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\begin{center}{\bf M953, Homework 5, due Friday, March 15, 2013}\end{center}
\noindent Instructions: Do any three problems.
\begin{itemize}
\item[(1)] Give and justify an example of a rational map $\phi: {\bf A}^1({\bf R})--\to {\bf A}^1({\bf R})$
which is not a morphism but for which $\operatorname{ind}(\phi)=\varnothing$.
\vskip\baselineskip
\item[(2)] Let $k$ be a field. Let $k[x]$ be a polynomial ring in an indeterminate $x$.
Show that every nonconstant element $f\in k(x)$ is transcendental over $k$
(i.e., show that there is no nonzero monic element $g\in k[t]$ having $f$ as a root).
\vskip\baselineskip
\item[(3)] Let $V\subseteq {\bf A}^m(k)$ and
$W\subseteq {\bf A}^n(k)$ be irreducible algebraic varieties over an algebraically
closed field $k$. Let $\phi=(\phi_1,\ldots,\phi_n): V--\to W$ be a rational map
defined by elements $a_i/b_i=\phi_i\in k(V)$,
for elements $a_i,b_i\in k[V]$ with $b_i\neq 0$.
Let $U$ be a nonempty
open subset of $V$ on which $\phi$ is defined (i.e., such that
none of the elements $b_i$ vanishes at any point of $U$).
If $\phi(U)$ is dense in the Zariski topology
on $W$, we say $\phi$ is \emph{dominant}.
\begin{itemize}
\item[(a)] If $\phi$ is dominant (i.e., if
$\phi(U)$ is dense in $W$ for some nonempty open subset $U\subseteq V$
disjoint from $V(b_1\cdots b_n)$), show that $\phi^*:k[W]\to k(V)$ is injective, and hence
induces a homomorphism $k(W)\to k(V)$.
\vskip\baselineskip
\item[(b)] Let $\phi$ and $U$ be as in (a).
If $\phi(U)$ is dense, show that $\phi(U')$ is dense for every
nonempty open subset $U'\subseteq V\setminus V(b_1\cdots b_n)$.
\vskip\baselineskip
\end{itemize}
\item[(4)] An irreducible algebraic variety $V\subseteq {\bf A}^n(k)$ such that $k(V)$
is a purely transcendental extension of $k$ is said to be \emph{rational}.
If $W$ is an irreducible algebraic variety for which
there is a dominant rational map $V--\to W$ with $V$ rational, we say $W$ is \emph{unirational}.
\begin{itemize}
\item[(a)] If $V$ is rational,
show that $V$ is birationally equivalent to ${\bf A}^d$, where
$d$ is the transcendence degree of $k(V)$ over $k$.
\vskip\baselineskip
\item[(b)] If $W$ is unirational such that $k(W)$ has transcendence degree $d$ over $k$,
show that there is a dominant rational map ${\bf A}^d\to W$. [Hint: Show there is a dominant
rational map ${\bf A}^n--\to W$ for some $n\geq d$. Then show there is a linear subspace
$L_d\subseteq {\bf A}^n$ of dimension $d$ such that the composition
$L_d\subseteq {\bf A}^n--\to W$ is defined and dominant.]
\vskip\baselineskip
\item[(c)] If $W$ is unirational such that $k(W)$ has transcendence degree 1 over $k$,
show that $W$ is rational. (You may look up and apply L\"uroth's Theorem without proof,
but state it if you do.)
\vskip\baselineskip
\end{itemize}
\item [(5)] Let $f,g\in k(V)$ for an irreducible
algebraic variety $V\subseteq {\bf A}^m(k)$.
Assume $f=\frac{a}{b}$ and $g=\frac{c}{d}$ for elements $a,b,c,d\in k[V]$.
If $f$ and $g$ are equal on a nonempty open subset $U\subseteq V\setminus V(bd)$,
then they are equal on $V\setminus V(bd)$ and they are equal as elements of $k(V)$.
\vskip\baselineskip
\item [(6)]
\begin{itemize}
\item[(a)] Let $D$ be a domain with field of fractions $K$. Let $f_1=\frac{a_1}{b_1},\ldots,f_r=\frac{a_r}{b_r}\in K$
where $a_1,\ldots,a_r\in D$ and $b_1,\ldots,b_r\in D\setminus \{0\}$ for all $i$.
If $f_i=f_j$ in $K$ for all $i$ and $j$ and if $(b_1,\dots,b_r)=(1)$ in $D$, prove that there is an element $c\in D$
such that $c=f_i$ in $K$ for all $i$.
\vskip\baselineskip
\item[(b)] Let $V\subseteq {\bf A}^m(k)$ be an irreducible algebraic variety with $k$ algebraically closed.
Assume we have rational maps $\phi_i: V--\to {\bf A}^1(k)$, $i=1,\ldots, r$,
where each $\phi_i$ is defined by $f_i\in k(V)$, where $f_i=a_i/b_i$ with $a_i,b_i\in k[V]$, and hence
$\operatorname{ind}(\phi_i)=V(b_i)$.
If $\cap_i \operatorname{ind}(\phi_i)=\varnothing$ and $\phi_i=\phi_j$ on
$(V\setminus\operatorname{ind}(\phi_i))\cap(V\setminus\operatorname{ind}(\phi_i))$
for all $i$ and $j$, show
there is a morphism $\phi:V\to {\bf A}^1(k)$ such that $\phi$ restricted to
$V\setminus\operatorname{ind}(\phi_i)$ is $\phi_i$
for each $i$.
\end{itemize}
\end{itemize}
\end{document}