\nopagenumbers
\def\Cat#1{\hbox{${\cal #1}$}}
\def\Shf#1{\hbox{${\cal #1}$}}
\def\Hom{\hbox{Hom}}
\def\Top#1{\hbox{$\hbox{${\cal T\!\scriptstyle\cal OP}$}(#1)$}}
\def\Set{\hbox{${\cal S\!\scriptstyle\cal ET}$}}
\def\invlim{\hbox{\hbox{lim}\hskip-12pt\lower5pt\hbox{$\leftarrow$ }}}
\def\dirlim{\hbox{\hbox{lim}\hskip-12pt\lower5pt\hbox{$\rightarrow$ }}}
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\hbox to\hsize{\hfil \bf Homework 3: Math 953 Spring 2005\hfil}
\vskip\baselineskip
\hbox to\hsize{\hfil Due February 7, 2005\hfil}
(1) Let {\bf R} have the usual topology. Let \Shf P be the presheaf of
bounded locally constant continuous functions on {\bf R}. I.e., for each open
set $U\subset {\bf R}$, let $\Shf{P}(U)$ be the set of all continuous functions
$f:U\to {\bf R}$ such that for each $f$ and $U$ there is an $M$ with
$|f(x)-f(y)|\le M$ for all $x$ and $y$ in $U$, and such that for each $x\in U$,
$U$ contains a neighborhood $V$ of $x$ such that $f(y)=f(x)$ for all
$y\in V$.
\hskip.2in{(a)} Show that evaluation defines an isomorphism $\Shf{P}_p\to {\bf R}$
for each stalk $\Shf{P}_p$ of \Shf P.
\hskip.2in{(b)} Show that \Shf P is not a sheaf.
\hskip.2in{(c)} Let \Shf L be the presheaf of locally constant continuous functions;
this is defined the same way as \Shf P but without requiring boundedness.
Show that \Shf L is a sheaf, and
define a morphism $\Shf P\to \Shf L$ which induces an isomorphism on stalks.
Conclude that \Shf L is (isomorphic to) the sheafification of \Shf P.
(2) There are various versions of the Nullstellensatz.
Here are four. (If we impose the condition in (c) and (d) that
$k$ is algebraically closed, then all four are equivalent.)
Pick one and show that it implies one of the others.
\hskip.2in(a) (Weak Nullstellensatz) Let $R=k[x_1,\ldots,x_n]$ be the polynomial ring
in $n$ variables over an algebraically closed field $k$, and let $J\subset R$
be an ideal. If $Z(J)\subset k^n$ is empty, then $J=R$. (When $n=1$, this just says that
any nonconstant polynomial has a root in $k$.)
\hskip.2in(b) (Strong Nullstellensatz) Let $R=k[x_1,\ldots,x_n]$ be the polynomial ring
in $n$ variables over an algebraically closed field $k$, and let $J\subset R$
be an ideal. Then $I(Z(J))=\sqrt{J}$.
\hskip.2in(c) (Ring version) Let $R=k[x_1,\ldots,x_n]$ be the polynomial ring
in $n$ variables over a field $k$, and let $J\subset R$
be an ideal. Let I be the intersection of all maximal ideals that contain
$J$. Then $I=\sqrt{J}$.
\hskip.2in(d) (Field version) Let $R=k[x_1,\ldots,x_n]$ be the polynomial ring
in $n$ variables over a field $k$, and let $J\subset R$ be an ideal such that
$K=R/J$ is a field. Then $K$ is an algebraic extension of $k$.
(3) Let $f_1:A_1\to A_3$ and $f_2:A_2\to A_3$ be maps of sets.
Regarding $f_1$ and $f_2$ as giving an
inverse system, $\invlim A_i$ turns out to be
the fiber product, denoted
$A_1\times_{A_3} A_2$. In the category
of sets, it is just
$\{(a_1,a_2)\in A_1\times A_2: f_1(a_1)=f_2(a_2)\}$.
(See Lang, for example, for fiber products, pullbacks, and inverse limits.)
\hskip.2in(a) Let $A_i$ for $i=1,2$ both be the parabola
$y=x^2$ in the plane. Let $f_1$ be the projection to the
$x$-axis and let $f_2$ be the projection to the $y$-axis.
Determine $A_1\times_{A_3}A_2$ as a subset of the usual product
$A_1\times A_2$.
\hskip.2in(b) We can also regard $f_i$ as a direct system.
If so, what do we get for $\dirlim A_i$?
(4) Let $f_2:A_1\to A_2$ and $f_3:A_1\to A_3$ be homomorphisms of rings.
The direct limit $\dirlim A_i$ in this case is just
the tensor product $A_2\otimes_{A_1} A_3$. (See Lang,
for example, for tensor products, pushouts, and direct limits.)
\hskip.2in(a) Let $f_i:{\bf Z}\to {\bf Z}/3i{\bf Z}$, for $i=2,3$.
Show that $\dirlim A_i$ is isomorphic to ${\bf Z}/m{\bf Z}$
for some $m$ and determine $m$ (using either what you know about
tensor products, or showing ${\bf Z}/m{\bf Z}$
satisfies the universal property of direct limits.)
\hskip.2in(b) We can also regard $f_i$ here as an inverse system.
If so, what do we get for $\invlim A_i$?
\bye