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\begin{document}
\centerline{\sc Bryden Cais. \\ Algebraic Geometry, HW 2.\\   \today}

\medskip\noindent
1.  Let $f:X\rightarrow Y$ be a morphism of schemes and $y\in Y$ a point.  Recall that $X_y$ is defined to be
the fibre product $X\times_{Y} \spec k(y)$, using the maps $f$ and the natural may $g:\spec k(y)\rightarrow Y$
given by $\p\mapsto y$ for all $\p\in\spec k(y)$.  Without loss of generality, we may assume that the maps
\begin{align*}
    &X\times_{Y}\spec k(y)\rightarrow \spec k(y),\\
    &X\times_{Y}\spec k(y)\rightarrow X
\end{align*}
are given by projection onto each component.  We therefore see that $X_y$ consists of all points
$(x,\p)\in X\times \spec k(y)$ such that $f(x)=g(\p)=y$, that is, the underlying topological space
is $f^{-1}(y)\subset X$.

\medskip\noindent
2. Let $f:\P^n\times \P^m\rightarrow \P^{nm+n+m}$ be given by
$$((X_0,\ldots X_n),(Y_0,\ldots,Y_m))\mapsto (X_0Y_0,X_1Y_0,\ldots,X_nY_0,X_0Y_1,\ldots,X_nY_m).$$
Observe first that $f$ is well defined since some $X_i\neq 0$ so that if $X_iY_j=0$ for all $j$ then
$Y_j=0$ for all $j$.  Moreover, it is not difficult to see that $f$ is injective.  Indeed, let $X_i,Y_j\neq 0$.
Then we set $Z_{kl}=X_kY_l$.  We thus have
$$(X_0,\ldots,X_n)=(Z_{0j}/Z_{ij},Z_{1j}/Z_{ij},\ldots,Z_{nj}/Z_{ij}).$$
Similarly, we can reconstruct the point $(Y_0,\ldots,Y_m)$ from the point $(Z_{00},\ldots Z_{nm})$
so that $f$ is indeed injective.  Next, observe that the image of $f$ lies in the closed subset $Z$ of
$\P^{nm+m+n}$ given by $(Z_{uv}Z_{st}-Z_{ut}Z_{sv})$ for $u,s=0,\ldots n$ and $t,v=0,\ldots ,m$.
We show that $f:\P^n\times\P^m\rightarrow Z$ is surjective.
Without loss of generality, let $(Z_{00},\ldots,Z_{nm})\in Z$ and suppose that $Z_{00}=1$.
Then we have $Z_{uv}=Z_{uv}Z_{00}=Z_{0v}Z_{u0}$ for all $u,v$.
It follows that
$$(Z_{00},\ldots,Z_{nm})=f((Z_{00},Z_{10},\ldots, Z_{n0}),(Z_{00},Z_{01},\ldots,Z_{0m}))$$
so that $f:\P^n\times\P^m\rightarrow Z$ is surjective.  It follows that $f$ gives a closed
embedding.  If $X,Y$ are projective varieties, then $X,Y$ are closed subsets of $\P^n,\P^m$
for some $n,m$.  Thus, $f(X\times Y)$ is an isomorphism or $X\times Y$ with a closed subset of
$\P^{nm+n+m}$ so that the product of two projective varieties is again projective.


\medskip\noindent
3.  Let $f:X\rightarrow Y$ be a finite morphism and let $U=\spec A$ be any open subset of $Y$ such that $f^{-1}(U)=\spec B$
with $B$ an integral extension of $\phi(A)$, where $\phi:A\rightarrow B$ is the ring homomorphism
corresponding to $f:\spec B\rightarrow\spec A$ (that is, a finitely generated $A$ algebra).  We first show that $f$ is a
closed map.  Indeed, let $V(I)$ be a closed set in $\spec B$.  Then we claim that $f(V(I))=V(\phi^{-1}(I))$.
Indeed, let $\p\in V(\phi^{-1}(I))$.  Then $\phi^{-1}(I)\subseteq \p$ so that $\bar{\p}=\p\bmod \phi^{-1}(I)$ is a prime ideal of $A/\phi^{-1}(I)$.
Now since $B$ is integral over $\phi(A)$, we see that $B/I$ is integral over $\phi(A/\phi^{-1}(A))$.  Then by the Going Up Theorem,
there exists a prime ideal $\bar{\q}$ of $B/I$ such that $\bar{\p}=\phi^{-1}(\bar{\q})$.  Using the correspondence between prime ideals of $B/I$
and prime ideals of $B$ containing $I$, we have a prime ideal $\q$ of $B$ with $\q\bmod I=\bar{\q}$ and $\phi^{-1}(\q)=\p$.
Hence, $\q\in V(I)$ and $f(\q)=\p$, as desired.

We now claim that the fibres of $f$ are finite sets.  Without loss of generality, we may suppose that $A\subset B$ (since
we can just replace $A$ by its image under $\phi$).  We claim that for every ideal $I\subset B$ we have
$\dim I\cap A=\dim I$.  Indeed, by the Going Up Theorem (since $B$ is integral over $A$), any sequence
$$I\cap A\subseteq\p_1\subseteq\p_2\subseteq\ldots\subseteq \p_n$$ of prime ideals in $A$ gives rise to a sequence
$$I\subseteq \q_1\subseteq\q_2\subseteq\ldots\subseteq \q_n$$ of prime ideals of $B$ with $A\cap \q_i=\p_i$ for each $i$.
This gives $\dim I\geq \dim I\cap A$.  The Going Down Theorem gives the reverse inequality, so we conclude that
$\dim I=\dim I\cap A$ for any ideal $I$ of $B$.  In particular, it follows that $\dim\spec A=\dim\spec B$.
Now let $\p$ be any prime ideal of $A$ and set $I=\p B$. Then since $I\cap A=\p$, we have $\dim \p B=\dim\p$.  Since
$A$ is noetherian, $\dim A <\infty $ from which it follows that there are only finitely many prime ideals of $B$ lying
over $\p\subset A$.  In other words, the fibres of $f$ are finite.

\medskip\noindent
4.  Let $Y=(xy+zt=0)\subset \A^4_{x,y,z,t}$ and let $X$ be the union of $U_1,U_2$ with
\begin{align*}
    U_1&=(y+z't=0)\subset \A^4_{u,y,z',t}\\
    U_2&=(x'y+t=0)\subset \A^4_{x',y,v,t}
\end{align*}
and let $f:X\rightarrow Y$ be given by
\begin{align*}
    (x,y,z',t)&\mapsto (u,y,uz',t)=(x,y,z,t)\\
    (x',y,v,t)&\mapsto (vx',y,v,t)=(x,y,z,t).
\end{align*}
Let $W=(y=z=0)\subset Y$.  It is not difficult to see that $\cdim(W,Y)=1$ (since $W\subset Y$, clearly, and $\cdim(W,\A^4)=2$ since
it is given by two non-redundant equations, while $\cdim(Y,\A^4)=1$ since it is given by 1 equation).  Moreover, we have
$$f^{-1}(W)=(uz'=y=v=0)\subset X.$$
We therefore let $Z$ be the component of $f^{-1}(W)$ given by $(z'=y=v=0)\subset X$.  We claim that $\cdim(Z,X)=2$.
Indeed, $Z\subset X$ is is given by the 2 non-redundant equations
$z'=v=0$ (which force $y=t=0$ when combined with the defining equations of $X$).  Thus, $2=\cdim(Z,X) > \cdim(W,Y)=1$.
Observe, of course, that $\overline{f(Z)}$ is just the line $y=z=t=0$, which is a proper subset of $W=(y=z=0)$.

\medskip\noindent
5.  Let $X,Y$ be varieties and $f_1,f_2:X\rightarrow Y$ two morphisms which agree on an open dense subset $U\subset X$.
We define $g:X\rightarrow Y\times Y$ by $g(x)=(f_1,f_2)(x)$.  Let $\Delta:Y\rightarrow Y\times Y$ be the usual
diagonal embedding; since $Y$ is a variety, it is separated so that $D=\Delta(Y)\subset Y\times Y$ is closed.
Certainly $Y=\Delta^{-1}(D)$.  Since $f_1,f_2$ agree on $U$, let $f=f_1|_U=f_2|_U$ and define
$\tilde{g}:U\rightarrow Y\times Y$ by $\tilde{g}=(f,f)=g|_U=\Delta\circ f$.
We then have
$$g^{-1}(D)\supseteq \tilde{g}^{-1}(D)=f^{-1}\circ\Delta^{-1}(D)=f^{-1}(Y)=U.$$
Since $D$ is closed and $g$ is a morphism, we see that $g^{-1}(D)\subseteq X$ is closed and
contains the open dense subset $U$; it follows that $g^{-1}(D)=X$, and hence that $g(X)\subseteq D$,
that is, $f_1,f_2$ agree on all of $X$.

\medskip\noindent
6. Let $X$ be a variety and suppose that $U,V\subset X$ are open affine subsets.  Since $X$ is a variety, it is separated,
and hence the image of $\Delta(X)$ in $X\times X$ is closed.  It follows that $\Delta(X)\cap (U\times V)$
is a closed subset of $U\times V$.  But $U\times V$ is affine;  Since
$\Delta(X)\cap (U\times V)\simeq U\cap V$ is a closed subset of an affine set, it is affine.

\medskip\noindent
7. Let $X$ be a proper variety, and let $f\in \O_X(X)$.  Consider the closed subset of $Z=(x,y)$ of $X\times \A^1$ given by $f(x)y=1$.
Since $X$ is proper, the projection $pr_2$ of $Z$ to $\A^1$ is closed.  Observe first that $pr_2(Z)$ is a proper subset of $\A^1$
since it does not contain the point $y=0$.  Since the only proper closed subsets of $\A^1$ are finite sets, it follows that
$pr_2(Z)$ is {\em finite}, so that $f$ takes only finitely many nonzero values on $X$.  It follows that $f$ must take on some value,
say $\alpha\in K$ on an open dense subset of $X$.  It then follows from Problem 5 that $f=\alpha$ identically on $X$, i.e.
that $\O_X(X)=k$.

\medskip\noindent
8. Let $X$ be a variety and $U\subset X$ open and proper.  Observe that the diagonal $D$ of $U\times X$ is a closed subset
of $U\times X$ (indeed, it consists of all points $(u,x)\in U\times X$ satisfying $f(u,x)=u-x=0$).  Since $X$ is proper, the
projection $pr_2:U\times X\rightarrow X$ is a closed map; it follows that $pr_2(D)=U$ is a closed subset of $X$.  Since $U$
is both open and closed, we must have $U=X$.


\end{document}