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\begin{document}
\smallskip
\centerline{\sc Math 632, Lecture 10\\ January 30, 2004}
\begin{theorem}
Let $X$ be an arbitrary scheme and $Y$ affine. Then there is a bijection
$$\Hom(X,Y)\longleftrightarrow \Hom(\O_Y(Y),\O_X(X)).$$
\end{theorem}
\newarrow{Corresponds}<--->
\begin{proof}
Set $Y=\Spec A$, let $\{U_i\}$ be an affine open cover of $X$ and for each $i,j$ let $\{U_{ijk}\}_{k\in K_{ij}}$
be an affine open cover of $X_i\cap X_j$. Then we the diagram
\begin{diagram}
\Hom(X,\Spec A) & \rTo & \prod \Hom(U_i,\Spec A) &\pile{\rTo^{p_1} \\ \rTo_{p_2}}& \prod_{(i,j)}\prod_{k\in K_{ij}} \Hom(U_{ijk},\Spec A)\\
\dTo & & \dCorresponds& & \dCorresponds\\
\Hom(A,\O_X(X)) & \rTo & \prod \Hom(A,\O_{U_i}(U_i)) &\pile{\rTo \\ \rTo}& \prod_{(i,j)}\prod_{k\in K_{ij}} \Hom( A,\O_{U_{ijk}}(U_{ijk}))\\
\end{diagram}
in which the two rows are left exact and the two right-hand columns are bijections. It follows that the first column
is also a bijection.
\end{proof}
\section{Examples}
\begin{enumerate}
\item Take $\A^n_{\Z}=\Spec \Z[T_1,\ldots,T_n]$. To give a map $X\rightarrow \A^n_{\Z}$
for any scheme $X$ is to give a ring map $\Z[T_1,\ldots,T_n]\rightarrow \O_X(X)$, which amounts
to picking an element of $\O_X(X)$ for the image of each $T_i$. Thus, we have the identifications
$$\Hom(X,\A^n_{\Z})\longleftrightarrow \Hom(\Z[T_1,\ldots,T_n],\O_X(X))\simeq \O_X(X)^{\oplus n}.$$
\item The natural map $A\rightarrow A_f$ for any ring $A$ and any $f\in A$
induces an open immersion $\Spec A_f\rightarrow \Spec A$ onto $X_f=\{x\in X : f_x\neq 0\ \text{in}\ k(x)\}$
(as any open in $X$ can be covered by sets of the form $X_{fg}$).
\item Let $k$ be a field and consider $\A^1_k= \Spec k[T]$. The point $(0)\in \A^1_k$ is open and dense
(since every prime ideal of $k[T]$ contains $0$), while all other points are closed and have the form
$(f)$ for $f\in k[T]$ a monic irreducible polynomial. The residue field at $(0)$ is $k(T)$
while the residue field at $f$ is $k[T]/(f)$, which is a finite field extension of $k$.
\item If $A$ has a unique minimal prime (for example, if $A$ is a domain) then $\Spec A$ has a unique point $z$,
open and dense in $\Spec A$.
\item Consider $\A^1_{\Z}=\Spec \Z[X]$. We have a natural mapping $\A^1_{\Z}\rightarrow \Spec \Z$.
The points of $\A^1_{\Z}$ are
\begin{itemize}
\item The unique open and dense point $(0)$. The residue field is $\Z[X]_{(0)}=\Q(X)$ and the
fiber over the point $(0)\in \Spec \Z$ is $\A^1_{\Q}$.
\item Prime ideals of the form $(f)$ with $f\in \Z[X]$ irreducible. Such points are open,
with closure the set of all prime ideals of the form $(p,f)$ where $p\in \Z$ is a prime
such that $f\bmod p$ is reducible. The residue field at $f$ is $\Frac(\Z[X]/(f))$,
and such $(f)$ lie in the fiber over $(0)$.
\item Maximal ideals of the form $(p,f)$ with $f$ irreducible modulo $p$. The residue field
at $(p,f)$ is $\F_p[X]/(f)$, and these points are in the fiber over $(p)$.
\end{itemize}
One pictures $\Spec \Z[X]$ as in Mumford's Red Book:
\item Consider $\A^2_k=\Spec k[x,y]$ with $k$ algebraically closed. The points are
\begin{itemize}
\item The unique open and dense point $(0)$. The residue field is $k(x,y)$.
\item Prime ideals of the form $(f)$ with $f\in k[x,y]$ irreducible. Such points are open,
with closure the set of all maximal ideals of the form $(x-a,y-b)$ with $f(a,b)=0$.
The residue field at $(f)$ is $\Frac(k[x,y]/(f))$, i.e. the function field of the irreducible
subvariety $f=0$.
\item Maximal ideals of the form $(x-a,y-b)$. The residue field is just $k$.
\end{itemize}
\end{enumerate}
\begin{definition}
Let $S$ be a scheme. Then an $S$-scheme $X$ is a scheme together with a map $X\rightarrow S$.
An $S$-map of $S$-schemes $X,Y$ is a commutative diagram
\begin{diagram}
X & \rTo & Y\\
& \rdTo & \dTo\\
& & S
\end{diagram}
\end{definition}
As an example, let $S=\Spec A$. By abuse of language, we will often refer to an $S$-scheme as an $A$-scheme.
An $A$-scheme $X$ is just a scheme together with a ring map $A\rightarrow \Gamma(X,\O_X)$, making $\O_X(U)$
an $A$-algebra for each open $U\subseteq X$. Thus, $\O_X$ becomes a sheaf of $A$-algebras, and maps
$f:X\rightarrow Y$ of $A$-schemes must have $f^{\#}$ a map of $A$-algebras.
In classical algebraic geometry, one studies $k$-schemes for an algebraically closed field
$k$. The base over which a scheme $X$ is considered can make a great difference in the properties and structure
of $X$. For example, consider $X=\Spec \C$. As a scheme over $\C$, $X$ has no nontrivial automorphisms.
Over $\R$, $\Aut(X)\simeq \Z/2\Z$, while over $\Z$, $\Aut(X)$ is uncountable. The moral is that algebraic
geometry must be developed with respect to an arbitrary base scheme.
\begin{theorem}
Let $f:X\rightarrow Y$ be a map of schemes. Then the following are equivalent:
\begin{enumerate}
\item For all open affines $\Spec B\subseteq Y$ and all open affines $\Spec A\subseteq f^{-1}(\Spec B)$
the ring $A$ is a finitely generated $B$-algebra.
\item There exists an open cover $\{\Spec B_i\}$ of $Y$ and an open cover $\{\Spec A_{ij}\}$
of $f^{-1}(\Spec B_i)$ for each $i$ such that each $A_{ij}$ is a finitely generated $B_i$ algebra
(for all $i$).
\end{enumerate}
\end{theorem}
\begin{definition}
Such a map $f$ is said to be {\em locally of finite type}, (with the ``locality'' referring to the source scheme).
\end{definition}
\begin{proof}
It suffices to prove that (2) implies (1). Since for $b_i\in B_i$ we have $f^{-1}(\Spec (B_i)_{b_i})$
covered by $\Spec (A_{ij})_{b_i}$, so the hypothesis is inherited by basic opens of $\Spec B_i$.
To reduce to the case $Y=\Spec B$ we must cover $\Spec B$ by basic opens $\Spec B_b$
that are also basic opens $\Spec (B_i)_{b_i}$ of $\Spec B_i\subseteq Y$. We will take this up next lecture.
\end{proof}
\end{document}