 40 }{\circle*{.05}}
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\multiput( -1 , -1 )( -0.001875 , 0.00174419 ){ 40 }{\circle*{.05}}
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\multiput( -1.675 , -0.597015 )( -0.001875 , 0.000639659 ){ 40 }{\circle*{.05}}
\multiput( -1.75 , -0.571429 )( -0.001875 , 0.000587084 ){ 40 }{\circle*{.05}}
\multiput( -1.825 , -0.547945 )( -0.001875 , 0.000540735 ){ 40 }{\circle*{.05}}
\multiput( -1.9 , -0.526316 )( -0.001875 , 0.000499667 ){ 40 }{\circle*{.05}}
\multiput( -1.975 , -0.506329 )( -0.001875 , 0.000463106 ){ 40 }{\circle*{.05}}
\multiput( -2.05 , -0.487805 )( -0.001875 , 0.000430416 ){ 40 }{\circle*{.05}}
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\multiput( -2.275 , -0.43956 )( -0.001875 , 0.000350713 ){ 40 }{\circle*{.05}}
\multiput( -2.35 , -0.425532 )( -0.001875 , 0.00032902 ){ 40 }{\circle*{.05}}
\multiput( -2.425 , -0.412371 )( -0.001875 , 0.000309278 ){ 40 }{\circle*{.05}}
\multiput( -2.5 , -0.4 )( -0.001875 , 0.000291262 ){ 40 }{\circle*{.05}}
\multiput( -2.575 , -0.38835 )( -0.001875 , 0.000274776 ){ 40 }{\circle*{.05}}
\multiput( -2.65 , -0.377358 )( -0.001875 , 0.00025965 ){ 40 }{\circle*{.05}}
\multiput( -2.725 , -0.366972 )( -0.001875 , 0.00024574 ){ 40 }{\circle*{.05}}
\multiput( -2.8 , -0.357143 )( -0.001875 , 0.000232919 ){ 40 }{\circle*{.05}}
\multiput( -2.875 , -0.347826 )( -0.001875 , 0.000221076 ){ 40 }{\circle*{.05}}
\multiput( -2.95 , -0.338983 )( -0.001875 , 0.000210113 ){ 40 }{\circle*{.05}}
\multiput( -3.025 , -0.330579 )( -0.001875 , 0.000199947 ){ 40 }{\circle*{.05}}
\multiput( -3.1 , -0.322581 )( -0.001875 , 0.0001905 ){ 40 }{\circle*{.05}}
\multiput( -3.175 , -0.314961 )( -0.001875 , 0.000181708 ){ 40 }{\circle*{.05}}
\multiput( -3.25 , -0.307692 )( -0.001875 , 0.000173511 ){ 40 }{\circle*{.05}}
\multiput( -3.325 , -0.300752 )( -0.001875 , 0.000165856 ){ 40 }{\circle*{.05}}
\multiput( -3.4 , -0.294118 )( -0.001875 , 0.000158697 ){ 40 }{\circle*{.05}}
\multiput( -3.475 , -0.28777 )( -0.001875 , 0.000151991 ){ 40 }{\circle*{.05}}
\multiput( -3.55 , -0.28169 )( -0.001875 , 0.000145702 ){ 40 }{\circle*{.05}}
\multiput( -3.625 , -0.275862 )( -0.001875 , 0.000139795 ){ 40 }{\circle*{.05}}
\multiput( -3.7 , -0.27027 )( -0.001875 , 0.00013424 ){ 40 }{\circle*{.05}}
\multiput( -3.775 , -0.264901 )( -0.001875 , 0.00012901 ){ 40 }{\circle*{.05}}
\multiput( -3.85 , -0.25974 )( -0.001875 , 0.00012408 ){ 40 }{\circle*{.05}}
\multiput( -3.925 , -0.254777 )( -0.001875 , 0.000119427 ){ 40 }{\circle*{.05}}
}}
\end{picture}
\vskip4\baselineskip
\caption{The hyperbola $\ZA{2}(xy-1)$ and its projective closure $\ZP{2}(xy-z^2)$.}\label{hyperbolaFig}
\end{figure}
\ \vskip5\baselineskip
\begin{figure}[h]
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\put(3.2,0){$x$}
\put(-3,0){\line(1,0){6}}
\put(1.2,1){$y=x^2$}
% AWK script for points on parabola: echo 1 | awk '{for(i=0;i<50;i++) {t=i*4/50-2; s=(i+1)*4/50-2; n=20; a=t; b=t^2; c=s; d=s^2; cc=(c-a)/n; dd=(d-b)/n; print "\\multiput(",a,",",b,")(",cc,",",dd,")\{",n,"\}\{\\circle*\{.05\}\}"}}'
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\multiput( -1.6 , 2.56 )( 0.004 , -0.01248 ){ 20 }{\circle*{.05}}
\multiput( -1.52 , 2.3104 )( 0.004 , -0.01184 ){ 20 }{\circle*{.05}}
\multiput( -1.44 , 2.0736 )( 0.004 , -0.0112 ){ 20 }{\circle*{.05}}
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\multiput( -1.04 , 1.0816 )( 0.004 , -0.008 ){ 20 }{\circle*{.05}}
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\multiput( 1.92 , 3.6864 )( 0.004 , 0.01568 ){ 20 }{\circle*{.05}}
}}
\hbox to0in{\hskip1.5in\hbox to0in{
% AWK sscript for points on proj closure of parabola: echo 1 | awk '{for(i=0;i<50;i++) {t=i*2*0.06284; s=(i+1)*2*0.06284; n=20; b=2+2*sin(t); a=cos(t)-.1*b; d=2+2*sin(s); c=cos(s)-.1*d; cc=(c-a)/n; dd=(d-b)/n; print "\\multiput(",a,",",b,")(",cc,",",dd,")\{",n,"\}\{\\circle*\{.05\}\}"}}'
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%\put(-.5,4.1){\line(1,-1){4.5}}
\multiput(-.5,4.4)(.0115,-.0115){400}{\circle*{.05}}
\put(2,2){$z=0$}
\put(.8,.6){$yz=x^2$}
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\multiput( -0.0625759 , 0.0038712 )( 0.0062791 , 0.00000767424){ 20 }{\circle*{.05}}
\multiput( 0.063006 , 0.00402468 )( 0.00607165 , 0.00158197 ){ 20 }{\circle*{.05}}
\multiput( 0.184439 , 0.035664 )( 0.00576843 , 0.00313131 ){ 20 }{\circle*{.05}}
\multiput( 0.299808 , 0.0982902 )( 0.00537422 , 0.00463125 ){ 20 }{\circle*{.05}}
\multiput( 0.407292 , 0.190915 )( 0.00489523 , 0.00605814 ){ 20 }{\circle*{.05}}
\multiput( 0.505197 , 0.312078 )( 0.00433901 , 0.00738946 ){ 20 }{\circle*{.05}}
\multiput( 0.591977 , 0.459867 )( 0.00371436 , 0.00860421 ){ 20 }{\circle*{.05}}
\multiput( 0.666264 , 0.631951 )( 0.0030311 , 0.00968324 ){ 20 }{\circle*{.05}}
\multiput( 0.726886 , 0.825616 )( 0.00230004 , 0.0106095 ){ 20 }{\circle*{.05}}
\multiput( 0.772887 , 1.03781 )( 0.00153269 , 0.0113684 ){ 20 }{\circle*{.05}}
\multiput( 0.803541 , 1.26518 )( 0.000741161 , 0.011948 ){ 20 }{\circle*{.05}}
\multiput( 0.818364 , 1.50414 )( -0.0000620576 , 0.0123391 ){ 20 }{\circle*{.05}}
\multiput( 0.817123 , 1.75092 )( -0.000864297 , 0.0125356 ){ 20 }{\circle*{.05}}
}}
\end{picture}
\vskip-\baselineskip
\caption{The parabola $\ZA{2}(y-x^2)$ and its projective closure $\ZP{2}(yz-x^2)$.}\label{parabolaFig}
\end{figure}
In the figures above, note that the projective closure of the hyperbola has equation $xy-z^2$ and
the projective closure of the parabola has equation $yz-x^2$;
these are the same equations after a permutation of the variables
(i.e., up to a projective change of coordinates the projective closures are the same).
In the affine plane they are different: the parabola has a single point at infinity,
while the hyperbola has two.
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex19.1}
Let $J\subseteq\CC[x_0,\ldots,x_n]$ be a homogeneous ideal.
Show that $\sqrt{J}$ is homogeneous and hence that
$\ZP{n}(\sqrt{J})$ is defined. Conclude that
$\ZP{n}(J)=\ZP{n}(\sqrt{J})$.
\end{exer}
\begin{proof}[Solution by Jason Hardin]
Let $F\in \sqrt{J}$ and let $F=F_0+F_1+\cdots+F_d$ be its decomposition into homogeneous components. By exercise 18.3, it suffices to show that $F_i\in\sqrt{J}$ for $0\leq i\leq d$.
Since $F\in\sqrt{J}$, we know $F^{m_1}\in J$ for some $m_1$. Observe that $(F^{m_1})_0=(F_0)^{m_1}$. Since $J$ is homogeneous, $(F_0)^{m_1}=(F^{m_1})_0\in J$ by exercise 18.3. Thus, $F_0\in\sqrt{J}$. This means that $G:=F-F_0=F_1+\cdots+F_d\in\sqrt{J}$, so $G^{m_2}\in J$ for some $m_2$. Observe that $(F_1)^{m_2}=(G^{m_2})_{m_2}\in J$ since $J$ is homogeneous, and thus $F_1\in\sqrt{J}$. So $H:=G-F_1=F_2+\cdots+F_d\in\sqrt{J}$. Continuing in this fashion, we see that at each step the smallest remaining non-zero homogeneous component of $F$ is in $\sqrt{J}$. This process terminates after $d+1$ steps, and we have $F_i\in\sqrt{J}$ for $0\leq i\leq d$.
Since $\sqrt{J}$ is homogeneous, $\ZP{n}(\sqrt{J})$ is defined. Moreover, we have $[p]\in\ZP{n}(J)$ if and only if $p\in\ZA{n+1}(J)=\ZA{n+1}(\sqrt{J})$ if and only if $[p]\in\ZP{n}(\sqrt{J})$. Hence, $\ZP{n}(J)=\ZP{n}(\sqrt{J})$.
\end{proof}
\begin{exer}\label{ex19.2}
Show that $\ZA{2}(f)$ is infinite for every degree 2 polynomial $f(x,y)\in\CC[x,y]$.
\end{exer}
\begin{proof}[Solution by Kat Shultis]
Let $f(x,y)\in\CC[x,y]$ be a degree 2 polynomial. Then
$f(x,y)=\alpha x^2+\beta y^2+\gamma xy+\delta x+\epsilon y +\eta$ and at least one of
$\alpha, \beta,\gamma$ is nonzero. We consider each case separately. If $\alpha\neq 0$,
then if we choose any $y_0\in\CC$, we get a quadratic polynomial $f(x,y_0)$ in one variable,
$x$, and since $\CC$ is algebraically closed, we know that there is an $x_0\in\CC$ such that
$f(x_0,y_0)=0$. Similarly, if $\beta\neq 0$, we can choose any $x_0\in\CC$, and get a quadratic
polynomial in $y$, namely $f(x_0,y)$. Hence there exists some $y_0\in\CC$ such that $f(x_0,y_0)\in\CC$.
Thus, in either of these two cases, $Z_{\AAA^2}(f)$ has the same cardinality of $\CC$, and hence is infinite.
Next, we assume that $\gamma\neq 0$. If
$x_0\in\CC\setminus\{0\}$, then $f(x_0,y)=\alpha x_0^2+\beta y^2+\gamma x_0 y +\delta x_0+\epsilon y +\eta$
is at least a linear polynomial in $y$, as $\gamma\neq 0$. Thus there exists some $y_0\in\CC$ such that
$f(x_0,y_0)=0$. Here, $Z_{\AAA^2}(f)$ is in bijection with $\CC\setminus\{0\}$ which is infinite.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{February 28, 2011}\label{lect20}
\noindent{\bf Projective changes of coordinates.}
To show rigorously that all irreducible conics are in some sense the same, we formalize the notion
of projective changes of coordinates. Let $M\in{\rm GL}_{n+1}(\CC)$, so $M=(m_{ij})$ is an $(n+1)\times(n+1)$
invertible matrix with complex entries. Let $T_M:A=\CC^{n+1}\to B=\CC^{n+1}$ be the linear transformation
defined by $M$ with respect to the standard basis ${\bf e}_0=(1,0,\ldots,0)^t, \ldots,{\bf e}_n=(0,\ldots,0,1)^t$.
If ${\bf a}=(a_0,\ldots,a_n)^t\in\CC^{n+1}$, then
$$T_M({\bf a})=M{\bf a}= \Big(\sum_jm_{0j}a_j,\ldots,\sum_jm_{nj}a_j \Big)\in\CC^{n+1},$$
so $T_M$ is an algebraic mapping $T_M=(f_0,\ldots,f_n)$ where, taking $\CC[A]=\CC[x_0,\ldots,x_n]$,
the coordinate functions are $f_i=\sum_jm_{ij}x_j$.
If we write $\CC[B]=\CC[y_0,\ldots,y_n]$, then we have $T_M^*:\CC[B]\to\CC[A]$, given by
$T_M^*(y_i)=f_i=y_i\circ T_M$.
We can regard $M$ as giving a change of coordinates on $\CC^{n+1}$ by taking
${\varepsilon}_i=(T_M)^{-1}({\bf e}_i)$. Given the coordinate vector ${\bf a}\in\CC^{n+1}$
of a point in $A$ with respect to the standard basis, then ${\bf b}=M{\bf a}$ is the coordinate
vector of the same point with respect to the basis $\{\varepsilon_i\}$, since
$$\sum_ib_i{\varepsilon}_i = \sum_ib_i(T_M)^{-1}({\bf e}_i)=(T_M)^{-1} \big(\sum_ib_i{\bf e}_i \big) =
(T_M)^{-1}({\bf b})=(T_M)^{-1}T_M({\bf a})={\bf a}=\sum_ia_i{\bf e}_i.$$
A linear change of coordinates $T_M$ on $\CC^{n+1}$ gives a
change of coordinates also on $\pr n$ (denoted $T_M^P$ and
well-defined since $T_M$ is linear and hence takes lines
through the origin to lines through the origin), since
$\pr n$ inherits its coordinates from $\CC^{n+1}$. We call this change of coordinates a
{\em projective change of coordinates}. The set of all such coordinate changes $T_M^P$
forms a group under composition,
the projective linear group, denoted ${\pr{}}GL_n(\CC)$. By Exercise \ref{ex20.1},
${\rm{\pr{}}GL}_n(\CC)\cong{\rm GL}_{n+1}(\CC)/\CC^*$, where we regard
the non-zero complex number $\CC^*$ as a subgroup of ${\rm GL}_{n+1}(\CC)$
by thinking of $c\in\CC^*$ as $cI_{n+1}$; this is in fact a normal subgroup.
Note that a projective change of coordinates on $\pr n$
is defined by picking a basis $\varepsilon_0,\ldots,\varepsilon_n$
of $\CC^{n+1}$. Picking the basis $\varepsilon_i$ is equivalent to picking
linear forms $y_j\in\CC[x_0,\ldots,y_n]$, $0\leq j\leq n$ (a {\em form} is just another word for
a homogeneous polynomial). Given the basis $\varepsilon_i$, the forms $y_j$
are those such that $y_j(\varepsilon_i)$ is 1 if $i=j$ and 0 otherwise, and given the forms
$y_j$, on can recover the basis vectors $\varepsilon_i$ by solving the equations
$y_j(\varepsilon_i)=\delta_{ij}$ (where here $\delta_{ij}$ is Kronecker's delta-function;
i.e., $(\delta_{ij})$ is the identity matrix). If the basis $\varepsilon_i$ is given by
$(T_M)^{-1}({\bf e}_i)$ for a matrix $M=(m_{ij})$, then the forms $y_i$
are $\sum_jm_{ij}x_j$. This is because if $[{\bf a}]$ is the coordinate vector
(with respect to the standard basis) of a
point in $\pr n$, then the coordinate vector with respect to the basis $\varepsilon_i$ is
$[{\bf b}]$, where ${\bf b}=M{\bf a}$. The $i$th coordinate of ${\bf b}$ is
$b_i$, but
$$y_i(p)=y_i(\sum_jb_j\varepsilon_j)=\sum_jb_jy_i(\varepsilon_j)=b_i=
\sum_jm_{ij}a_j=\sum_jm_{ij}x_j(p),$$
hence $y_i=\sum_jm_{ij}x_j$.
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex20.1}
Let $M_1,M_2\in{\rm GL}_{n+1}(\CC)$. Show that $T_{M_1}^P=T_{M_2}^P$ if and only if
$M_1=cM_2$ for some non-zero scalar $c\in\CC$.
\end{exer}
\begin{proof}[Solution by Ashley Weatherwax]
$\Leftarrow$ Suppose $M_1 = cM_2$ for a scalar $c \in \CC$. Then
$$T_{M_2}^P ([a]) = [M_2 a] = [cM_2 a] = T_{cM_2}^P ([a]) = T_{M_1}^P ([a])$$
$\Rightarrow$ Suppose $T_{M_1}^P ([a]) = T_{M_2}^P ([a])$. Then we can compose
both sides with $T_{M_2^{-1}}^P$ to get
$T_{M_2^{-1}M_1}^P = T_I^P$. Thus it suffices to show that $T_M^P = T_I^P$
implies $M = cI$ for some constant $c$.
Let $[a] \in \pr n$. Then $T_M^P([a]) = T_I^P([a])$ implies
$[Ma] = [a]$, and so there exists a $c_a$ so that
$$Ma = c_a a$$
But notice that this says that $a$ is an eigen-vector, and as $a$
was arbitrary, every vector is an eigen-vector. Let also
$[b] \neq [a] \in \pr n$. Then there is an eigen-value $c_b$ for $b$ so that $Mb = c_b b$.
But as every vector
is an eigen-vector, $a+b$ is an eigen-vector. By linearity we have
$$M(a+b) = Ma + Mb = c_a a + c_b b$$
Thus $a+b$ is an eigen-vector iff $c_a = c_b=c$, and thus as $a,b$ were arbitrary
every eigen-vector has the same
eigen-value. Thus for some $c \in \CC$ we have $M = cI$.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{March 2, 2011}\label{lect21}
\noindent{\bf Classifying projective conics.}
Now that we have the notion of a projective change of coordinates,
we will classify homogeneous forms of degree 2 in 3 variables;
i.e., we will classify projective plane conics.
The result is that up to a projective change of coordinates,
a form $0\neq F\in\CC[x,y,z]$ of degree 2 is either:
\begin{itemize}
\item[(a)] $F=xy-z^2$;
\item[(b)] $F=xy$; or
\item[(c)] $F=x^2$
\end{itemize}
These are exactly the three cases asserted in Lecture \ref{19:conicclasses}.
We will do this first under the following assumptions.
Let $F\in\CC[X,Y,Z]$ be a non-zero form of degree 2.
First, $\ZP{2}(F)$ is infinite (this is because of Exercise \ref{ex19.2}).
Second, for each point $p\in\ZP{2}(F)$, there is a form $L_p$ defining a line containing $p$
such that $p$ is the only point of $\ZP{2}(F)$ on that line. (We will justify this later.)
So, first assume that $F$ is not irreducible; i.e., $F$ is reducible.
Since $F$ has degree 2, then, by Exercise \ref{ex21.1}, $F=AB$ for linear forms $A$ and $B$.
If $\ZP{2}(A)=\ZP{2}(B)$, then $A=cB$ for a nonzero constant $c$, and we can choose
a coordinate system specified by linear forms $x,y, $ and $z$
in which $x=\sqrt{c}B$, and in which $y$ and $z$ are any two additional
linear forms such that $\{x,y,z\}$ are linearly independent. Then $F=AB=cB^2=x^2$.
If $\ZP{2}(A)\neq\ZP{2}(B)$, then $A$ is not a non-zero multiple of $B$, so $A$ and $B$
are linearly independent. This time we pick $x=A$, $y=B$ and extend this to a basis
$\{x,y,z\}$ of the linear forms. Now $F=xy$.
Finally, assume $F$ is irreducible. Since $\ZP{2}(F)$ is infinite, we can pick distinct points
$p,q\in\ZP{2}(F)$. Let $x=L_p$, $y=L_q$, and let $z=0$ be the equation of the line through $p$ and $q$.
Let $r$ be the point where $L_p$ and $L_q$ meet. Since by assumption,
$L_p\cap \ZP{2}(F)=\{p\}$ and $L_q\cap \ZP{2}(F)=\{q\}$, we see that
$r\not\in\ZP{2}(F)$ and we also see that neither of the points $p$ and $q$
is on both lines $L_p=0$ and $L_q=0$. Thus $L_p=0$ and $L_q=0$ are different lines
and hence $p,q$ and $r$ are 3 distinct points, and not collinear.
If $x$, $y$ and $z$ were dependent, there would be constants
$a,b,c$, not all 0, such that $cx+by+cz=0$. Evaluating at $p$ gives $a0+by(p)+c0=0$,
since $p\in\{x=0\}\cap\{z=0\}$. But $p\not\in\{y=0\}$,
so $b=0$. Evaluating at $q$ gives $a=0$, and evaluating at $r$ gives $c=0$.
This shows that $x,y$, and $z$ are linearly independent and hence define a coordinate system on $\pr2$.
The coordinates of $p$, $q$ and $r$ in this coordinate system are, respectively,
$[(0,1,0)]$, $[(1,0,0)]$, and $[(0,0,1)]$.
Therefore we can write $F$ as a linear combination $F=ax^2+bxy+cy^2+dxz+eyz+fz^2$.
We have $0=F(p)=F(0,1,0)=c$, $0=F(q)=F(1,0,0)=a$, and $0\neq F(r)=f$.
Thus $F=bxy+dxz+eyz+fz^2$.
Now use the fact that the only common zero of $F$ and $y$ is $q$. Solving $F=0$ and $y=0$
simultaneously gives $F=dxz+fz^2$, which has $(-f,0,d)$ as a root. Since
$L_q\cap \ZP{2}(F)=\{q\}$, we see that $[(-f,0,d)]=[(1,0,0)]$, hence $d=0$.
Likewise, solving $F=0$ and $x=0$ simultaneously gives $e=0$.
Thus $F=bxy+fz^2$. Since $F$ is irreducible we cannot have $b=0$. Replacing $x$ by $x/b$
and $z$ by $iz/\sqrt{f}$ gives $F=xy-z^2$.
We still have to justify that for each point $p\in\ZP{2}(F)$, there is a form $L_p$ defining a line containing $p$
such that $p$ is the only point of $\ZP{2}(F)$ on that line.
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex21.1}
If $F\in\CC[x_0,\ldots,x_n]$ is a non-zero homogeneous polynomial and $G\in\CC[x_0,\ldots,x_n]$
is a factor of $F$, show that $G$ is homogeneous.
\end{exer}
\begin{proof}[Solution by Katie Morrison]
Since $G$ is a factor of $F$, there exists an $H \in \CC[x_0,\ldots,x_n]$ such that $F=GH$.
Let $m_G$ be the degree of the homogeneous component of $G$ of least degree, and let
$m_H$ be the degree of the homogeneous component of $H$ of least degree. Then the
component of least degree in $GH$ will have degree $m_Gm_H$, and the product of other
components of $G$ and $H$ will have strictly larger degree. But $F=GH$ is homogeneous
so there is only one component and it must have degree $m_Gm_H$. Thus, $G$ and $H$
can only have one component each as well, and so $G$ is homogeneous.
\end{proof}
\mysection{March 4, 2011}\label{lect22}
\noindent{\bf Classifying projective conics (cont.).}
Note: Class began with jason presenting Problem \ref{ex18.3}.
To finish our classification of conics, given an irreducible form $F\in\CC[x,y,z]$ of degree 2,
we need to verify for each point $p\in Z(F)\subset\CC^3$ that there is a linear form $L_p$ such
$\ZP{2}(F,L_p)=\{[p]\}$. (In fact, $L_p$ is the form defining the tangent line to $\ZP{2}(F)$ at $p$.)
So let $[p]\in\ZP{2}(F)$.
To define $L_p$ (without reference to tangent lines), pick any two linearly independent linear forms
$u$ and $v$ which both vanish at $[p]$. Pick any third linear form $w$ that does not vanish at $[p]$.
Then $u,v$ and $w$ are linearly independent. (If not we have scalars $a,b,c$, not all 0, such that
$au+bv+cw=0$, but evaluation at $[p]$ gives $0=au(p)+bv(p)+cw(p)=cw(p)$, hence
$c=0$ since $w(p)\neq0$. Thus $au+bv=0$, but we chose $u$ and $v$ to be independent.)
Using this coordinate system, we can write $F=au^2+buv+cv^2+duw+evw+fw^2$ for some scalars
$a,\ldots,f$. Since $F(p)=0$, and sicne $u(p)=0=v(p)$, we have $f=0$. If $d=e=0$, then
$F$ factors (no form in 2 variables is ever irreducible unless it's linear).
Thus either $d\neq0$ or $e\neq0$. Let $L_p=du+ev$, and assume $e\neq0$; the argument in case
$d\neq0$ is similar.
Now solve the system $F=0, L_p=0$. Since we assume $L_p=0$, we can write $v=-du/e$.
Substitute into $F$ to get $0=F=(a-(bd/e)+c(d/e)^2)u^2$. If $a-(bd/e)+c(d/e)^2=0$, then
$\ZP{2}(L_p)\subseteq \ZP{2}(F)$, hence by the projective Nullstellensatz $F\in \sqrt{(L_p)}=(L_p)$,
so $L_p|F$, contradicting having $F$ be irreducible. Thus $u^2=0$, so $u=0=v$, hence
the only solution is $[p]$; i.e., $\ZP{2}(F,L_p)=[p]$.
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex22.1}
Let $F\in\CC[x,y,z]$ be an irreducible form of degree 2 and let $[p]\in\ZP{2}(F)$.
Show that, up to multiplication by scalars, $L_p$ is the only linear form vanishing at $[p]$
such that $\ZP{2}(F,L_p)=[p]$.
\end{exer}
\begin{proof}[Solution]
Pick a coordinate system $x,y,z$ such that $L_p=x$ and
let $[p]=\ZP{2}(x,y)$. Let $L=ax+by+cz$ be a linear form
which is not a scalar multiple of $L_p$
and write $F=dx^2+exy+fy^2+gxz+hyz+iz^2$. Since $F([p])=0$ we have
$i=0$. The restriction of $F=0$ to the line $x=0$
gives $fy^2+hyz=0$, and since $\ZP{2}(F,L_p)=[p]$,
we see that $h=0$. Thus $F=dx^2+exy+fy^2+gxz$ and since $F$ is irreducible,
then $f\neq0$ and $g\neq0$. In order for $L([p])=0$ we need $c=0$.
Since $L$ is not a scalar multiple of $L_p$, we must have $b\neq0$.
Dividing $L$ by $b$ allows us to reduce to the case that
$L=ax+y$. Solving $F=0$ with $L=0$ gives $y=-ax$ and
$dx^2-aex^2+a^2fx^2+gxz=0$ and hence either $x=0$ (which gives us
$[p]\in\ZP{2}(F,L)$) or $(d-ae+a^2f)x+gz=0$ which has the solution
$[(1,-a,0)]$ if $d-ae+a^2f=0$ or $[(-g,ag,d-ae+a^2f)]$ if $d-ae+a^2f\neq0$.
Either way, $\ZP{2}(F,L)$ is not just $[p]$.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{March 7, 2011}\label{lect23}
\noindent{\bf Maps of projective algebraic sets.}
It is natural to mimic what we did for affine algebraic sets in trying to define algebraic
maps of projective algebraic sets. If we do this we would want to say that a map
$f:V\to W$ of projective algebraic sets $V\subseteq\pr N$ and $W\subseteq\pr M$
is the restriction of $V$ of a mapping $F:\pr N\to \pr M$, where $F=(F_0,\ldots,F_M)$
in which each $F_i$ is a polynomial in $\CC[\pr N]$.
In order for $F$ to be well-defined, we need: each $F_i$ to be homogeneous;
we need the degree of each $F_i$ which is not the 0 polynomial to be the same;
and we need $\ZP{N}(F_0,\ldots,F_M)=\varnothing$.
But it turns out this is not enough to get a good notion of maps of projective algebraic sets,
as the next two examples will help us to see.
\begin{example}\label{eg23.1}
Let $\CC[\pr1]=\CC[a,b]$ and let $\CC[\pr2]=\CC[x_0,x_1,x_2]$.
Consider $F:\pr1\to\pr2$ where $F=(F_0,F_1,F_2)$ with
$F_0=ab$, $F_1=a^2$ and $F_2=b^2$. Note that
the $F_i$ all are homogeneous of degree 2, and that
if $F_1=0$ and $F_2=0$, then $a=0$ and $b=0$ so
$\ZP{N}(F_0,F_1,F_2)$ is empty, and $F$ gives a well-defined
homomorphism of $\pr1$ into $\pr2$.
In addition, $F$ defines a homomorphism
$F^*:\CC[\pr2]\to\CC[\pr1]$ in the usual way, by sending
$F_i$ to $x_i$. The kernel of $F^*$ is $(x_0^2-x_1x_2)$, and
$x_0^2-x_1x_2$ is in fact
the equation of the image of $\pr1$ under $F$.
The map $F$ is bijective to its image $C=F(\pr1)$
and we'll eventually see that $C$ is isomorphic to $\pr1$, but
the homomorphism $\overline{F^*}:\CC[\pr2]/I(C)\to\CC[\pr1]$
induced by $F^*$ is not an isomorphism.
Thus the connection between homogeneous coordinate rings
and projective algebraic sets is less direct than is the connection
between coordinate rings and affine algebraic sets.
\end{example}
\begin{example}\label{eg23.2}
Again let $\CC[\pr1]=\CC[a,b]$ and let $\CC[\pr2]=\CC[x_0,x_1,x_2]$.
Consider another map $F:\pr1\to\pr2$ where this time
$F=(F_0,F_1,F_2)$ with
$F_0=a$, $F_1=b$ and $F_2=0$. Note that
the non-zero $F_i$ both are homogeneous of degree 1, and that
if $F_0=0$ and $F_1=0$, then $a=0$ and $b=0$ so
$\ZP{N}(F_0,F_1,F_2)$ is again empty, and $F$ gives a well-defined
homomorphism of $\pr1$ into $\pr2$.
The equation of the image $C=F(\pr1)$ of $F$ is $x_2=0$, and
$(x_2)$ is the kernel of the homomorphism
$F^*:\CC[\pr2]\to\CC[\pr1]$. In fact we get an isomorphism
$\CC[\pr2]/I(C)\cong\CC[\pr1]$, so we might expect
$C$ and $\pr1$ to be isomorphic, and once we define things correctly,
they will be.
\end{example}
The curves $C$ in the preceding two examples turn out both to be isomorphic
to $\pr1$, but the first example shows that the homogeneous coordinate ring
is not enough to detect this, and in both examples there is no well-defined map
$G:\pr2\to\pr1$ defined by homogeneous polynomials
whose restriction to $C$ is the inverse of $F$.
In fact, the only well-defined map $G:\pr2\to\pr1$ defined by homogeneous
polynomials are the constant maps.
We now justify this. Certainly if $G=(G_0,G_1)$ where $G_i\in\CC$ for both $i$
(and where the $G_i$ are not both 0) is a well-defined map $\pr2\to\pr1$,
but it is the constant map that sends every point of $\pr2$ to the point
$(G_0,G_1)$. Suppose $G_0$ or $G_1$ is not a constant.
Suppose for example, that $G_0$ is not a constant. Then $G_0$ is a homogeneous
polynomial of some degree $d>0$. As noted above, for $G$ to be well-defined,
$G_1$ is either the 0 polynomial or is homogeneous of degree $d$.
If $G_1$ is the 0 polynomial, then for each $p\in\pr2$ where $G_0(p)\neq0$,
$G(p)=[(1,0)]$, and for each $p\in\ZP{2}(G_0)$, $G(p)=[(0,0)]$, which is undefined.
I.e., at those points where $G$ is defined, it's constant. So
we would be in the worst of all possible worlds: $G$ is both constant and (since
by the projective Nulstellensatz $\ZP{2}(G_0)\neq \varnothing$) not well-defined!
Thus $G_1$ is not a constant, hence $G_1$ is a homogeneous polynomial of degree $d$.
If $G_0$ and $G_1$ have a non-constant common factor $F$, then
$F$ is homogeneous by Exercise \ref{ex21.1} and we have
$\varnothing\neq\ZP{2}(F)\subseteq \ZP{2}(G_0,G_1)$, so $G$ is not well-defined.
More generally, since $(0,0,0)\in Z_{\AAA^3}(G_0,G_1)$, it follows that
$Z_{\AAA^3}(G_0,G_1)\neq\varnothing$, and we will see that every irreducible component
of $Z_{\AAA^3}(G_0,G_1)$ has dimension at least 1 (see Exercise \ref{ex38.5}).
Since for every point $p\in Z_{\AAA^3}(G_0,G_1)$, the line $L_p$ through the origin which also
goes through $p$ is contained in $Z_{\AAA^3}(G_0,G_1)$, it follows that
$L_p$ represents a point of $\ZP{2}(G_0,G_1)$, and hence
$Z_{\AAA^3}(G_0,G_1)\neq\varnothing$.
Thus $G$ is not well-defined.
\begin{rem}
For any two homogeneous polynomials $G_0$ and $G_1$,
B\'ezout's Theorem states that the number of points of $\ZP{2}(G_0,G_1)$ is infinite
if $G_0$ and $G_1$ have a non-constant common factor, and otherwise
$\ZP{2}(G_0,G_1)$ consists of exactly $\deg(G_0)\deg(G_1)$ points, if the points
are counted with multiplicity. In either case B\'ezout implies that
$\ZP{2}(G_0,G_1)$ is non-empty unless one of the $G_i$ is a non-zero constant.
Thus, if we allow ourselves to use B\'ezout's Theorem,
we get a shorter way to show that no non-constant map $G:\pr2\to\pr1$
is well-defined.)
\end{rem}
Since there is no map $G:\pr2\to\pr1$, we cannot get an isomorphism
$C\to\pr1$ using only maps defined by homogeneous polynomials.
The resolution to the problem is to define a more flexible notion of algebraic map.
We do this by working locally.
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex23.1}
Consider the map $F$ of Example \ref{eg23.1}.
Show that ${\rm Im}(F)=\ZP{2}(x_0^2-x_1x_2)$ and that
${\rm ker}(F^*)=(x_0^2-x_1x_2)$, but that
$\overline{F^*}$ is not an isomorphism.
\end{exer}
\begin{proof}[Solution]
First, ${\rm Im}(F)\subseteq\ZP{2}(x_0^2-x_1x_2)$,
since $F([(a,b)])=[(ab,a^2,b^2)]$ satisfies $(ab)^2-a^2b^2=0$,
which also shows that $(x_0^2-x_1x_2)\subseteq {\rm ker}(F^*)$.
Conversely, say $[(c,d,e)]\in\ZP{2}(x_0^2-x_1x_2)$.
If $d=0$, then $c^2-de=0$ implies $c=0$, and we have
$[(c,d,e)]=[(0,0,1)]=F([(0,1)])$. If $d\neq0$, let $b=c/d$ and let $a=1$.
Then $F([(a,b)])=F([(1,b)])=[(b,1,b^2)]=[(c,d,db^2)]$,
but $c^2=de$ implies $e=c^2/d=b^2d$, so $[(c,d,db^2)]=[(c,d,e)]$.
Thus ${\rm Im}(F)=\ZP{2}(x_0^2-x_1x_2)$.
Note that $F^*(x_i)$ is homogeneous of degree 2 for each $i$.
Thus if $H\in\CC[x_0,x_1,x_2]$ and we write $H$ as a sum $H=H_0+\cdots+H_d$
of its homogeneous components, then for each non-zero term $H_j$,
$F^*(H_j)$ is homogeneous of degree $2\deg(H_j)$. Thus $H\in{\rm ker}(F^*)$
if and only if $H_j\in {\rm ker}(F^*)$ for each $j$. In particular,
${\rm ker}(F^*)$ is a homogeneous ideal. Thus to show
${\rm ker}(F^*)=(x_0^2-x_1x_2)$, given that we already have seen
$(x_0^2-x_1x_2)\subseteq {\rm ker}(F^*)$,
it suffices to show $F^*(H)=0$ implies $x_0^2-x_1x_2$ divides $H$
whenever $H$ is homogeneous.
But $F^*(H)=0$ implies $\ZP{2}(x_0^2-x_1x_2)={\rm Im}(F)\subseteq\ZP{2}(H)$,
hence by Theorem \ref{ProjNullstellensatz}, the projective Nullstellensatz,
that $H\in \sqrt{(x_0^2-x_1x_2)}$, but
$\sqrt{(x_0^2-x_1x_2)}=(x_0^2-x_1x_2)$ since $x_0^2-x_1x_2$ is irreducible.
(To see irreducibility, note that any factor of $x_0^2-x_1x_2$ is homogeneous
by Exercise \ref{ex21.1}. Thus were $x_0^2-x_1x_2$ not to be irreducible, it would be
a product of two linear forms, say $x_0^2-x_1x_2=LM$. But $\ZP{2}(L,M)$ consists of the
point $p$ where the lines defined by $L$ and $M$ cross. Also note that the gradient
of $LM$ is $\nabla(LM)=L\nabla(M)+M\nabla(L)$, hence vanishes at $p$.
But $\nabla(x_0^2-x_1x_2)=(2x_0,-x_2,-x_1)$ never vanishes. Thus
$x_0^2-x_1x_2$ cannot be the product of two linear forms and so
is irreducible. In language we'll see later on,
$\ZP{2}(x_0^2-x_1x_2)$ is smooth, but a reducible conic $\ZP{2}(LM)$ is not.)
Finally, $\overline{F^*}$ is not an isomorphism, since the image of each variable
is homogeneous of degree 2. Thus neither $a$ nor $b$ is in the image,
so $\overline{F^*}$ is not onto.
\end{proof}
\begin{exer}\label{ex23.2}
If $I\subset\CC[\pr N]$ is a homogeneous ideal and if
$Q$ is a minimal prime ideal containing $I$,
show that $Q$ is homogeneous. [Hint: apply
Exercise \ref{ex6.4}.]
\end{exer}
\begin{proof}[Solution]
Let $Q$ be minimal among all prime ideals containing $I$; i.e., if $I\subseteq P\subseteq Q$
with $P$ prime, then $P=Q$.
Let $H$ be the ideal generated by all homogeneous elements of $Q$.
Then $H$ is homogeneous and $I\subseteq H\subseteq Q$.
Say $FG\in H$. Then $Q$ being prime tells us either $F$ or $G$ is in $Q$; say $F\in Q$.
Then every homogeneous component of $F$ is in $Q$ hence in $H$, so $F\in H$.
Thus $H$ is prime, hence $H=Q$.
[Note: We don't need the hint; I don't remember now
what I had in mind to use the hint for. One thing it does do
is tell us that $\sqrt{I}$ is the intersection of the minimal primes containing
$I$, of which there are finitely many. Clearly, $\sqrt{I}\subseteq Q$,
for each minimal prime $Q$ containing $I$.
By Exercise \ref{ex6.4}, $\sqrt{I}$ is a finite intersection of prime ideals, hence by
Lemma \ref{primelem}, each $Q$ is one of these ideals.]
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{March 9, 2011}\label{lect24}
To define a more flexible notion of algebraic map, we recall the notion of localization.
\noindent{\bf Localization.}
Let $R$ be a commutative
ring with $1\neq0$, let $S\subseteq R$ be a {\em multiplicatively closed} subset;
i.e., if $a,b\in S$, then so is $ab$. For simplicity, we also assume $1\in S$.
Define $S^{-1}R$ to be equivalence classes of all fractions of the form
$\frac{r}{s}$, such that $r\in R$ and $s\in S$, where we say
$$\frac{r_1}{s_1}\sim\frac{r_2}{s_2}$$
if for some $s\in S$ we have $s(r_1s_2-s_1r_2)=0$.
Then $S^{-1}R$ is a ring under the usual addition and multiplication
operations on fractions, and we have a canonical homomorphism
$R\to S^{-1}R$ defined by $r\mapsto\frac{r}{1}$.
\begin{example}\label{eg24.1}
If $f\in R$, let $S=\{1,f,f^2,\ldots\}$. Then $S^{-1}R$ is denoted $R_f$
and we have an isomorphism $R[t]/(tf-1)\cong S^{-1}R$ induced by
$r\mapsto \frac{r}{1}$ for all $r\in R$ and by $t\mapsto \frac{1}{f}$.
\end{example}
\begin{example}\label{eg24.2}
Let $P\subset R$ be a prime ideal.
Let $S=R\setminus P$. Then $S^{-1}R$ has a unique maximal ideal
(which comes from $P$); we denote $S^{-1}R$ by $R_P$.
(A ring with a unique maximal ideal is sometimes called a {\em local ring}.
N.B.: Some people use local ring only for Noetherian rings with a unique
maximal ideal.)
\end{example}
If $P$ is a maximal ideal and $R=\CC[V]$ for an
affine algebraic set $V\subseteq\AAA^n$, then
the elements of $R_P$ can be regarded as {\em germs} of functions.
\begin{defn}
Given a function $f:X\to Y$ of topological spaces and a point $x\in X$,
the germ of $f$ at $x$ is an equivalence class of functions defined on
neighborhoods of $x$. If $U_1$ and $U_2$ are open neighborhoods of
$x$, and if $g_i:U_i\to Y$ are functions, then we say $g_1$ and $g_2$
{\em represent the same germ} (or
{\em have the same germ})
if for some open neighborhood $U_3\subseteq U_1\cap U_2$ of
$x$ we have $g_1|_{U_3}=g_2|_{U_3}$.
\end{defn}
\begin{example}\label{eg24.4}
Let $f:{\mathcal R}\to{\mathcal R}$ be the function $f(x)=|x|$.
For each $\epsilon>0$, let $g_\epsilon:{\mathcal R}\to{\mathcal R}$ be
\begin{equation*}
g_\epsilon(x)=
\begin{cases}
|x| & \text{if $|x|\leq\epsilon$,}\\
\epsilon & \text{if $|x|\geq\epsilon$.}
\end{cases}
\end{equation*}
Then $f$ and $g_\epsilon$ for every $\epsilon>0$ represent (or have) the same germ at $x=0$.
\end{example}
\begin{example}\label{eg24.5}
If $0\in S$, then $S^{-1}R=\{0\}$. This is because we can use
$s=0$ in $s(r_1s_2-s_1r_2)=0$, and hence get
$\frac{r_1}{s_1}\sim\frac{r_2}{s_2}$ for any two fractions; in particular,
every fraction is equivalent to $\frac{0}{1}$.
\end{example}
\begin{example}\label{eg24.6}
If $R$ is a domain, then $R_{(0)}$ is a field. if $R=\CC[V]$ for some irreducible affine
algebraic set $V$, then we denote $R_{(0)}$ by $\CC(V)$, and refer to it as the
{\em function field} of $V$.
\end{example}
\begin{example}\label{eg24.7}
It is easy to see that the homomorphism $R\to S^{-1}R$ is injective if $R$ is a
domain and $0\not\in S$, but otherwise injectivity can fail. For example,
Let $V=Z(xy)\subset\AAA^2$, so $V$ is the union of the coordinate axes.
Let $\overline{x}$ be the image of $x$ under the quotient $\CC[\AAA^2]\to\CC[V]$,
where $\CC[AAA^2]=\CC[x,y]$ and $\CC[V]=\CC[x,y]/(xy)$ (note that
$I(V)=(xy)$). Then $\CC[V]_{\overline{x}}\cong\CC[x]_x\cong\CC[x,\frac{1}{x}]$
and $(\overline{y})$ is the kernel of $\CC[V]\to\CC[V]_{\overline{x}}$.
\end{example}
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex24.1}
Justify the claims made in Example \ref{eg24.1}.
\end{exer}
\begin{proof}[Solution by Ashley Weatherwax]
We'll start by refreshing what example 24.1.1 was. Let $f\in R$ and let $S = \{1, f, f^2, ... \}.$ Then $S^{-1}R$ is
denoted $R_f$ and we have an isomorphism $R[t]/(tf-1) \cong S^{-1}R$ induced by $r \mapsto \frac{r}{1}$ for all
$r \in R$ and by $t \mapsto \frac{1}{f}$.
Define as above $h: R[t] \rightarrow S^{-1}R$ by $r \mapsto \frac{r}{1}$ and $t\mapsto \frac{1}{f}$. Then we
get an induced map $\phi: R[t]/(tf-1) \rightarrow S^{-1}R$. Clearly, $h$ is a homomorphism, and we know this induced map is well defined, as $(tf-1) \subset
\text{ ker } h$.
Now define a map $g: R \rightarrow R[t]/(tf-1)$, defined by $r \mapsto r$. Then
$g(f) = f$, which is a unit in $R[t]/(tf-1)$. So by the universal property of localization, there is a unique
homomorphism $\psi: S^{-1}R \to R[t]/(tf-1)$, which sends $\frac{r}{f^k} \mapsto g(r)g(f^k)^{-1}=rt^m$.
\noindent \textbf{Claim}: $\phi \circ \psi = id_{S^{-1}R}$ and $\psi \circ \phi = id_{R[t]/(tf-1)}$
Let $\frac{r}{f^k} \in S^{-1}R$. Then
$$\phi \circ \psi(\frac{r}{f^k})=g(r)g(f^k)^{-1} = \phi (rf^{-k}) = rf^{-k} = \frac{r}{f^k}$$
Let $[a_0 + a_1 t+ a_2 t^2 + ... + a_k t^k] \in R[t]/(tf-1)$. Then
$$\psi \circ \phi ([a_0 + ... + a_k t^k]) = \psi (a_0 + \frac{a_1}{f} + ... + \frac{a_k}{f^k})
= a_0 + a_1f^{-1} + ... + a_k f^{-k}$$
But as $f^{-1} = t$, this is equal to
$$\psi \circ \phi ([a_0 + ... + a_k t^k]) = a_0 + a_1 t + ... + a_k t^k$$
Then as $\phi$ has an inverse, it is an isomorphism.
\end{proof}
\begin{exer}\label{ex24.2}
Suppose we defined $$\frac{r_1}{s_1}\sim\frac{r_2}{s_2}$$
by the condition that $r_1s_2-s_1r_2=0$. Show that
this need not be an equivalence relation by giving two examples.
In one example, let $0\in S$ for any $R$ and $S$ of your choice.
For the other example, take $R=\CC[x,y]/(xy)$ and $S=\{1,x,x^2,\ldots\}$.
\end{exer}
\begin{proof}[Solution by Anisah Nu'Man]
Let $R={\mathbb Z}$ and $S=\{0, 1, a, a^2, a^3, \dots\}$ for any $a\in {\mathbb Z}$ with $a\not=0$.
Observe we have $\displaystyle \frac{1}{0} \sim \frac{0}{0}$ since $1(0)=0(0)=0$
and $\displaystyle \frac{0}{0} \sim \frac{0}{a}$ since $0(a)-0(0)=0$. Yet
$\displaystyle \frac{1}{0} \nsim \frac{0}{a}$ since $1(a)-0(0)=a\not=0$.
Now let $R=\CC[x,y]/(xy)$ and $S=\{1,x,x^2,\dots\}$. Observe we have
$\displaystyle \frac{y^{2}}{x} \sim \frac{0}{x}$ since $y^2(x)-x(0)=y(yx)-0=0$ and
$\displaystyle \frac{0}{x} \sim \frac{y}{1}$ since $0(1)-x(y)=0-xy=0-0=0$. Yet
$\displaystyle \frac{y^2}{x} \nsim \frac{y}{1}$ since $y^2(1)-xy=y^2-0=y^2 \not=0$.
\end{proof}
\begin{exer}\label{ex24.3}
Show that $(\overline{y})$ is the kernel of the homomorphism
$\CC[V]\to\CC[V]_{\overline{x}}$ in Example \ref{eg24.7}
and that $\CC[V]_{\overline{x}}\cong\CC[x]_x$.
\end{exer}
\begin{proof}[Solution by Zheng Yang]
First, $\CC[x,y]/(xy)_{\overline{x}}\cong\CC[x,y,t]/(xy,tx-1)\cong
\CC[x,y,t]/(xy,tx-1,txy)\cong\CC[x,y,t]/(xy,tx-1,y)\cong\CC[x,t]/(tx-1)\cong\CC[x]_x$.
The kernel of the localization map $R\to S^{-1}R$ is $\{r:rs=0\}$
for some $s\in S$. Here $S=\{1,x,x^2,\ldots\}$, so the kernel is
$(\overline{y})$.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{March 11, 2011}\label{lect25}
\noindent{\bf The structure sheaf (or the sheaf of regular functions).}
Let $V\subseteq\AAA^n$ be an irreducible closed subset.
For each non-empty open subset $U$ of $V$
define $\OO_V(U)$ to be $\cap_{p\in U}\CC[V]_{I_V(p)}\subset\CC(V)$.
For the empty set, define $\OO_V(\varnothing)=0$.
Elements of $\CC(V)$ are called rational functions. If $p\in V$ and if
a rational function $f$
is in $\CC[V]_{I_V(p)}$, we say $f$ is {\em regular} at $p$.
The elements of
$\OO_V(U)$ are precisely the rational functions regular at each point $p\in U$,
so we call $\OO_V(U)$ the ring of regular functions on $U$ (or the ring of functions
regular on $U$).
It is not in general easy to determine $\OO_V(U)$, given an open subset
$U\subseteq V$. There are important cases where $\OO_V(U)$ is known.
\begin{lem}
Let $V\subseteq\AAA^n$ be an irreducible closed subset.
Let $f\in\CC[V]$ and let $U_f=V\setminus Z_V(f)$. Then
$\OO_V(U_f)=\CC[V]_f$.
\end{lem}
\begin{proof}
This is true by definition if $f=0$, so assume $f\neq 0$.
Since $p\in U_f$ implies $f(p)\neq0$, we have $f\not\in I_V(p)$.
Thus $\frac{g}{f^m}\in\CC[V]_{I_V(p)}$ for all $g\in\CC[v]$ and all $m>0$.
Hence $\CC[V]_f\subseteq \OO_V(U_f)$.
Now say $h\in\OO_V(U_f)$. Thus for each $p\in U_f$ we have
$h=\frac{g_p}{f_p}$, where $g_p\in\CC[V]$ and $f_p\not\in I_V(p)$.
Thus $Z_V(\{f_p:p\in U_f\})\cap U_f=\varnothing$, so
$Z_V(\{f_p:p\in U_f\})\subseteq Z_V(f)$, hence
$f\in\sqrt{((\{f_p:p\in U_f\})}$. Therefore, $f^m=a_{p_1}f_{p_1}+\cdots a_{p_r}f_{p_r}$
for some elements $a_{p_i}\in \CC[V]$ and points $p_i\in U_f$,
so $f^mh=a_{p_1}f_{p_1}h+\cdots a_{p_r}f_{p_r}h=
a_{p_1}f_{p_1}\frac{g_{p_1}}{f_{p_1}}+\cdots a_{p_r}f_{p_r}\frac{g_{p_r}}{f_{p_r}}=
a_{p_1}g_{p_1}+\cdots a_{p_r}g_{p_r}$. Call this $g$; then $g\in\CC[V]$ and
$h=\frac{g}{f^m}\in \CC[V]_f$, so $\OO_V(U_f)\subseteq \CC[V]_f$.
\end{proof}
\begin{example}
Let $V$ be an irreducible closed subset of $\AAA^n$.
Then $\OO_V(V)=\CC[V]$. Just take $f=1$, using the fact that
$\CC[V]_1=\CC[V]$.
\end{example}
\begin{example}
Let $U=\AAA^2\setminus\{(0,0)\}$.
Then $\OO_{\AAA^2}(U)=\CC[\AAA^2]=\CC[x,y]$. By definition,
$\CC[\AAA^2]\subseteq \OO_{\AAA^2}(U)$, so let $h\in \OO_{\AAA^2}(U)$.
Then $h|_{U_x}\in\CC[\AAA^2]_x$ and $h|_{U_y}\in\CC[\AAA^2]_y$,
hence in $\CC(\AAA^2)$ we can write $h=\frac{f}{x^m}=\frac{g}{y^n}$
for some $f,g\in\CC[\AAA^2]$ and some $m,n\geq0$.
Thus $y^nf=x^mg$, so by unique factorization we have
$f=\phi x^m$ and $g=\gamma y^n$ and $h=\phi=\gamma\in\CC[\AAA^2]$.
\end{example}
\begin{example}
It can happen that $\OO_V(U)$ is not a finitely generated $\CC$-algebra
(see A. Neeman, {\em Steins, affines and Hilbert's fourteenth problem}, Ann. of Math.
127 (1988), 229--244, for example) or even Noetherian.
One of the most famous examples is one of (if not the) first,
due to D. Rees ({\em On a problem of Zariski}, Illinois J. Math. 2 (1958), 145--149).
Rees constructed a ring $A=\CC[V]$ and an open subset $U=V\setminus Z_V(I)$
such that $B=\OO_V(U)$, which is of the form (in his notation) $\cup_nI^{-n}$,
is not a finitely generated $\CC$-algebra, or even a Noetherian ring.
% Check this over, then uncomment:
% A similar but lower dimensional example was suggested by Bill Heinzer
% (over dinner with Roger and Sylvia Wiegand the other day).
% Consider the subring $R = k[x,t^2(t-1), t(t-1), x]$
% of the polynomial ring $k[t,x]$ over a field $k$.
% Then $R=\CC[V]$ for $V=Z_V(z^3-y(y-z))\subset\AAA^3$, where $\CC[\AAA^3]\CC[x,y,z]$.
% Notice that the integral closure of $R$ is $k[t,x]$.
% Let $P*$ be the prime ideal $(x-t)k[t,x]$ and let $P$ denote the contraction of $P*$ to $R$.
% Notice that $R$ localized at $P$ is equal to $k[t,x]$
% localized at $P*$. The claim now is that $\OO_V(U)$ for $U=V \setminus Z_V(P)$ is not Noetherian.
% This ring is the intersection of the localizations $R_Q$ as
% $Q$ varies over the height-one primes of $R$ other than $P$. Call this ring $S$.
% The complete integral closure of $S$ is the ring $k[x,t][1/(x-t)] = S*$. To show
% $S$ is not Noetherian, it suffices to show that $S*$ is
% not integral over $S$, but this follows because $Pk[t,x]$ has the maximal ideal
% $(t-1,x)k[t,x]$ as a minimal prime.
\end{example}
\newpage
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex25.1}
A topological space $X$ is said to be {\em quasi-compact}
if every open cover has a finite subcover.
(Some people reserve the use of the word compact to mean
quasi-compact and Hausdorff.)
Show that every open subset of an algebraic set is quasi-compact.
\end{exer}
\begin{proof}[Solution by Jason Hardin]
Let $V\subseteq\mathbb{A}^n$ be an algebraic set and $U\subseteq V$ be an open subset. Let $\{U_i\}_{i\in I}$ be an open cover for $U$. So $U\subseteq\cup_{i\in I}U_i$. This means that $\cap_{i\in I}(V\backslash U_i)\subseteq(V\backslash U)$. Since $V\backslash U_i$ is a closed subset of $V$, by the definition of the subspace topology it has the form $V\backslash U_i=Z(J_i)\cap V$, where $J_i\subseteq\mathbb{C}[x_1,\ldots,x_n]$ is an ideal. Now we have
\begin{eqnarray*}
\cap_{i\in I}(V\backslash U_i) &=& \cap_{i\in I}(Z(J_i)\cap V)\ =\ \cap_{i\in I}(Z(J_i))\cap V\ =\ Z((\cup_{i\in I}J_i))\cap V\\
&=& Z(f_{i_1},\ldots,f_{i_t})\cap V,
\end{eqnarray*}
where $f_{i_j}\in J_{i_j}$. Let $I'=\{i_1,\ldots,i_t\}\subseteq I$. Then
\begin{eqnarray*}
V\backslash U &\supseteq& \cap_{i\in I}(V\backslash U_i)\ =\ Z(f_{i_1},\ldots,f_{i_t})\cap V\ =\ \cap_{i\in I'}(Z(J_i)\cap V)\ =\ \cap_{i\in I'}(V\backslash U_i),
\end{eqnarray*}
i.e., $U\subseteq\cup_{i\in I'}U_i$. So $\{U_i\}_{i\in I'}$ is a finite subcover of $\{U_i\}_{i\in I}$, and hence $U$ is quasi-compact.
\end{proof}
\begin{exer}\label{ex25.2}
Show that the open subsets of the form $U_f$ give a basis for the
Zariski topology on an algebraic set $V$.
\end{exer}
\begin{proof}[Solution 1, by Philip Gipson]
Since the Zariski topology is formed by declaring $\{Z(J):J\subseteq\CC[V]\}$ to be the
closed sets and $U_f=V\setminus Z_V(f)$ it is enough to show that the closed
sets $\{U_f^c\}=\{Z_V(f)\}$ generates $\{Z(J):J\subseteq\CC[V]\}$.\\
To that end let $J$ be any ideal in $\CC[V]$. Since $\CC[V]$ is noetherian $J$ is
finitely generated $J=\langle f_1,...,f_n\rangle$. Thus $Z_V(J)=Z_V(f_1,...,f_n)=\bigcap Z_V(f_i)$
and so we conclude that $\{Z_V(f)\}$ generates $\{Z(J)\}$ and thus is a basis for the topology.
\end{proof}
\begin{proof}[Solution 2, by Kat Shultis]
We first check that the collection of sets of the form $U_f$ form a basis. Notice that the polynomial 0 is
zero everywhere, and so $U_0=\emptyset$. Also, the polynomial 1 is zero nowhere, so for every
$v\in V$, $v\in U_1=V$. Now, let $v\in U_f\cap U_g$. This means that $f(v)\neq 0$ and $g(v)\neq 0$.
Thus, $(fg)(v)\neq 0$. If $w\in U_{fg}$ then $(fg)(w)\neq 0$ so that $f(w)\neq 0$ and $g(w)\neq 0$,
so that $v\in U_{fg}\subseteq U_f\cap U_g$. Thus, the sets $U_f$ form a basis of some topology.
We must now check that the basis topology given by the sets of the form $U_f$ is the Zariski
topology. Let $U$ be a set that is open in the basis topology. Then $U=\cup_\alpha U_{f_\alpha}$
for some collection $\{f_\alpha\}$. By definition, each $U_f$ is open in the Zariski topology, and so
each open set in the basis topology is also open in the Zariski topology. Now, let $U$ be an open
set in the Zariski topology. Thus, $U$ is of the form $V\setminus Z_V(T)$ for some collection of
polynomials $T$. However, we can write this as $U=\cup_{s\in U} U_{f_s}$ where $f_s$ does
not vanish at $s$. Thus, every open set in the Zariski is a union of basis elements, and so, we
have that Zariski open sets are also open in the basis topology. Hence, the basis topology is the
Zariski topology.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{March 14, 2011}\label{lect26}
\noindent{\bf What is a sheaf?}
Let $X$ be a topological sapce. We can regard the topology as specifying a category
whose objects are the open subsets and whose arrows are the inclusion maps.
If you are comfortable with category theory, a sheaf of abelian groups is then a
contravariant functor to the category of abelian groups, but the functor must satify
some conditions which essentially say that the ``sections'' are determined ``locally''.
Thus a sheaf $\SSS$ (of abelian groups) on $X$ is a certain assignment of an abelian group
$\SSS(U)$ for every open subset $U\subseteq X$. We refer to the elements of $\SSS(U)$
as {\em sections} of $\SSS$ over $U$, or {\em global sections} if $U=X$.
For each inclusion $U_2\subseteq U_1$
we have the ``restriction'' homomorphism $\rho_{12}:\SSS(U_1)\to\SSS(U_2)$.
If $f\in\SSS(U_1)$, it is common to write $f|_{U_2}$ for $\rho_{12}(f)$. (This is just
formal notation, motivated by the example of sheaves of the type $\OO_V$
where sections over open sets $U\subseteq V$ are actual functions
$U\to\CC$, since it makes sense to restrict a function on an open set to
an open subset.) Moreover,
if $U_2=U_1$, then $\rho_{12}$ is the identity. If $U_3\subseteq U_2$, then
$\rho_{23}\rho_{12}=\rho_{13}$. If $U=\varnothing$, then $\SSS(U)=0$.
So far, this just says that $\SSS$ is a contravariant functor.
In addition, we require that sections be determined locally. I.e.,
if $\{U_i\}$ is an open cover of an open set $U$, and if $f,g\in\SSS(U)$
such that $f|_{U_i}=g|_{U_i}$ for all $i$, we require that $f=g$,
and if there are sections $f_i\in\SSS(U_i)$ for all $i$, such that
$f_i|_{U_i\cap U_j}=f_j|_{U_j\cap U_i}$ for all $i$ and $j$, then we require that
there is a section $f\in\SSS(U)$ such that $f|_{U_i}=f_i$ for all $i$.
\begin{example}
If $V$ is an affine algebraic set, it's not hard to check that $\OO_V$
is a sheaf.
\end{example}
We can now enlarge the class (or category) objects we can deal with.
\begin{defn}
An irreducible Zariski-closed subset of $\AAA^n$ is called an affine {\em variety}.
A {\em quasi-affine} variety is a non-empty Zariski-open subset of
an affine variety.
\end{defn}
Now that we have quasi-affine varieties, we need to define their morphisms.
\begin{defn}\label{morphdef}
Let $V_1$ and $V_2$ be affine varieties and let $U_i\subseteq V_i$ be quasi-affine varieties.
A morphism $\phi:U_1\to U_2$ is a continuous map such that for each point $p\in U_1$
there are affine open subsets $\phi(p)\in U_{\phi(p)}\subseteq V_2$ and
$p\in U_p\subseteq\phi^{-1}(U_{f(p)})\subseteq V_1$ such that $\phi^*$ (i.e., composition with $\phi$)
defines a $\CC$-homomorphism $\OO_{V_2}(U_{\phi(p)})\to\OO_{V_1}(U_p)$.
\end{defn}
\renewcommand{\thethm}{\arabic{section}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{March 16, 2011}\label{lect27}
\noindent{\bf Morphisms of quasi-affine varieties.}
\begin{example}
Let $\phi:V\to W$ be an algebraic map of irreducible affine algebraic sets.
Then $\phi$ is a morphism. By Proposition \ref{contProp}, $\phi$ is continuous.
For each $p\in V$, take $U_p=V$ and $U_{\phi(p)}=W$.
Since $\phi^*$ defines a $\CC$-homomorphism $\CC[W]\to\CC[V]$,
we see that $\phi$ is a morphism.
\end{example}
\begin{example}
Let $\phi:V\to W$ be a morphism of irreducible affine algebraic sets.
Then $\phi$ is an algebraic map. Since $\phi$ is a morphism
there are open subsets $p\in U_p\subseteq \phi^{-1}(U_{\phi(p)})$
and $\CC$-homomorphisms $\phi^*:\OO_W(U_{\phi(p)})\to\OO_V(U_p)$.
Since $U_p$ is affine, there is an element $f_p\in\CC[V]$ such that
$\OO_V(U_p)=\CC[V]_{f_p}$. Let $i_p:\CC[W]\subseteq \OO_W(U_{\phi(p)})$
be the canonical inclusion. Then we have $\CC$-homomorphisms
$\CC[W]\xrightarrow{\ i_p\ }\OO_W(U_{\phi(p)})\xrightarrow{\ \phi^*\ }\OO_V(U_p)=\CC[V]_{f_p}$
which for simplicity we will refer to as $\phi^*_p$.
Thus for any $h\in\CC[W]$, we have $\phi^*_p(h)=\frac{g_p}{f^{m_p}_p}$ for elements
$g_p\in\CC[V]$. Note that
$$\frac{g_p}{f^{m_p}_p}\Big|_{U_p\cap U_q}=\phi^*_p(h)\big|_{U_p\cap U_q}=\frac{g_q}{f^{m_q}_q}\Big|_{U_q\cap U_p}$$
so in $\CC(V)$ we have $\frac{g_p}{f^{m_p}_p}\frac{g_q}{f^{m_q}_q}$ for all $p$ and $q$.
Since $\{U_p\}$ is a cover of $V$, we have
$Z_V(\{f_p:p\in V\})=\varnothing$, so $1=\sum_pa_pf^{m_p}_p$, where each $a_p\in\CC[V]$
but only finitely many are non-zero. Let $\eta=\sum_pa_pg_p$. Then in $\CC(V)$ we have
$$\eta=\sum_pa_pg_p=\sum_pa_pf^{m_p}_p\frac{g_p}{f^{m_p}_p}=
\Big(\sum_pa_pf^{m_p}_p\Big)\frac{g_q}{f^{m_q}_q}=\frac{g_q}{f^{m_q}_q}=\phi^*_p(h),$$
and so $\phi^*_p(h)=\eta\in\CC[V]$. Thus $\phi^*$ is a $\CC$-homomorphism $\CC[W]\to\CC[V]$, so
by Theorem \ref{mapsVShomoms}, $\phi$ is an algebraic map.
\end{example}
\begin{example}
Let $U$ be the quasi-affine variety $U=\AAA^1\setminus\{0\}$ and let $W$ be
the affine variety $W=Z(xy-1)\subset\AAA^2$.
Then $U$ and $W$ are isomorphic. Let $f:U\to W$ be $f:s\mapsto (s,\frac{1}{s})$
and let $g:W\to V$ be $g:(a,b)\mapsto a$. It is easy to see that $f$ and $g$ are bijective,
inverse to each other and continuous (since the $U$ has the finite complement topology,
being a subspace of $\AAA^1$, as does $W$, by Exercise \ref{ex10.2}).
If we take $\CC[\AAA^1]=\CC[t]$ and $\CC[W]=\CC[x,y]/(xy-1)\cong\CC[x]_x$,
it is also easy to see that $f^*:\CC[W]\to\OO_{\AAA^1}(U)=\CC[\AAA^1]_t$
is the $\CC$-homomorphism $x\mapsto t$
and $g^*:\OO_{\AAA^1}(U)=\CC[\AAA^1]_t \to \CC[W]$
is the $\CC$-homomorphism $t\mapsto x$,
and hence $f$ and $g$ are morphisms and $U$ and $W$ are therefore isomorphic.
\end{example}
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}[The Universal Property of Localization]\label{ex27.1}
Given rings $R$ and $T$, a multiplicatively closed subset $S\subset R$ and a homomorphism
$f:R\to T$ such that $f(s)$ is a unit for every $s\in S$, show that there is a unique
homormorphism $S^{-1}f:S^{-1}R\to T$ such that $f$ factors as
$$R\xrightarrow{r\mapsto \frac{r}{1}} S^{-1}R\xrightarrow{S^{-1}f}T.$$
[Hint: consider $S^{-1}f:\frac{r}{s}\mapsto f(r)(f(s))^{-1}$.]
\end{exer}
\begin{proof}[Solution by Becky Egg]
Define $S^{-1}f:S^{-1}R \rightarrow T$ by $S^{-1}f(r/s)=f(r)f(s)^{-1}$.
% check that it's well-defined
If $\frac{r}{s} \sim \frac{r'}{s'}$, then there exists $t \in S$ with $t(rs'-r's)=0$. So $trs'=tr's$, and hence
\[f(t)f(r)f(s')=f(trs')=f(tr's)=f(t)f(r')f(s).\]
Since $f(t),f(s),f(s')$ are units of $T$, we have $f(r)f(s)^{-1}=f(r')f(s')^{-1}$,
that is, $S^{-1}f(\frac{r}{s})=S^{-1}f(\frac{r'}{s'})$. So $S^{-1}f$ is well-defined. \\
Also note that,
\begin{eqnarray*}
S^{-1}f(\frac{r}{s} \cdot \frac{r'}{s'})&=&f(rr')f(ss')^{-1}\\
&=&f(r)f(r')[f(s)f(s')]^{-1} \\
&=&f(r)f(s)^{-1}f(r')f(s')^{-1} \\
&=&S^{-1}f(\frac{r}{s})S^{-1}f(\frac{r'}{s'})
\end{eqnarray*}
and
\begin{eqnarray*}
S^{-1}f(\frac{r}{s}+\frac{r'}{s'})&=&S^{-1}f(\frac{rs'+r's}{ss'}) \\
&=&f(rs'+r's)f(ss')^{-1} \\
&=&[f(r)f(s')+f(r')f(s)]f(s)^{-1}f(s')^{-1} \\
&=&f(r)f(s)^{-1}+f(r')f(s')^{-1} \\
&=&S^{-1}f(\frac{r}{s})+S^{-1}f(\frac{r'}{s'}),
\end{eqnarray*}
so $S^{-1}f$ is in fact a homomorphism.
Then, given $r \in R$, we have
\[S^{-1}f(\frac{r}{1})=f(r)f(1)^{-1}=f(r),\]
i.e., $f$ factors through $S^{-1}R$. \\
To see that $S^{-1}f$ is unique, suppose that $\phi:S^{-1}R \rightarrow T$
is a homomorphism such that such that $\phi(r/1)=f(r)$ for all $r \in R$. We have
\[\phi(\frac{r}{s})=\phi(\frac{r}{1})\phi(\frac{1}{s})=f(r)\phi(\frac{1}{s}).\]
However,
\[\phi(\frac{1}{s})\phi(\frac{s}{1})=\phi(\frac{1}{1})=1,\]
that is, $\phi(\frac{1}{s})$ is invertible, with $[\phi(\frac{1}{s})]^{-1}=\phi(\frac{s}{1})$.
Thus we have $f(r)=\phi(\frac{r}{s})\phi(\frac{s}{1})$, and so
\begin{eqnarray*}
Sf^{-1}(\frac{r}{s})&=&f(r)f(s)^{-1} \\
&=& \phi(\frac{r}{s})\phi(\frac{s}{1})f(s)^{-1} \\
&=&\phi(\frac{r}{s})f(s)f(1)^{-1}f(s)^{-1} \\
&=& \phi(\frac{r}{s}).
\end{eqnarray*}
So the homomorphism $S^{-1}f$ is the unique map such that $f$ factors through $S^{-1}R$.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{March 18, 2011}\label{lect28}
\noindent{The structure sheaf for projective varieties.}
To define the structure sheaf for projective varieties we mimic the definitions
for affine varieties. The function field $\CC(\pr n)$ is defined to be
all fractions $\frac{F}{G}$ such that $F$ and $G$ are homogeneous elements
of $\CC[\pr n]$ with $G$ not the zero polynomial and with $\deg(F)=\deg(G)$.
Now let $p\in\pr n$ and define $\OO_{\pr{n},p}\subset\CC(\pr n)$ to be
those elements of $\CC(\pr n)$ regular at $p$; i.e., all
$\frac{F}{G}\in\CC(\pr n)$ with $G(p)\neq0$.
For any open subset $U\subseteq\pr n$, define $\CC_{\pr n}(U)$ to be the ring
of functions regular on $U$; i.e., $\CC_{\pr n}(U)=\cap_{p\in U}\OO_{\pr{n},p}$.
Open sets of particular interest are those of the form $U_G=\pr{n}\setminus Z_{\pr n}(G)$
where $G\in\CC[\pr n]$ is a homogeneous polynomial, but not the zero polynomial.
\begin{prop}\label{usualaffopeninpn}
Let $\CC[\pr n]=\CC[x_0,\ldots,x_n]$. Then
$\OO_{\pr n}(U_{x_i})=\CC[\frac{x_0}{x_i},\ldots,\frac{x_n}{x_i}]=\CC[\AAA^n]$.
\end{prop}
\begin{proof}
Since $\frac{x_0}{x_i},\ldots,\frac{x_{i-1}}{x_i},\frac{x_{i+1}}{x_i},\ldots,\frac{x_n}{x_i}$
are algebraically independent, $\CC[\frac{x_0}{x_i},\ldots,\frac{x_n}{x_i}]$ is just the polynomial ring in
$n$ indeterminates, hence $\CC[\frac{x_0}{x_i},\ldots,\frac{x_n}{x_i}]=\CC[\AAA^n]$.
So now we show $\OO_{\pr n}(U_{x_i})=\CC[\frac{x_0}{x_i},\ldots,\frac{x_n}{x_i}]$.
By definition, $\frac{x_j}{x_i}\in\OO_{\pr n}(U_{x_i})$, so
$\CC[\frac{x_0}{x_i},\ldots,\frac{x_n}{x_i}]\subseteq \OO_{\pr n}(U_{x_i})$.
Let $q\in U_{x_i}$ and let
$$\widetilde{q}=
(\frac{q_0}{q_i},\ldots,\frac{q_{i-1}}{q_i},\frac{q_{i+1}}{q_i},\ldots,\frac{q_n}{q_i})\in\AAA^n.$$
Then $\OO_{\pr n,q}=\CC[\frac{x_0}{x_i},\ldots,\frac{x_n}{x_i}]_{I_{\AAA^n}(\widetilde{q})}$.
To see this, let $F/G\in \OO_{\pr n,q}$, where $F$ and $G$ are homogeneous, $G\neq0$,
$d=\deg(G)$ and either $d=\deg(F)$ or $F=0$.
Then $F/G=(F/x_i^d)/(G/x_i^d)=f/g$, where $f=F(x_0/x_i,\ldots,x_n/x_i)$
and $g=G(x_0/x_i,\ldots,x_n/x_i)$. But $G(q)\neq0$ implies
$g(\widetilde{q})\neq0$, so $f/g\in \CC[\frac{x_0}{x_i},\ldots,\frac{x_n}{x_i}]_{I_{\AAA^n}(\widetilde{q})}$.
Conversely, if $f/g\in\CC[\frac{x_0}{x_i},\ldots,\frac{x_n}{x_i}]_{I_{\AAA^n}(\widetilde{q})}$
for $f,g\in\CC[\frac{x_0}{x_i},\ldots,\frac{x_n}{x_i}]$ with
$g\not\in I_{\AAA^n}(\widetilde{q})$, then either $f=0$ and so $f/g=0\in\OO_{\pr n,q}$ or
$f\neq0$ and we have
$f/g=F/G$ for $F=x_i^df$ and $G=x_i^dg$, where $d=\max(\deg(f),\deg(g))$.
Then $F$ and $G$ are homogeneous of the same degree with $G(q)=g(\widetilde{q})\neq0$,
so $f/g=F/G\in \OO_{\pr n,q}$.
Therefore we have
$\OO_{\pr n}(U_{x_i})=\cap_{q\in U_{x_i}}\OO_{\pr n,q}
=\cap_{\widetilde{q}\in\AAA^n}\CC[\AAA^n]_{I_{\AAA^n}(\widetilde{q})}
=\OO_{\AAA^n}(\AAA^n)=\CC[\frac{x_0}{x_i},\ldots,\frac{x_n}{x_i}]$.
\end{proof}
\begin{example}
Let $\CC[\pr n]=\CC[x_0,\ldots,x_n]$.
Here is an alternative way to evaluate $\OO_{\pr n}(U_{x_i})$.
By Exercise \ref{ex28.1}, we have
$\OO_{\pr n}(U_{x_i})=R_{x_i}$.
But for any homogeneous $F$ of degree $m$,
$F/x_i^m=F(x_0/x_i,\dots,x_n/x_i)\in\CC[x_0/x_i,\ldots,x_n/x_i]$,
and for any $f(x_0/x_i,\dots,x_n/x_i)\in\CC[x_0/x_i,\ldots,x_n/x_i]$
of degree $d$ we have $f=F/x_i^d\in R_{x_i}$; note that $F=x_i^df$ is homogeneous of degree $d$.
Thus $R_{x_i}=\CC[\frac{x_0}{x_i},\ldots,\frac{x_n}{x_i}]$.
\end{example}
\begin{example}
Using Exercise \ref{ex28.1}, we see that the
global sections $\OO_{\pr n}(\pr n)$ of $\OO_{\pr n}$ are just constants;
$\OO_{\pr n}(\pr n)=R_1=\CC$.
\end{example}
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex28.1}
Let $G\in\CC[\pr n]$ be homogeneous. Show that
$\OO_{\pr n}(U_G)$ is the ring $R_G$ of all fractions of the form $\frac{F}{G^m}\in\CC(\pr n)$
where $F$ is either 0 or homogeneous and $\deg(F)=m\deg(G)$.
\end{exer}
\begin{proof}[Solution by Becky Egg]
We have that
\[{\mathcal O}_{\pr n}(U_G)=\bigcap_{p \in U_G} {\mathcal O}_{\pr n,p} \]
where $U_G$ is the set of points in $\pr n$ at which $G$ is not 0,
and elements of ${\mathcal O}_{\pr n,p}$ are of the form $\frac{F}{H}$,
where $H(p) \neq 0$ and $\deg(F)=\deg(H)$.
First, consider $\frac{F}{G^m} \in R_G$ with $\deg(F)=m \deg(G)$.
For any $p \in U_G$, we have that $G(p) \neq 0$, and hence $G^m(p) \neq 0$.
So $\frac{F}{G^m} \in {\mathcal O}_{\pr n,p}$ for all $p \in U_G$, and
thus $\frac{F}{G^m} \in {\mathcal O}_{\pr n}(U_G)$.
Now suppose that $\frac{F}{H} \in {\mathcal O}_{\pr n}(U_G)$.
Note that $H(p) \neq 0$ for all $p \in U_G$, so $q \in Z_{\pr n}(H)$
implies that $q \notin U_G$, and hence $G(q)=0$. So we have
\[G \in I(Z_{\pr n}(H))=\sqrt{(H)},\]
where equality of the ideals follows from the projective Nullstellensatz.
So $G^m \in (H)$ for some $m$, i.e., $G^m=F'H$ for some $F' \in \CC[\pr n]$. Thus we have
\[\frac{F}{H}=\frac{F'F}{F'H}=\frac{F'F}{G^m} \in R_G.\]
Note also that $\deg(F)=\deg(H)$, and so
\[\deg(F'F)=\deg(F'H)=\deg(G^m)=m \deg(G).\]
Thus ${\mathcal O}_{\pr n}(U_G)$ is the ring $R_G$ of all fractions of the
form $\frac{F}{G^m} \in \CC(\pr n)$ where $F$ is either 0 or homogeneous
and $\deg(F)=m\deg(G)$.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{March 28, 2011}\label{lect29}
\noindent{\bf Quasi-projective varieties.}
\begin{defn}
A {\em projective variety} is a closed irreducible subset of $\pr n$.
A {\em quasi-projective variety} is a non-empty open subset of a projective variety.
Given a projective variety $V\subseteq \pr n$, the homogeneous coordinate ring
of $V$ is $\CC[V]=\CC[\pr n]/I_{\pr n}(V)$ and the function field of $V$
is the field $\CC(V)$ of all fractions $F/G$ such that $F,G\in\CC[V]$ are homogeneous,
and where $G\neq0$ and $F$ is either 0 or $\deg(F)=\deg(G)$.
For $p\in V$, we define $\OO_{V,p}$ to be the ring of functions regular at $p$;
i.e., all fractions $F/G$ such that $F,G\in\CC[V]$ are homogeneous,
and where $G(p)\neq0$ and $F$ is either 0 or $\deg(F)=\deg(G)$.
Finally, we set $\OO_V(\varnothing)=0$ and for a non-empty open subset
$U\subseteq V$, we set $\OO_V(U)=\cap_{p\in U}\OO_{V,p}\subseteq \CC(V)$.
\end{defn}
Now we establish some notation. Let $V\subset\pr n$ be a projective variety.
Let $0\neq G\in\CC[V]$ be homogeneous of degree $d=\deg(G)>0$.
Let $U_G=V\setminus Z_V(G)$.
\begin{example}
Let $\CC[\pr 3]=\CC[x,y,z,w]$ and let $V\subset\pr 3$ be the zero locus of $xw-zy$.
There is a map $\pr1\times\pr1\to V$ called the Segre embedding
which we will eventually be able to see is an isomorphism,
defined by $([(a,b)],[(u,v)])\mapsto [(au,av,bu,bv)]$, so the parametric equations
for the image are $x=au$, $y=av$, $z=bu$ and $w=bv$, and substituting into
$xw-zy$ gives $aubv-buav=0$. Figure \ref{hyperboloid} shows a graph
of the map $\AAA^1\times\AAA^1\to\AAA^3$ given by
$(a,u)\mapsto (au,a,u)$. This is just the an affine image of the graph of $\pr1\times\pr1\to \pr3$;
i.e., the image in the affine complement $\pr3\setminus Z_{\pr3}(w)$ of the locus $w=0$ in $\pr3$.
The grid lines are the affine images of the two $\pr1$ factors, called rulings. The locus
$Z_V(y,w)$ is one of those lines, so the open subset $U=V\setminus Z_V(y,w)\subset V$
is the complement of the grid line defined by $y=0=w$.
Note that $U=U_w\cup U_y$.
Thus $\OO_V(U)=\cap_{p\in U}\OO_{V,p}=(\cap_{p\in U_w}\OO_{V,p})\cap(\cap_{p\in U_y}\OO_{V,p})$.
Now, in $\CC(V)$ we have $\frac{\overline{z}}{\overline {w}}=\frac{\overline {x}}{\overline{y}}$, since
$\overline{x}\overline{w}-\overline{z}\overline{y}=0$, where $\overline{\ }:\CC[\pr3]\to\CC[V]$
is the quotient homomorphism. But for $q=[(0,1,0,0)]\in U_y$, $\frac{\overline{z}}{\overline {w}}$
does not appear to meet the definition for being in $\OO_{V,q}$ since $\overline{w}(q)=0$.
It is only when we rewrite $\frac{\overline{z}}{\overline {w}}$ as $\frac{\overline{x}}{\overline{y}}$
that we see that it is an element of $\OO_{V,q}$.
And for $q=[(0,0,1,0)]\in U_w$, $\frac{\overline{x}}{\overline{y}}$
does not appear to meet the definition for being in $\OO_{V,q}$ since $\overline{y}(q)=0$.
It is only when we rewrite $\frac{\overline{x}}{\overline{y}}$ as $\frac{\overline{z}}{\overline{w}}$
that we see that it is an element of $\OO_{V,q}$.
Thus $\frac{\overline{x}}{\overline{y}}$, as written, is regular at the points of $U_y$
but not at all points of $U_w$, and $\frac{\overline{z}}{\overline{w}}$, as written, is
regular at the points of $U_w$ but not at all points of $U_y$. However,
$\frac{\overline{x}}{\overline{y}}=\frac{\overline{z}}{\overline{w}}$ on
$U_y\cap U_w$, so we have an element regular at all points of $U=U_y\cup U_w$.
It turns out that there is no non-constant homogeneous $G\in \CC[\pr3]$
such that $Z_{\pr3}(G)\cap U=\varnothing$, so we cannot express this element
by a single fraction $\frac{\overline{F}}{\overline{G}}$ such that $G$ is non-zero
at all points of $U$.
\end{example}
\ \vskip-1in
\setcounter{figure}{0}
\begin{figure}[h]
\vskip1in
\caption{Hyperboloid $xw-zy$ defined parametrically in $\AAA^3$ by $x=au$, $y=a$, $z=u$, $w=1$}\label{hyperboloid}
\vbox to0in{\includegraphics[width=6in, height=3in]{hyperboloid.png}\vss}
\vskip3in
\end{figure}
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex29.1}
Let $V\subset\pr n$ be a projective variety.
Let $0\neq G\in\CC[V]$ be homogeneous of degree $d=\deg(G)>0$.
Show that $\OO_V(U_G)$ is the ring
of all fractions $F/G^m$ such that $m\geq0$, $F\in\CC[V]$ is homogeneous,
and either $F$ is 0 or $\deg(F)=m\deg(G)$. (We will later see in Exercise \ref{ex33.1} that
$U_G$ is isomorphic to a closed subset of $\AAA^n$ for some $n$, and hence is
referred to as an {\em open affine subset} of $V$.)
\end{exer}
\begin{proof}[Solution]
Clearly, all $F/G^m\in \OO_V(U_G)$, so assume $h\in \OO_V(U_G)$.
For each $p\in U_G$ we can write $h=F_p/G_p$ where $F_p$ and $G_p$ are homogeneous
of the same degree but $G_p(p)\neq 0$. Since
$$Z_V(\{G_p:p\in U_G\})\cap U_G=\varnothing,$$
we have $Z_V(\{G_p:p\in U_G\})\subseteq Z_V(G)$.
By the Projective Nullstellensatz, Theorem \ref{ProjNullstellensatz},
for some $m\geq0$ we have
$G^m\in I(\{G_p:p\in U_G\})$, so as elements of $\CC[V]$ we have
$G^m=\sum_pH_pG_p$ for homogeneous $H_p\in\CC[V]$ of
degree $\deg(H_p)=m\deg(G)-\deg(G_p)$ for those $H_p$ which are not 0
(all but finitely many of the $H_p$ are in fact 0 since the sum is a finite sum).
Now we see that $h=((\sum_pH_pG_p)/G^m)h
=(\sum_pH_phG_p)/G^m
=(\sum_pH_p(F_p/G_p)G_p)/G^m
=(\sum_pH_pF_p)/G^m$ has the required form.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{March 30, 2011}\label{lect30}
\noindent{\bf Morphisms of quasi-projective varieties.}
Let $U$ be a quasi-projective variety. Thus there is some
projective variety $V\subseteq \pr n$ and $U$ is a nonempty open subset of $V$.
The structure sheaf $\OO_U$ is defined; for any open subset $U'\subseteq U$ we have
$\OO_U(U')=\OO_V(U')$. We can also speak of affine open subsets of $U$: for
$U'$ to be an open affine subset of $U$ just means it is an open affine subset of $V$.
\begin{defn}
Let $U_1$ and $U_2$ be quasi-affine projective varieties. A map $\phi:U_1\to U_2$
is a {\em morphism} if $\phi$ is continuous in the Zariski topology, such that for each point
$p\in U_1$ there are open affine neighborhoods $V_p\subseteq U_1$ and $W_p\subseteq U_2$
such that $\phi(p)\in W_p$, $p\in V_p$, $\phi(V_p)\subset W_p$ and $\phi^*$ induces
a $\CC$-homomorphism
$\phi^*:\OO_{U_2}(W_p)\to \OO_{U_1}(V_p)$.
\end{defn}
\begin{example}
Let $\CC[\pr n]=\CC[x_0,\ldots,x_n]$. Then $U_{x_i}$ is isomorphic to $\AAA^n$.
For simplicity, we'll consider the case that $i=0$, so let $U=U_{x_0}$. Define $\phi:U\to\AAA^n$
by $\phi([(a_0,\ldots,a_n)])=(\frac{a_1}{a_0},\ldots,\frac{a_n}{a_0})$; note that
$[(a_0,\ldots,a_n)]\in U_{x_0}$ means that $a_0\neq0$, so division by
$a_0$ is allowed. Since also
$(\frac{a_1}{a_0},\ldots,\frac{a_n}{a_0})=(\frac{ca_1}{ca_0},\ldots,\frac{ca_n}{ca_0})$
for any representative $(ca_0,\ldots,ca_n)$, $0\neq c\in\CC$, of
$[(a_0,\ldots,a_n)]$, $\phi$ is well-defined. Let $\psi:\AAA^n\to U$
be defined by $(b_1,\ldots,b_n)\mapsto [(1,b_1,\ldots,b_n)]$.
It is easy to check that $\phi$ and $\psi$ are inverses of each other.
By Proposition \ref{usualaffopeninpn}, $\OO_U(U)=\CC[\frac{x_1}{x_0},\ldots,\frac{x_n}{x_0}]$.
Let's use $y$ for the variables on $\AAA^n$, so $\OO_{\AAA^n}(\AAA^n)=\CC[y_1,\ldots,y_n]$.
Given any $f(y_1,\ldots,y_n)\in\OO_{\AAA^n}(\AAA^n)$, we have by composition
the map $\phi^*(f)=f\circ\phi$. For any $[(a_0,\ldots,a_n)]\in U$, note that
$(\phi^*(f))([(a_0,\ldots,a_n)])=f(\phi([(a_0,\ldots,a_n)])=f(\frac{a_1}{a_0},\ldots,\frac{a_n}{a_0})$
we see $\phi^*$ is just the $\CC$-homomorphism
$\CC[y_1,\ldots,y_n]\to \CC[\frac{x_1}{x_0},\ldots,\frac{x_n}{x_0}]$ given
by $y_j\mapsto \frac{x_j}{x_0}$ for each $j$. Similarly, it is easy to check that
$\psi^*$ is the inverse $\CC$-homomorphism.
This also shows that $\phi$ and $\psi$ are continuous. Consider $\phi$; it suffices
to show that $\phi^{-1}(U_f)$ is open for open subsets of the form $U_f$
where $f\in\CC[y_1,\ldots,y_n]$. But $\phi^{-1}(U_f)=U_{\phi^*(f)}$, which is open.
Thus $\phi$ and $\psi$ are inverse morphisms, hence $\phi$ is an isomorphism.
\end{example}
\begin{rem}
If we identify $U_{x_i}$ with $\AAA^n$, we thus have
$\OO_{\AAA^n}(U)=\OO_{U_{x_i}}(U)=\OO_{\pr n}(U)$
for any open subset $U\subseteq\AAA^n$.
From this point-of-view, we can regard
the coordinate ring of $\AAA^n=U_{x_i}$ as being
$\OO_{\AAA^n}(\AAA^n)=\OO_{U_{x_i}}(U_{x_i})=\OO_{\pr n}(U_{x_i})
=\CC[\frac{x_0}{x_i},\ldots,\frac{x_n}{x_i}]$.
\end{rem}
\begin{rem}
We now see that any quasi-affine variety is isomorphic to a quasi-projective variety.
If $U$ is quasi-affine, then $U$ is a non-empty open subset of an affine variety
$V$; i.e., $U\subseteq V\subseteq\AAA^n\cong U_{x_0}\subset\pr n$.
If we identify $U$ with its isomorphic image in $U_{x_0}\subset\pr n$, we can regard
the Zariski-closure of $U$ in $\pr n$ as a projective variety (see Exercise \ref{ex30.1}).
\end{rem}
\begin{example}
If $X$ is an affine variety, then we can recover $X$ up to isomorphism
from its coordinate ring $\CC[X]$. In particular, $\CC[X]$ is a finitely generated
$\CC$-algebra, so for some $n$ we have a surjective homomorphism $h:\CC[\AAA^n]\to\CC[X]$.
The kernel of $h$ is a prime ideal. Let $Y=Z(P)\subseteq\AAA^n$.
Then $X\cong Y$ since $\CC[X]\cong\CC[Y]$.
However, with projective varieties this is not true: it is quite common
to have projective varieties $X\cong Y$ such that the homogeneous
coordinate rings $\CC[X]$ and $\CC[Y]$ are not isomorphic.
For example, consider the map $\nu_2:\pr1\to\pr2$ given by
$\nu_2([(a,b)])=[(a^2,b^2,ab)]$, known as the 2-uple Veronese embedding of
$\pr1$ in $\pr2$. Then $\nu_2$ is an isomorphism of $X=\pr1$ to its image
$Y=Z_{\pr2}(xy-z^2)$, but $\CC[\pr1]=\CC[s,t]$, while $\CC[Y]=\CC[x,y,z]/(xy-z^2)$.
The map $\nu_2$ induces a homomorphism
$\nu_2^*:\CC[x,y,z]\to\CC[s,t]$ given by $x\mapsto s^2$, $y\mapsto b^2$ and
$z\mapsto st$. This is not surjective so the induced homomorphism
$\CC[x,y,z]/(xy-z^2)\to\CC[s,t]$ is certainly not an isomorphism.
In fact, there is no isomorphism. One way to see this is that
$\CC[x,y,z]/(xy-z^2)$ is not a regular ring, but $\CC[s,t]$ is. We'll discuss this in more detail later,
but here's how this difference manifests itself.
Let $M$ be the maximal ideal \
$(\overline{x},\overline{y},\overline{z})\subset \CC[\overline{x},\overline{y},\overline{z}]=\CC[x,y,z]/(xy-z^2)$.
Then $M/M^2$ is a 3-dimensional vector space, spanned by the images of $\overline{x},\overline{y}$, and
$\overline{z}$ in $M/M^2$. However, every maximal ideal of $\CC[s,t]$ is of the form $(s-a,t-b)$
for $a,b\in \CC$, but $(s-a,t-b)/(s-a,t-b)^2$ is a 2-dimensional vector space,
spanned by the images of $s-a$ and $t-b$. If there were an isomorphism
$\CC[x,y,z]/(xy-z^2)\to\CC[s,t]$, then $\CC[s,t]$ would have some maximal ideal
$I$ such that $I/I^2$ were 3-dimensional.
\end{example}
The 2-uple Veronese embedding is one of several classical morphisms it's worth knowing about.
More generally, for each $d$ and $n$ we have the $d$-uple Veronese embedding
$\nu_d:\pr n\to \pr N$, where $N=\binom{n+d}{d}-1$. There is also the Segre embedding
$\sigma:\pr n\times\pr m\to \pr N$, where $N=(n+1)(m+1)-1$. And there is the Pl\"ucker embedding
$\rho:{\rm Gr}(r,n)\to\pr N$ of the Grassmann variety ${\rm Gr}(r,n)$
of $r$-planes in $n$-space to $\pr N$, where $N=\binom{n}{r}-1$.
More on these in the next lecture.
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex30.1}
Let $U$ be a quasi-affine variety. I.e., $U$ is a non-empty open subset
of some closed irreducible subset $V\subseteq\AAA^n$.
Thinking of $\AAA^n=U_{x_0}\subset\pr n$ as an open subset
of $\pr n$, we refer to the Zariski closure of $U$ in $\pr n$ as its
{\em projective closure}.
Show that the projective closure of $U$
is irreducible, hence a projective variety.
\end{exer}
\begin{proof}[Solution]
Let $C=V\setminus U$.
First, $U$ is an irreducible topological space. For if $A$ and $B$
are closed subsets of $U$ with $U=A\cup B$ so $V=A\cup(B\cup C)$.
Since $V$ is irreducible, either $V=A$ (and hence $U=A$) or $V=B\cup C$
(and hence $U=B$), so $U$ is irreducible.
Let $W$ be the closure of $U$ in $\pr n$ and let $D$ and $E$ be closed subsets of $W$
such that $W=D\cup E$. Since $U$ is irreducible, either $U\subseteq D$ or $U\subseteq E$;
assume the former, the argument being the same if $U\subseteq E$.
If $U\subseteq D$, then the closure $W$ of $U$ is also contained in $D$,
hence $W$ is irreducible.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{April 1, 2011}\label{lect31}
\noindent{\bf The Segre embedding.}
Let $s$ and $t$ be positive integers.
The Segre embedding is a map of $\sigma:\pr s\times\pr t\to\pr N$, where $N=(s+1)(t+1)-1$.
Given a point $([{\bf a}],[{\bf b}])=[(a_0,\ldots,a_s)],(b_0,\ldots,b_t)]\in\pr s\times \pr t$, think of the points
of $\pr N$ as the set of non-zero $(s+1)\times(t+1)$ matrices, modulo scalar multiplication by
non-zero scalars. Then $\sigma(([{\bf a}],[{\bf b}]))=[(a_ib_j)]$, where $(a_ib_j)$ is the $(s+1)\times(t+1)$ matrix
whose entries are $a_ib_j$. Note that we can write this as a matrix multiplication: ${\bf a}^T{\bf b}=(a_ib_j)$,
where we think of ${\bf a}$ and ${\bf b}$ as row vectors. Moreover, the map is well-defined:
${\bf a}^T{\bf b}=(a_ib_j)$ is zero if and only if both ${\bf a}$ and ${\bf b}$ are zero,
and replacing ${\bf a}$ and ${\bf b}$ by possibly different representatives of the same points
just replaces $(a_ib_j)$ by a non-zero scalar multiple, which represents the same point of $\pr N$.
The product $\pr s\times\pr t$ does not have an intrinsic structure of algebraic variety according to our
definitions, but in fact $\sigma$ is injective and the image $\sigma(\pr s\times\pr t)$ is closed and irreducible, so
we can regard $\pr s\times\pr t$ as an algebraic variety by identifying it with its image.
First we check that $\sigma$ is injective. Let $(a_ib_j)$ be a representative of
$\sigma(([{\bf a}],[{\bf b}]))$. Pick any non-zero entry $a_ib_j$.
Let ${\bf c}=(a_0b_j,\ldots,a_sb_j)$ be the transpose of column $j$ of the matrix $(a_ib_j)$
and let ${\bf d}=(a_ib_0,\ldots,a_ib_t)$ be row $i$.
Then $[{\bf a}]=[{\bf c}]$, and $[{\bf b}]=[{\bf d}]$, and so we can recover
$([{\bf a}],[{\bf b}])$ given $[(a_ib_j)]$, hence $\sigma$ is injective.
To show that $\sigma(\pr s\times\pr t)$ is closed, note that every matrix representing
a point in the image of $\sigma$
has rank 1, since every row of ${\bf a}^T{\bf b}$ is a multiple of ${\bf b}$, and the rows are not all ${\bf 0}$.
Conversely, given any $(s+1)\times(t+1)$ rank 1 matrix $M$, it has some non-zero row, say ${\bf b_j}$,
and every other row is a multiple of ${\bf b_j}$; say row 1 is $a_0{\bf b_j}$, row 2 is $a_1{\bf b_j}$, etc.
Thus $M={\bf a}^T{\bf b}$ for ${\bf a}=(a_0,\ldots,a_s)$, so $[M]$ is in the image of $\sigma$.
In particular, the image of $\sigma$ is precisely the set of $(s+1)\times(t+1)$ rank 1 matrices.
Given an $(s+1)\times(t+1)$ rank 1 matrix $M$, every rectangular submatrix has rank at most 1.
In particular, every $2\times 2$ submatrix has determinant 0. If we denote the homogeneous coordinates
on $\pr N$ by $x_{ij}$, corresponding to the $ij$ entry of an $(s+1)\times(t+1)$ matrix, let $I$ be
the homogeneous ideal generated by all
polynomials of the form $x_{ij}x_{kl}-x_{il}x_{kj}$, with $i,j,k,l$ distinct.
Then we see that $Z_{\pr N}(I)$ contains the image of $\sigma$.
But any non-zero $(s+1)\times(t+1)$ matrix in $Z_{\pr N}(I)$ must have rank 1.
To see this, suppose $M$ is a non-zero $(s+1)\times(t+1)$ matrix which does not have rank 1.
Then two columns of $M=(m_{ij})$ are linearly independent. Say the columns are columns $j$ and $l$.
Similarly, two rows of these two columns must be linearly independent;
say the rows are rows $i$ and $k$. Thus the square matrix whose entries are in rows $i$ and $k$
and in columns $j$ and $l$ has rank 2, so its determinant $m_{ij}m_{kl}-m_{il}m_{kj}\neq0$,
and therefore $M$ is not in the zero locus of $I$. Thus the zero locus of $I$ is precisely
the image of $\sigma$.
Finally, we want to see that $\sigma(\pr s\times\pr t)$ is irreducible.
Let $\CC[\pr s]=\CC[y_0,\ldots,y_s]$ and let $\CC[\pr t]=\CC[z_0,\ldots,z_t]$.
Consider $U_{x_{ij}}\subset\pr N$. Then $\sigma^{-1}(U_{x_{ij}})=U_{y_i}\times U_{z_j}$.
Thus $\sigma$ gives an algebraic map (and hence a morphism) of affine varieties
$\AAA^{s+t}=\AAA^s\times\AAA^t=U_{y_i}\times U_{z_j}\to U_{x_{ij}}=\AAA^N$, since the
map is given by polynomial functions.
But the inclusion of $U_{x_{ij}}=\AAA^N$ into $\pr N$ is also a morphism,
hence $U_{y_i}\times U_{z_j}\to \pr N$ is a morphism of quasi-projective varieties
by Exercise \ref{ex31.1}.
Since $\AAA^{s+t}=U_{y_i}\times U_{z_j}$ is irreducible, so is the closure of its image
in $\pr N$ under $\sigma$ (by Exercise \ref{ex31.2}).
But the closure of any non-empty open subset of $\AAA^{s+t}$ is all of
$\AAA^{s+t}$, since $\AAA^{s+t}$ is irreducible.
Consider the open subset $\cap_{i,j}U_{y_i}\times U_{z_j}$,
which is just $U_{y_0\cdots y_sz_0\cdots z_t}$ inside
$\AAA^{s+t}$, where we take the $y$'s and $z$'s as the coordinate variables on
$\AAA^{s+t}$; i.e., $\CC[\AAA^{s+t}]=\CC[y_0,\cdots,y_s,z_0,\cdots, z_t]$.
Thus the closure of $U_{y_0\cdots y_sz_0\cdots z_t}$ includes the closure
of $\AAA^{s+t}=U_{y_i}\times U_{z_j}$ for all $i$ and $j$, hence
the closure of $\sigma(U_{y_0\cdots y_sz_0\cdots z_t})$ includes
$\sigma(U_{y_i}\times U_{z_j})$ for all $i$ and $j$. Thus the closure of
$\sigma(U_{y_0\cdots y_sz_0\cdots z_t})$ includes the union
$\cup_{i,j}\sigma(U_{y_i}\times U_{z_j})$. But $\sigma(U_{y_i}\times U_{z_j})=
\sigma(\pr s\times \pr t)\cap U_{x_{ij}}$, and the open sets $U_{x_{ij}}$ cover $\pr N$,
so $\cup_{i,j}\sigma(U_{y_i}\times U_{z_j})=\pr s\times \pr t$.
In particular, $\sigma(\pr s\times \pr t)$ is the closure of the image of
$\AAA^{s+t}=U_{y_i}\times U_{z_j}$ for any fixed $i$ and $j$.
Thus $\sigma(\pr s\times \pr t)$ is irreducible by Exercise \ref{ex31.2}.
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex31.1}
Let $U\to V$ and $V\to W$ be morphisms of quasi-projective varieties.
Show that the composition $U\to W$ is a morphism.
\end{exer}
\begin{proof}[Solution by Anisah Nu'man, with some changes]
Let $\phi: U \rightarrow V$ and $\psi: V\rightarrow W$ be morphisms of quasi-projective varieties.
We will show $(\psi \circ \phi): U \rightarrow W$ is a morphism. First note $(\psi \circ \phi)$ is
continuous in the Zariski topology since the composition of continuous functions is continuous.
Let $p\in U$. Since $\phi(p)\in V$ and $\psi: V\rightarrow W$ is a morphism,
there exist affine open subsets
$V_{\phi(p)} \subseteq V$ and $W_{\phi(p)} \subseteq W$ such that
$\psi(\phi(p))\in W_{\phi(p)}$, $\phi(p)\in V_{\phi(p)}$, $\psi(V_{\phi(p)}) \subseteq W_{\phi(p)}$,
and $\psi^{*}: \OO_W(W_{\phi(p)}) \rightarrow \OO_V(V_{\phi(p)})$ is a $\CC$-homomorphism.
Since $\phi$ is a morphism, there exist affine open subsets $U_p \subseteq U$ and
$V_p \subseteq V$ such that $\phi(p)\in V_p$, $p\in U_p$, $\phi(U_p) \subseteq V_p$, and
$\phi^{*}: \OO_V(V_p) \rightarrow \OO_U(U_p)$ is a $\CC$-homomorphism.
Now pick an open affine neighborhood $V'_p$ of $\phi(p)$ in $V_p\cap V_{\phi(p)}$
and pick an open affine neighborhood $U'_p$ of $p$ in $U_p\cap \phi^{-1}(V'_p)$.
Since $U_p$ and $V_p$ are affine (and thus have a basis of open sets
of the form $\{(U_p)_f:f\in\OO_U(U_p)\}$ and $\{(V_p)_g:g\in\OO_V(V_p)\}$, respectively)
we may assume that $V'_p=(V_p)_g$ for some $g\in\OO_V(V_p)$, and that
$U'_p=(U_p)_f$ for some $f\in\OO_U(U_p)$. The homomorphism
$\OO_V(V'_p)\to\OO_U(U'_p)$ is the $\CC$-homomorphism induced
from $\OO_V(V_p)\to\OO_U(U_p)$ by inverting $f$ and $g$.
Thus $(\psi \circ \phi)$ induces a $\CC$-homomorphism
$(\psi \circ \phi)^*:\OO_W(W_{\phi(p)})\to \OO_V(V_{\phi(p)})\to \OO_V(V'_p)\to \OO_U(U'_p)$,
as required.
\end{proof}
\begin{exer}\label{ex31.2}
Let $f:U\to V$ be a morphism of quasi-projective varieties.
Show that the Zariski-closure of $f(U)$ in $V$ is irreducible.
\end{exer}
\begin{proof}[Solution]
Say $\overline{f(U)}=A\cup B$ for closed subsets $A$ and $B$.
Thus $f^{-1}(A)$ and $f^{-1}(B)$ are closed subsets whose union is $U$,
hence (since $U$ is irreducible) either $U=f^{-1}(A)$ or $U=f^{-1}(B)$,
so either $\overline{f(U)}=A$ or $\overline{f(U)}=B$, hence
$\overline{f(U)}$ is closed.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{April 4, 2011}\label{lect32}
\noindent{\bf The Veronese embedding.}
We begin with a lemma.
\begin{lem}
The number of monomials in $\CC[\pr r]=\CC[x_0,\ldots,x_r]$ of degree $d$ is $\binom{r+d}{r}$.
\end{lem}
\begin{proof}
There is a bijection between permutations of $r$ bars and $d$ stars
and monomials $x_0^{n_0}\cdots x_r^{n_r}$ such that $d=n_0+\cdots+n_r$.
For example, if $r=3$ and $d=2$, the permutations ``$\ |\ |\ *\ *\ |\ $" gives
the monomial $x_0^0x_1^0x_2^2x_3^0=x_2^2$. But there are
$\binom{r+d}{r}$ permutations of $r$ bars and $d$ stars.
\end{proof}
We now define a map $\nu_d:\pr r\to\pr N$ where $N=\binom{r+d}{r}-1$.
Let $\CC[\pr r]=\CC[x_0,\ldots,x_r]$ and
enumerate the monomials of degree $d$ in the variables $x_0,\ldots,x_r$
as $G_0,\ldots, G_N$. For each point $p=[(a_0,\ldots,a_r)]\in\pr r$,
define $\nu_d(p)=[(G_0(a_0,\ldots,a_r),\ldots,G_N(a_0,\ldots,a_r)]\in\pr N$.
It is easy to check that this is well-defined.
Note that we can also regard $\nu_d$ as an algebraic map $\AAA^{r+1}\to \AAA^{N+1}$.
The corresponding $\CC$-homomorphism
$$\nu_d^*:\CC[\pr N]=\CC[\AAA^{N+1}]=\CC[y_0,\ldots,y_N]\to\CC[x_0,\ldots,x_r]
=\CC[\AAA^{r+1}]=\CC[\pr r]$$
is given by $y_i\mapsto G_i$.
\begin{rem}
A polynomial ring $R=\CC[x_0,\ldots,x_r]$ is a graded ring in the sense that
there is a canonical ring isomorphism $h:R_0\oplus R_1\oplus\cdots\to R$,
where $R_i$ is the $\CC$-vector space span of the homogeneous polynomials
of degree $i$. The homomorphism $h$ is induced by the inclusions $R_i\subset R$
by the universal property of direct sums. It is typical for people to just
identify $R_0\oplus R_1\oplus\cdots$ with $R$.
The subring $R_0\oplus R_d\oplus R_{2d}\oplus\cdots\subset R$ is called a
Veronese subring of $R$. It is exactly the image of $\nu_d^*$.
[Let $S$ be a ring. Let $I$ be some index set, and
let $A_i$, $i\in I$, be a family of $S$-modules. Recall that the direct product $\Pi_iA_i$
is the set of all maps $f:I\to\cup_iA_i$ such that $f(i)\in A_i$. For example,
$A_1\times A_2=\{(a_1,a_2):a_1\in A_1, a_2\in A_2\}$ is really
the set of all maps $f:\{1,2\}\to A_1\cup A_2$ such that $f(1)\in A_1$ and $f(2)\in A_2$.
Such a map $f$ can be thought of as the element $(f(1),f(2))$.
The direct sum $\oplus_i A_i$ is the subset
of the direct product $\Pi_iA_i$ of all maps $f\in\Pi_iA_i$ such that
$f(i)$ is 0 for all but finitely many $i$. Note for each $j$ that there is a natural inclusion
of $\phi_j:A_j\subseteq \oplus_iA_i$.
Now let $B$ be an $S$-module.
The {\em universal property of direct } is the fact that
if we are given $S$-module homomorphisms $h_i:A_i\to B$ for all $i$,
then there is a unique $S$-module homomorphism $h:\oplus_i A_i\to B$
such that $h\circ\phi_j=h_j$.]
\end{rem}
To show that $\nu_d:\pr r\to\pr N$ is a morphism, it's convenient to pick an enumeration
$G_0,\ldots,G_N$ such that $G_0=x_0^d$, $G_1=x_1^d$, $\ldots$, $G_r=x_r^d$,
with the rest of the monomials enumerated in whatever way the reader desires.
Let $0\leq i\leq r$ and let $p\in\pr r$. If $\nu_d(p)\in U_{y_i}$, then
$y_i(\nu_d(p))\neq0$, hence $G_i(p)=(\nu_d^*(y_i))(p)\neq0$, but since $G_i=x_i^d$,
we see $G_i(p)\neq0$ if and only if $x_i(p)\neq0$. Conversely,
if $x_i(p)\neq0$ then $G_i(p)=(nu_d^*(y_i))(p)\neq0$, so $\nu_d(p)\in U_{y_i}$.
Thus $\nu_d^{-1}(U_{y_i})=U_{G_i}=U_{x_i}$ and we see that
$\nu_d|_{U_{x_i}}:\AAA^r\cong U_{x_i}\to U_{y_i}\cong\AAA^N$
is an algebraic mapping whose corresponding $\CC$-homomorphism
$\nu_d|_{U_{x_i}}^*:\OO_{\pr N}(U_{y_i})\to\OO_{\pr r}(U_{x_i})$
is the homomorphism $\CC[y_0/y_i,\ldots,y_N/y_i]\to\CC[x_0/x_i,\ldots,x_r/x_i]$
given by $y_j/y_i\mapsto G_j/G_i=G_j/x_i^d=G_j(x_0/x_i,\ldots,x_r/x_i)$.
Since $\nu_d|_{U_{x_i}}$ is an algebraic mapping, it is continuous for each $i$.
Since the open sets $U_{x_i}$ give an open cover of $\pr r$, it follows that
$\nu_d:\pr r\to\pr N$ is continuous and since the induced homomorphisms
$(\nu_d|_{U_{x_i}})^*:\OO_{\pr N}(U_{y_i})\to\OO_{\pr r}(U_{x_i})$
are $\CC$-homomorphisms, $\nu_d$ is itself a morphism.
We next want to show that $\nu_d$ is an isomorphism to its image, which is closed
in $\pr N$.
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex32.1} Let $d$ be a positive integer and
let $G_0,\ldots,G_t\in\CC[\pr s]=\CC[x_0,\ldots,x_s]$ be homogeneous such that
for each $i$, $G_i$ is either 0 or has degree $i$. Also assume that $G_i$ is not zero for some $i$.
Let $U\subseteq\pr r$ be the open subset $U=\pr r\setminus Z_{\pr r}(G_0,\ldots,G_N)$.
Show that $f:[(a_0,\ldots,a_r)]\mapsto [G_0(a_0,\ldots,a_r),\ldots,G_N(a_0,\ldots,a_r)]$
defines a morphism $f:U\to \pr N$ as quasi-projective varieties and explain why
$f$ is not defined on $Z_{\pr r}(G_0,\ldots,G_N)$.
[The closed set $Z_{\pr r}(G_0,\ldots,G_N)$ is called the {\em indeterminacy locus} of $f$.
It is also called the {\em base locus} of $G_0,\ldots,G_N$.]
\end{exer}
\begin{proof}[Solution by Katie Morrison]
First note that $f$ is well-defined because
\[
\begin{split}
f([ca_0, \ldots, ca_r)]) &=
[c^d G_0(a_0,\ldots,a_r),\ldots,G_N(a_0,\ldots,a_r)] \\
&= [G_0(a_0,\ldots,a_r),\ldots,G_N(a_0,\ldots,a_r)] = f([a_0, \ldots, a_r)]),
\end{split}
\]
and so the value of $f$ is independent of the choice of representative at
which $f$ is evaluated. To see that $f$ is a morphism, consider the following.
Let $p \in U$ then since $p \notin Z_{\pr r}(G_0, \ldots, G_N)$, there exists a $G_i$
such that $G_i(p)\neq 0$. Without loss of generality, say $i=0$. Let $V_p=U_{G_0} \subseteq U$
be an affine neighborhood of $p$ and $W_p=U_{y_0}\subseteq \pr N=\CC[y_0, \ldots, y_N]$ be
an affine neighborhood of $f(p)$. By Exercise \ref{ex29.1}, $\OO_U(V_p)=
\{F/G_0^m~:~m\geq 0, F\in\CC[V] \text{ is homogeneous, and either } F=0 \text{ or }\deg(F)=
m\deg(G)\}$ and $\OO_{\pr N}(W_p) =\CC\left[\frac{y_1}{y_0}, \ldots, \frac{y_N}{y_0}\right]$.
For any $H\left(\frac{y_1}{y_0}, \ldots, \frac{y_N}{y_0}\right) \in \OO_{\pr N}(W_p)$,
$H\circ f = H\left(\frac{G_1}{G_0}, \ldots, \frac{G_N}{G_0}\right)$, and so $H\circ f$ will
have the form $F/G_0^m$ where $m$ is the degree of $H$. Thus, the image of $f^*$
is contained in $\OO_U(V_p)$ and it is clear that $f^*$ is a $\CC$-homomorphism into this image.
Thus $f$ is a morphism. Finally $f$ is not defined on $Z_{\pr r}(G_0, \ldots, G_N)$
because for any $a \in Z_{\pr r}(G_0, \ldots, G_N)$, $f(a)=[(0, \ldots, 0)]$ which is not
a valid point in projective space, thus $f$ cannot be defined on this set.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{April 6, 2011}\label{lect33}
\noindent{\bf The Veronese embedding (cont.).}
First we show that $\nu_d$ is injective.
Let $\overline{a}=(a_0,\ldots,a_r)$ and $\overline{b}=(b_0,\ldots, b_r)$
represent points in $\pr r$.
Suppose that $\nu_d([\overline{a}])=\nu_d([\overline{b}])$.
Since $\overline{a}\neq \overline{0}$, we know $a_i\neq0$ for some $i$.
But $[(a_0^d,\ldots,a_r^d,\ldots)]=\nu_d([\overline{a}])=\nu_d([\overline{b}])=[(b_0^d,\ldots,b_r^d,\ldots)]$,
so if $a_i\neq0$ then also $a_i^d\neq0$ and $b_i^d\neq0$ and hence $b_i\neq0$.
Thus we can divide by either $a_i$ or $b_i$ and from
$\nu_d([\overline{a}])=\nu_d([\overline{b}])$ we see
$G_j(\overline{a})/a_i^d=G_j(\overline{b})/b_i^d$ for all $j$.
But among the monomials $G_j$ we have $x_i^{d-1}x_k$ for each $k$, so
$$\frac{a_k}{a_i}=\frac{a_i^{d-1}a_k}{a_i^d}=\frac{b_i^{d-1}b_k}{b_i^d}=\frac{b_k}{b_i}.$$
Thus
$$\Big(1,\frac{a_1}{a_i},\ldots,\frac{a_r}{a_i}\Big)=\Big(1,\frac{b_1}{b_i},\ldots,\frac{b_r}{b_i}\Big)$$
hence $[\overline{a}]=[\overline{b}]$, so $\nu_d$ is injective.
Now in fact it is a fundamental result that a morphism of projective varieties is a closed map;
i.e., the image of a closed set is always closed. We will not prove this, at least not now.
Instead we'll show that the image of $\nu_d$ is closed.
Note that $(\nu_d|_{U_{x_i}})^*:\OO_{\pr N}(U_{y_i})\to\OO_{\pr r}(U_{x_i})$ is the homomorphism
$$\CC\Big[\frac{y_0}{y_i},\ldots,\frac{y_N}{y_i}\Big]\to\CC\Big[\frac{x_0}{x_i},\ldots,\frac{x_r}{x_i}\Big]$$
given by $\frac{y_j}{y_i}\mapsto\frac{G_j}{x_i^d}$.
This is surjective, since for each $k$ there is a $j$ such that $G_j=x_i^{d-1}x_k$,
and hence $\frac{x_k}{x_i}=G_j/x_i^{d-1}$ is in the image of $(\nu_d|_{U_{x_i}})^*$.
We can identify $U_{y_i}$ with $\AAA^N$ by regarding a point $[(c_0,\ldots,c_N)]\in U_{y_i}$
As being the point $(c_0/c_i,\ldots,c_{i-1}/c_i,c_{i+1}/c_i,\ldots,c_N)]\in \AAA^N$.
Let $I\subseteq \CC[\frac{y_0}{y_i},\ldots,\frac{y_N}{y_i}]$ be the kernel of $(\nu_d|_{U_{x_i}})^*$ so
$$\CC\Big[\frac{y_0}{y_i},\ldots,\frac{y_N}{y_i}\Big]/I\cong\CC\Big[\frac{x_0}{x_i},\ldots,\frac{x_r}{x_i}\Big].\eqno{(\dagger)}$$
Thus for any point $p\in U_{y_i}=\AAA^N$ in the zero locus of $I$, we have a maximal ideal
$M_q\subset \CC[\frac{x_0}{x_i},\ldots,\frac{x_r}{x_i}]$ corresponding under the isomorphism
$(\dagger)$ to $M_p/I$. Thus $((\nu_d|_{U_{x_i}})^*)^{-1}(M_q)=M_p$ and hence
$p=(\nu_d|_{U_{x_i}})(q)$; i.e., $\nu_d$ maps ${U_{y_i}})(U_{x_i})$ onto the closure of its image.
Since $\nu_d$ is injective, closed and continuous, it has a continuous
inverse, and since $\nu_d$ restricted to each open set $U_{x_i}\subset\pr r$
is an isomorphism to its image, the inverse map is a morphism,
so $\nu_d$ is an isomorphism.
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex33.1}
Let $V\subset\pr N$ be a projective variety.
Let $0\neq G\in\CC[V]$ be homogeneous of degree $d=\deg(G)>0$.
Show that $U_G$ is isomorphic to a closed subset of $\AAA^n$ for some $n$.
\end{exer}
\begin{proof}[Solution]
We give a sketch of the proof.
Let $W=\nu_d(\pr N)\subseteq \pr r$, where $r=\binom{N}{d}-1$. Thus $W$
is closed in $\pr r$ and hence so is $\nu_d(V)$. Note that $G$ is a linear combination
of monomials of degree $d$. Taking the same linear combination but with the corresponding
variables in $\pr r$ replacing the monomials we get a linear form $L\in\CC[\pr r]$ such that
$\nu_d^*(L)=G$ and hence such that $\nu_d^{-1}(\ZP{r}(L))=\ZP{N}(G)$.
Thus $\nu_d$ gives an isomorphism from $U_G$ to $W\cap U_L$,
while $U_L\subset\pr r$ is isomorphic to $\AAA^r$, so
$U_G$ is isomorphic to $W\cap U_L$ which is itself isomorphic to a closed subset of $\AAA^r$.
\end{proof}
\renewcommand{\thethm}{\arabic{section}.\arabic{thm}}
\setcounter{thm}{0}
\mysection{April 8, 2011}\label{lect34}
\noindent{\bf Grassmanians.}
Let $V$ be a $\CC$-vector space of dimension $n$ and let $r$ be an integer
with $0\leq r\leq n$. Let $\Gr(r,V)$, or $\Gr(r,n)$, denote the set of $r$-dimensional $\CC$-vector
subspaces of $V$. We call $\Gr(r,n)$ the {\em Grassmannian} of $r$-planes of $V$.
So far $\Gr(r,n)$ is just a set, but we will soon see how to give it topological and geometric
structure, so we think of the elements of $\Gr(r,n)$ as points.
\begin{example}
If $r=0$ or $r=n$, then $\Gr(r,n)$ consists of a single point.
Thus as a space it is natural to regard $\Gr(r,n)$ to be 0-dimensional
in these cases.
If $r=13$ we have
$t^i-1=(t^i-t^2)+(t^2-1)=t^2(t^{i-2}-1)+(t^2-1)\in(t^2-1,t^3-1)$.
Thus each term of $f$ in $(*)$ is in $(t^2-1,t^3-1)$ so
$f\in (t^2-1,t^3-1)$.
Thus $\lambda$ induces a surjective homomorphism
$(t^2-1,t^3-1)/(t^2-1,t^3-1)^2\to(t-1)/((t-1)^2)$.
Aside: This shows that $(t^2-1,t^3-1)=\Lambda^{-1}(\overline{t}-1)$.
We might be tempted at this point to claim that
therefore $(t^2-1,t^3-1)^2$ is the kernel of $\Lambda$
since $((\overline{t}-1)^2)=(\overline{0})$, and hence that
$\lambda$ is injective but
this is not sound reasoning; see Exercise \ref{ex40.1}.
So we still need to check that $\lambda$ is injective.
But since $\lambda:\CC[t^2,t^3]/(t^2-1,t^3-1)^2\to\CC[t]/((t-1)^2)$
is surjective and $\CC[t]/((t-1)^2)$ has $\CC$-vector space dimension
2 (with basis given by the images of 1 and $t-1$), we see
that $\CC[t^2,t^3]/(t^2-1,t^3-1)^2$ has $\CC$-vector space dimension
at least 2. On the other hand,
$(t^2-1,t^3-1)^2=((t^2-1)^2, (t^2-1)(t^3-1),(t^3-1)^2)=
(t^4-2t^2+1,t^5-t^3-t^2+1,t^6-2t^3+1)$. Moreover,
$(2+t^2)(t^4-2t^2+1)-(t^6-2t^3+1)=2t^3-3t^2+1\in(t^2-1,t^3-1)^2$.
Thus, modulo
$(t^2-1,t^3-1)^2$, every polynomial in $\CC[t^2,t^3]$
can be reduced to a polynomial of degree at most 2 and hence of the form
$a+bt^2$. Thus $\CC[t^2,t^3]/(t^2-1,t^3-1)^2$ has $\CC$-vector space dimension
at most 2, and hence exactly 2.
Thus $\lambda$, being surjective, is also injective as a vector space homomorphism, and thus
it is injective, as we wanted to show.
\end{example}
Here are a few additional facts about smoothness for a closed subset $X\subseteq\AAA^n$.
Let $p\in X$. Then we have the following facts.
\begin{itemize}
\item [(a)] The maximum $r$ of the dimensions of the irreducible components of $X$
containing $p$ is a lower bound on the dimension of the
Zariski tangent space of $X$ at $p$:
$$\dim_{\CC}I_X(p)/I_X(p)^2\geq r.$$
(See Atiyah-Macdonald, Corollary 11.15, p.\ 121 for a version of this.)
\item [(b)] If $X$ is smooth at $p$, then $\CC[X]_{I_X(p)}$ is a domain.
(See Theorem I.5.4A of Hartshorne's {\em Algebraic Geometry}.)
\item [(c)] If $p$ lies on more than one component of $X$, then $X$
is singular at $p$.
(See Exercise \ref{ex40.2}.)
\end{itemize}
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex40.1}
Let $f:A\to B$ be a homomorphism of rings.
Let $I\subseteq B$ be an ideal and
let $J=f^{-1}(I)$. Show by example
that it need not be true that $J^2=f^{-1}(I^2)$.
\end{exer}
\begin{proof}[Solution]
Let $f:\CC[x]\to \CC=\CC[x]/(x)$ be the quotient map.
Let $I=(0)$ so $J={\rm ker}(f)=(x)$.
Then $I^2=I=(0)$, so $f^{-1}(I^2)=J=(x)\neq(x^2)=J^2$.
\end{proof}
\begin{exer}\label{ex40.2}
Consider a closed subset $X\subseteq\AAA^n$.
If $p$ lies on more than one component of $X$, show that $X$
is singular at $p$. [Hint: show that $\CC[X]_{I_X(p)}$ is not a domain.]
\end{exer}
\begin{proof}[Solution]
Let $C_p$ be an irreducible component of $X$ containing $p$ and let $D_p$
be a different irreducible component of $X$ containing $p$.
Let $C$ be the union of $C_p$ with all irreducible components of $X$ except $D_p$
and let $D$ be the union of $D_p$ with all irreducible components of $X$ except $C_p$.
Thus $X=C\cup D$ (and hence $I(X)\subseteq I(C)\cap I(D)$),
where $p\in C\cap D$ and where $C$ and $D$ are
closed, proper subsets of $X$. Thus there are elements $f\in I(C)\setminus I(X)$ and
$g\in I(D)\setminus I(X)$ such that $fg\in I(X)$. Since $I(C)\subseteq I(p)$
and $I(D)\subseteq I(p)$, we see $f,g\in I(p)$. Modding out by $I(X)$ gives
$\overline{0}\neq\overline{f}\in I_X(C)$ and $\overline{0}\neq\overline{g}\in I_X(D)$
but $\overline{f}\overline{g}=\overline{0}$.
Let ${}^*:\CC[X]\to\CC[X]_{I_X(p)}$ be the localization homomorphism.
Then $\overline{f}^*\neq\overline{0}^*$, for if $\overline{f}^*=\overline{0}^*$, there would be
an element $s\in \CC[\AAA^n]$ such that
$\overline{s}\not\in I_X(p)$ (and hence $s\not\in I(p)$)
but $\overline{s}\overline{f}=\overline{0}$
in $\CC[X]$. Thus $sf$ vanishes on all of $X$. However, $f\in I(C)\setminus I(X)$
implies that $f$ does not vanish on $D_p$ so $s$ vanishes on $D_p$, which implies
$s(p)\neq0$ and therefore contradicts $s\not\in I(p)$.
Similarly, $\overline{g}^*\neq\overline{0}^*$, yet $\overline{f}^*\overline{g}^*=\overline{0}^*$,
so $\CC[X]_{I_X(p)}$ is not a domain and hence $X$ is not smooth at $p$.
\end{proof}
\begin{exer}\label{ex40.3}
Let $A$ be a ring, $M$ a maximal ideal of $A$ and $I$ an $M$-primary ideal.
Let $B$ be the localized ring $A_M$ and $J=IB$ the ideal generated by $I$
in $B$.
Show that the canonical homomorphism $F:A\to B$ (induced by $a\mapsto\frac{a}{1}$)
induces an isomorphism $f:A/I\to B/J$. [Note that therefore modding out by I
has the effect of inverting all elements not in $M$.]
\end{exer}
\begin{proof}[Solution]
Given $a\in A$, let $\overline{a}$ denote the image of $a$ in $A/I$, and likewise
Given $b\in B$, let $\overline{b}$ denote the image of $b$ in $B/J$.
Let $S=A\setminus M$. Then $\overline{s}$ is a unit in $A/I$ for every $s\in S$.
To see this, consider $I+(s)$. Since $I$ is $M$-primary, $\sqrt{I}=M$, so
$\sqrt{I+(s)}=\sqrt{I}+\sqrt{(s)}=M+\sqrt{s}$ contains both $M$ and $s$
hence equals $(1)$. Thus $1^n=1\in I+(s)$, so
$1=i+us$ for some $u\in A$ and $i\in I$ and hence $\overline{1}=\overline{u}\overline{s}$,
and therefore $\overline{s}$ is a unit as claimed.
Clearly $F(I)\subseteq J$, so $F$ induces a homomorphism $f:A/I\to B/J$.
Say $f(\overline{a})=\overline{0}$. Then $F(a)\in J$, so
there is an element $s\in A\setminus M$ and an element $c\in I$ such that
$\frac{a}{1}=F(a)=\frac{c}{s}$, hence there is a $t\in A\setminus M$
such that $t(sa-c)=0$ in $A$. Thus
$\overline{t}\overline{s}\overline{a}=\overline{t}\overline{c}=\overline{0}\in A/I$,
and since $\overline{t}\overline{s}$ is a unit in $A/I$, we see
$\overline{a}=\overline{0}$, so $f$ is injective.
Now consider any $\overline{b}\in B/J$. Then $b=\frac{a}{s}$ for some
$a\in A$ and $s\in S$. Let $\overline{u}$ be the inverse of $\overline{s}$.
Then $f(\overline{s}^{-1}\overline{a})=f(\overline{u}\overline{a})=
\overline{F(au)}=\overline{\frac{a}{1}}\overline{\frac{u}{1}}=
\overline{\frac{a}{s}}=b$, so $f$ is surjective and hence an isomorphism.
\end{proof}
\mysection{April 27, 2011}\label{lect41}
We can extend the definition of smoothness to closed subsets of projective space.
Let $p\in X\subseteq\pr N$, where $X$ is closed in $\pr N$.
Pick an $i$ such that $p\in U_{x_i}$ and regard $p\in X\cap U_{x_i}\subseteq U_{x_i}\cong\AAA^N$.
If $p$ is a smooth point of $X\cap U_{x_i}$ regarded as a closed subset of $\AAA^N$,
we say $p$ is a smooth point of $X$.
Consider $X=\ZP{N}(F)\subset\pr N$, where
$F\in\CC[\pr N]=\CC[x_0,\ldots,x_N]$ is non-constant and homogeneous.
Let $d=\deg(F)$, so $d>0$. Let $p\in\pr N$.
If $\frac{\partial F}{\partial x_i}(p)=0$ for all $i$, then
$F(p)=(\sum_ix_i\frac{\partial F}{\partial x_i}(p))/d=0$
so $p\in X$. By Exercise \ref{ex41.1}, $p$ is a singular point of $X$.
\begin{example}
Let $F=zy^2-x^3\in\CC[\pr2]$. Consider $X=\ZP{2}(F)\subset\pr2$.
Then $\nabla(F)=(-3x^2,2zy,y^2)$. Thus $\nabla(F)(p)=\overline{0}$
if and only if $x=y=0$, so the singular locus of $X$ consists of a single point,
$p=(0,0,1)$.
\end{example}
\noindent{\bf Divisors}: Let $X$ be a smooth, irreducible, quasi-projective variety.
A {\em prime divisor} is a closed irreducible subset of $X$ of codimension 1.
Define $\Div(X)$ to be the free abelian group on the prime divisors of $X$.
Each prime divisor $D\subset X$ determines a discrete valuation $\nu_D$.
Recall that a {\em discrete valuation} on a field $K$ is a
surjective map $\nu:K\to {\mathbb Z}\cup\{\infty\}$
such that the only element of $K$ mapping to $\infty$ is $0\in K$. The map $\nu$
must satisfy $\nu(ab)=\nu(a)+\nu(b)$ and $\nu(a+b)\geq\min(\nu(a),\nu(b))$.
The set $R$ of all $a\in K$ with $\nu(a)\geq0$ is a Noetherian ring, called the {\em valuation ring}
of $\nu$, and $K$ is the field of fractions of $R$. A {\em discrete valuation ring} (or DVR) is
a ring $R$ occurring in this way. (See Atiyah-Macdonald, p.\ 94, for background on DVRs.)
Before explaining where the discrete valuation associated to a prime divisor
comes from in general, we do an example.
\begin{example}
Let $X=\AAA^1=\CC$, so $\CC[X]=\CC[x]$. Then the prime divisors on $X$ are just the single points of $X$.
Thus an element of $\Div(X)$ is an expression of the form $\sum_{p\in X}m_pp$,
where $m_p\in{\mathbb Z}$ for all $p\in\AAA^1$, and such that $m_p=0$ for
all but finitely many $p$. Let $K=\CC(X)$ so $K=\CC(x)$ is the function field of $X$.
A point $p\in X$ is just an element of $\CC$.
Given $p\in X$, the associated discrete valuation $\nu_p$ is defined as follows.
For any $f\in\CC[X]=\CC[x]$, let $\nu_p(f)$ be the order of vanishing of $f$ at $p$;
i.e., the largest $m$ such that $(x-p)^m$ divides $f$. If $f$ is the 0 element,
we set $\nu_p(f)=\infty$. If $f,g\in \CC[x]$ with $g$ not the 0 element,
we define $\nu_p(f/g)=\nu_p(f)-\nu_p(g)$; this extends $\nu_p$ to all of $K$.
The valuation ring $R$ in this case is the localization $\CC[x]_{(x-p)}$ of $\CC[x]$ at the maximal
ideal $(x-p)$, and for any element $0\neq f\in R$we can regard
$\nu_p(f)$ as the largest $m$ such that $f\in M^m$, where $M$ is the unique maximal ideal
of $R$ (i.e., the ideal generated in $R$ by $x-p$).
The valuation $\nu_p$ specifies the order of zero (or pole) that an element
$f/g\in K$ has at $p$. If $\nu_p(f/g)=-m<0$, then we can say that
$f/g$ has a zero of order $-m$, but it's more common to say that $f/g$ has
a {\em pole} of order $m$ at $p$.
Since $\CC[x]$ is a UFD, it's easy to see that $\nu_p(fg)=\nu_p(f)+\nu_p(g)$
for $f,g\in\CC[x]$ and this extends to $\CC(x)$.
The fact that $\nu_p(f+g)\geq\min(\nu_p(f),\nu_p(g))$ comes from the
fact that you can get cancellation when you add, but when you drop terms
the order of zero at a point can go up. For example,
if $f=(x-p)^5-7(x-p)^2$ and $g=(x-p)^8+7(x-p)^2$, then
$\nu_p(f)=2$ and $\nu_p(g)=2$, but $\nu_p(f+g)=5$.
The valuations of $f$ and $g$ come from their terms of least degree
(when regarded as polynomials in $x-p$; i.e., as polynomials centered at $p$),
but adding $f$ to $g$ cancels these least degree terms, leading to $f+g$
having an increased valuation. When cancellation doesn't occur, such as
for $f=(x-p)^5-7(x-p)^2$ and $g=(x-p)^8+7(x-p)^3$, the term of $f$ of least degree
survives in $f+g$ to become the term of $f+g$ of least degree,
and the valuation of $f+g$ comes from this term of least degree,
which in this case gives $\nu_p(f+g)=\min(\nu_p(f),\nu_p(g))=\min(2,3)=2$.
\end{example}
\newpage
\vskip\baselineskip
\noindent{\huge\bf Exercises:}
\vskip\baselineskip
\begin{exer}\label{ex41.1}
Let $F\in\CC[\pr N]$ be non-constant and homogeneous.
Show that $p$ is a singular point of $\ZP{N}(F)$ if and only if
$(\nabla F)(p)=\overline{0}$.
\end{exer}
\begin{proof}[Solution]
Assume $(\nabla F)(p)=\overline{0}$; then as we saw above $p\in X$.
Say $F$ is the polynomial $F(x_0,\ldots,x_N)$, where $\CC[\pr N]=\CC[x_0,\ldots,x_N]$.
For some $i$ we have that $p\in U_{x_i}$.
Thus $p\in\ZP{N}(F)\cap U_{x_i}=Z_{\AAA^N}(F(x_0/x_i,\ldots,x_N/x_i))$,
and $p$ is singular if
$(\partial F/\partial (x_j/x_i))(p)=0$ for all $j\neq i$.
But we have $(\partial F/\partial (x_j/x_i))(p)=0$ if and only if
$(\partial F/\partial x_j)(p)=0$, hence
if $(\nabla F)(p)=\overline{0}$, then $p$ is a singular point of $X$.
Conversely, if $p$ is a singular point of $X$, then $p\in X$, so
$F(p)=0$. As before, $p\in U_{x_i}$ and hence
$(\partial F/\partial (x_j/x_i))(p)=0$ for all $j\neq i$,
so $(\partial F/\partial x_j)(p)=0$ for $j\neq i$. But
$F(p)=(\sum_lx_l\frac{\partial F}{\partial x_l}(p))/d=0$
while $x_i(p)\neq0$ and $\frac{\partial F}{\partial x_l}(p)=0$
for $l\neq i$, so $\frac{\partial F}{\partial x_i}(p)=0$, hence
$(\nabla F)(p)=\overline{0}$.
\end{proof}
\begin{exer}\label{ex41.2}
Recall that a valuation ring is a domain $R$ such that for each element
$r$ of the field of fractions $K$ of $R$ we have either $x\in R$ or $x^{-1}\in R$
(see Atiyah-Macdonald, p.\ 65).
Show that a DVR is a valuation ring.
\end{exer}
\begin{proof}[Solution]
See Atiyah-Macdonald for the proof.
\end{proof}
\mysection{April 29, 2011}\label{lect42}
Given a smooth quasi-projective variety $X$ and a prime divisor $D\subset X$,
let $U\subseteq X$ be any open affine subset of $X$ such that $U\cap D\neq\varnothing$.
Let $\OO_{X,D}$ be the localization of $\OO_X(U)$ at the prime ideal $I_U(D)$.
Then $\OO_{X,D}$ turns out to be a DVR; denote the corresponding valuation
on $\CC(X)$ by $\nu_D$. Given any $f\in\CC(X)\setminus \{0\}$,
there are only finitely many prime divisors $D$ such that $\nu_D(f)\neq0$
(see Hartshorne's {\em Algebraic Geometry}, Lemma II.6.1, p.\ 131). Thus
the sum ${\rm div}(f)=\sum_D\nu_D(f)D$ over all prime divisors $D$ on $X$ is itself a divisor.
A divisor coming in this way from a non-trivial rational function $f$ is called a {\em principal}
divisor.
Since ${\rm div}(fg)={\rm div}(f)+{\rm div}(g)$, the set of principal divisors is a subgroup
${\rm PrDiv}(X)$ of ${\rm Div}(X)$. The quotient group ${\rm Div}(X)/{\rm PrDiv}(X)$
is called the {\em divisor class group} of $X$, denoted ${\rm Cl}(X)$. If $D_1$ and $D_2$ are divisors
whose images $[D_i]$ in ${\rm Cl}(X)$ are the same (i.e., if
there is a rational function $f$ such that $D_1=D_2+{\rm div}(f)$), we say that
$D_1$ and $D_2$ are {\em linearly equivalent}.
\begin{example} Let $X=\AAA^N$. Then there is a bijection between
the irreducible polynomials in $\CC[\AAA^N]$ (modulo scalar multiples)
and prime divisors; given an irreducible $F\in\CC[\AAA^N]$,
the corresponding prime divisor is $Z(F)$. Choose such an $F$
for each prime divisor $D$ and denote it $F_D$.
Given any divisor $m_1D_1+\cdots+m_rD_r$, where each $D_i$ is a prime divisor,
$f=\Pi_iF_{D_i}^{m_i}\in\CC(X)$ is a rational function with
${\rm div}(f)=m_1D_1+\cdots+m_rD_r$. Thus
${\rm PrDiv}(X)={\rm Div}(X)$ and ${\rm Cl}(X)=0$.
\end{example}
\begin{example} Let $X=\pr N$. Then there is a bijection between
the irreducible homogeneous polynomials in $\CC[\pr N]$ (modulo scalar multiples)
and prime divisors; given an irreducible homogeneous $F\in\CC[\pr N]$,
the corresponding prime divisor is $\ZP{N}(F)$.
If we define the degree $\deg(D)$ of $D$ to be the degree of $F$,
we get a homomorphism $\deg:{\rm Div}(X)\to {\mathbb Z}$
whose kernel is ${\rm PrDiv}(X)$; i.e., ${\rm PrDiv}(X)$ is precisely the subgroup of
divisors of degree 0. (See Hartshorne's {\em Algebraic Geometry}, Proposition II.6.4, p.\ 132.)
\end{example}
Let $X$ be a smooth quasi-projective variety. Let $D=m_1D_1+\cdots+m_rD_r$
be a divisor where $D_1,\ldots,D_r$ are distinct prime divisors. We say $D$ is
{\em effective} if $m_i\geq 0$ for all $i$. For any divisor $D$, effective or not,
we write $|D|$ for the set of all effective divisors linearly equivalent to $D$,
and we write $L(D)$ for the set of rational functions $f$ such that either $f=0$
or $D+{\rm div}(f)$ is effective. Since for any prime divisor $P$ we have
$\nu_P(f+g)\geq\min(\nu_P(f),\nu_P(g))$, it follows that $L(D)$ is a $\CC$-vector space,
and we can regard $|D|$ as the associated projective space (i..e., as the 1-dimensional subspaces
of $L(D)$).
A fundamental fact is that if $X$ is a smooth projective variety, $L(D)$ is a finite dimensional
vector space (see Hartshorne's {\em Algebraic Geometry}, Theorem III.5.2, p.\ 228).
Studying the dimension of $L(D)$ for various $D$ is a major issue in algebraic geometry.
One of the most important theorems in algebraic geometry is the theorem of Riemann-Roch,
which bounds the dimension of $L(D)$ by numerical data related to $X$ and $D$.
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