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\begin{center}{\bf M953, Homework 6, due Friday, March 29, 2013}\end{center}
\noindent Instructions: Do any three problems.
\begin{itemize}
\item[(1)] Let $k$ be a field and let $\varnothing\neq S\subseteq {\bf A}^m(k)$.
Recall that $\operatorname{affspan}(S)=V(\operatorname{Lin}(S))$, where
$\operatorname{Lin}(S)=\{a_1x_1+\cdots+a_mx_m+a_0\in k[x_1,\ldots,x_m]=
k[{\bf A}^m] : S\subseteq V(a_1x_1+\cdots+a_mx_m+a_0)\}$.
For $p\in\operatorname{affspan}(S)$, let $W_p(S)=\{v-p:v\in\operatorname{affspan}(S)\}$.
Show that $W_p(S)$ is a vector subspace of $k^m={\bf A}^m(k)$ and that this subspace
is independent of $p$ (i.e., $W_p(S)=W_q(S)$ for all $q\in \operatorname{affspan}(S)$).
\vskip\baselineskip
\item[(2)] Let $k$ be a field and let $\varnothing\neq S\subseteq {\bf A}^m(k)$.
We define $\dim \operatorname{affspan}(S)$ to be
the vector space dimension of $W_p(S)$ for any $p\in\operatorname{affspan}(S)$
(this is well-defined by Problem (1)). Let $d=\dim \operatorname{affspan}(S)$.
Show that there exist $d+1$ points $p_0,\ldots,p_d\in S$ such that
$\operatorname{affspan}(\{p_0,\ldots,p_d\})=\operatorname{affspan}(S)$.
\vskip\baselineskip
\item[(3)] Let $k$ be a field and let $\phi=(\phi_1,\ldots,\phi_m):{\bf A}^1(k)\to {\bf A}^m(k)$
be a morphism. If $\deg(\phi_i)\leq d$ for all $i$, show that
$\dim\operatorname{affspan}(\operatorname{Im}(\phi))\leq d$.
\vskip\baselineskip
\item[(4)] Let $k$ be a field. Let $S\subseteq {\bf A}^m(k)$ and let $V$ be the Zariski closure of $S$.
Show that $\operatorname{affspan}(S)=\operatorname{affspan}(V)$.
\vskip\baselineskip
\item[(5)] Let $\phi=(\phi_1,\ldots,\phi_m):{\bf A}^1(k)\to {\bf A}^m(k)$ be a morphism such that
$m>1$ and $\deg\phi_i\leq 2$ for all $i$, not all constant.
Let $C$ be the closure of $\phi({\bf A}^1(k))$.
Show that $\dim\operatorname{affspan}(C)\leq 2$.
Conclude that $\operatorname{Sec}_2(C)$ is contained in a 2-dimensional plane.
[Aside: This shows that $\dim\operatorname{Sec}_2(C)\leq 2$,
and thus that rational curves of degree at most 2 in ${\bf A}^m(k)$ with $m>2$
are always defective.
One can also show that any curve of degree at most 2 is rational, hence
curves of degree at most 2 are always defective.]
\vskip\baselineskip
\item[(6)] Let $L_1\subset {\bf A}^3(k)$ be the line $V(x,y+1)$ and
let $L_2\subset {\bf A}^3(k)$ be the line
$V(z,y-1)$, where $k[{\bf A}^3]=k[x,y,z]$.
Let $P_1$ be the plane $V(y+1)$ and let
$P_2$ be the plane $V(y-1)$. Let $V=L_1\cup L_2$.
Let $\sigma_2:V^2\times \Delta_2\to {\bf A}^3(k)$,
so $\operatorname{Sec}_2(V)$ is the closure of the image
of $\sigma_2$. Show that $\operatorname{Im}(\sigma_2)=
({\bf A}^3(k)\setminus(P_1\cup P_2))\cup (L_1\cup L_2)$.
[Hint: If $q\in {\bf A}^3(k)\setminus(P_1\cup P_2)$,
consider the intersections of the planes $Q_1$ and $Q_2$
where $Q_i$ contains $q$ and $L_i$.]
\vskip\baselineskip
\item[(7)] Let $k$ be a field of characteristic not equal to 2
and let $V$ be the image of $\nu:{\bf A}^2(k)\to{\bf A}^5(k)$
defined by $\nu((a,b))=(a^2,ab,b^2,a,b)$ (so the component functions of $\nu$ are the nontrivial
monomials of degree at most 2); $V$ is known as the Veronese variety.
\begin{itemize}
\item[(a)] Show that $V$ is closed and that $\nu$ is an isomorphism to its image.
\item[(b)] Show that the set of points $(p',q')\in V\times V$
such that $p'\neq q'$ and such that the line $\ell_{pq}\subset {\bf A}^2(k)$
through $p$ and $q$ is neither vertical nor horizontal and does not go through the origin, where
$\nu(p)=p'$ and $\nu(q)=q'$, is a nonempty open subset $U'\subseteq V\times V$.
\item[(c)] Recall that the secant variety $\operatorname{Sec}_2(V)$
is the closure of the image of $\sigma_2: V^2\times\Delta_2\to {\bf A}^5(k)$, and hence the closure of the image of
$f:{\bf A}^5(k)\to {\bf A}^5(k)$ defined as $f(a,b,c,d,e)=e(a^2,ab,b^2,a,b)+(1-e)(c^2,cd,d^2,c,d)$.
Show in fact that $\operatorname{Sec}_2(V)$ is contained in the closure of the image
of $g:{\bf A}^4(k)\to {\bf A}^5(k)$ defined by $g(a,b,c,d)=c\nu((a,0))+d\nu((0,b))+(1-c-d)\nu((a/2,b/2))$.
[Hint: for points $p\neq q\in {\bf A}^2(k)$, let $p'=\nu(p)$ and $q'=\nu(q)$.
Show that the secant line $L_{pq}\subset {\bf A}^5(k)$ through $p'$ and $q'$
lies in the affine span of the image $\nu(\ell_{pq})$ of the line $\ell_{pq}\subset {\bf A}^2(k)$
through $p$ and $q$. Now use Problem (5).]
(Aside: This implies that $\dim \operatorname{Sec}_2(V)\leq 4$ and hence that $\operatorname{Sec}_2(V)$
is defective. In fact, $\dim \operatorname{Sec}_2(V)= 4$, so
$\operatorname{Sec}_2(V)$ is defined by a single polynomial equation on ${\bf A}^5(k)$.)
\end{itemize}
\vskip\baselineskip
\item[(8)] Let $k$ be a field of characteristic not 2 or 3, and let $C$ be the image of $\tau:{\bf A}^1(k)\to{\bf A}^3(k)$,
defined as $\tau(t)=(t,t^2,t^3)$. For any $a\in k$, the line $L_p$ through $p=\tau(a)$
with direction vector $(1,2a,3a^2)$ is called the tangent line to $C$ at $p$.
Let $k[{\bf A}^1]=k[t]$ and $k[{\bf A}^3]=k[x,y,z]$. Let $f\in k[x,y,z]$ be a nonzero polynomial of degree 1.
\begin{itemize}
\item[(a)] Show that $C$ is closed and that $\tau$ is an isomorphism to its image.
(In fact, the image of ${\bf A}^1(k)$ under a morphism is always closed, but this is harder to show.)
\item[(b)] Show that $\tau^*(f)$ has at most 3 roots, counted with multiplicity.
\item[(c)] Let $\tau(a)=p\in C$. If $L_p\subset V(f)$, show that $t=a$ is a root of multiplicity at least 2
(i.e., show that $(t-a)^2$ divides $\tau^*(f)$). [Aside: if $t=a$ is a root of multiplicity 3, we say $V(f)$ is an osculating plane
for $C$ at $p$.]
\item[(d)] Show that if $q$ is on a tangent line but not on $C$,
then $q$ is on no secant line of $C$.
Conclude for any $q\neq p\in C$, if $q\in L_p$, then $q$ is not in the image of
$\sigma_2:C^2\times\Delta_2\to {\bf A}^3(k)$.
\end{itemize}
\end{itemize}
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