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\begin{document}
\smallskip
\centerline{\sc Math 632, Lecture 11\\ February 2, 2004}
Recall the following Theorem from last lecture:
\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{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 $\Spec B_i\subseteq Y$.
Let us prove the more general result that if $X$ is a scheme and $\Spec R,\Spec R'$
are affine opens in $X$ with $x\in\Spec R\cap\Spec R'$ then there exists an open affine
$U\ni x$ in $\Spec R\cap \Spec R'$ that is a basic open set of $\Spec R$ and $\Spec R'$.
Since $\Spec R\cap \Spec R'$ is open in $\Spec R'$, we can find a basic open set $U'=\Spec R'_{r'}$
contained in $\Spec R\cap \Spec R'$ and containing $x$. Since any basic open of $U'$ is of the form
$\Spec R'_{r' s}$, we see that any basic open of $U'$ is also a basic open in $\Spec R'$. Thus we can replace
$\Spec R'$ by $\Spec R'_{r'}$ so that we must now prove:
Given $\Spec R'\subseteq \Spec R$ there exists a basic open $\Spec R_r$ of $\Spec R$ containing $x$ and lying inside
$\Spec R'$ that is also a basic open in $\Spec R'$. However, the containment $\Spec R'\subseteq \Spec R$
gives a map $\varphi: R\rightarrow R'$. Pick $r$ so that $\Spec R_r$ is contained in $\Spec R'$. Then we claim that
$\Spec R_r=\Spec R'_{\varphi(r)}$. But we have
$$\Spec R_r=\{x\in\Spec R : r(x)\neq 0 \ \text{in}\ k(x)\}=\{x\in\Spec R' : \varphi(r)(x)\neq 0 \ \text{in}\ k(x)\}=\Spec R'_{\varphi(r)}$$
simply by virtue of the containments $\Spec R_r\subseteq \Spec R'\subseteq \Spec R$.
Returning to the proof of the theorem, we have just shown that we can cover $\Spec B$
by basic opens $\Spec B_b=\Spec (B_i)_{b_i}$ that are also basic opens of $\Spec B_i$, so we have reduced
to the case $Y=\Spec B$. Now let $\Spec A\subseteq f^{-1}(\Spec B)$. We have assumed that
we have a covering $\Spec A_i$ of $f^{-1}(\Spec B)$ with each $A_i$ a finitely generated $(B_i)_{b_i}\simeq B_b$
algebra, and hence a finitely generated $B$-algebra. But by the above trick, we obtain a covering
of $\Spec A$ by basic open affines $\Spec A_{a_i} = \Spec (A_i)_{\alpha_i}$ such that each
$A_{a_i}$ is a finitely generated $B$-algebra (since each $(A_i)_{\alpha_i}=A_i[1/\alpha_i]$ is). We are thus reduced to
the following:
Let $B\rightarrow A$ is a map of rings with $a_1,\ldots,a_n$ in $A$ generating the unit ideal. If each $A_{a_i}$
is a finitely generated $B$-algebra, so is $A$. To see this, let $x_1,\ldots,x_n\in A$ be such that
$\sum x_i a_i = 1$ and let $A_{a_i}=B[z_{i,1}/a_i^N ,\ldots, z_{i,n_i}/a_i^N]$.
Then $A'=B[a_i,x_j, z_{k,l}]$ is a finitely generated $B$ sub-algebra of $A$ with the evident property
that $A_{a_i}=A'_{a_i}$. For any prime ideal $\p$ of $A$, there exists some $a_i$ with $a_i\not\in\p$
so that $A_{\p}$ is a further localization of $A_{a_i}$, from which it follows that $A_{\p}=A'_{\p}$
for all $\p\in\Spec A$. Hence $A=A'$, as required.
\end{proof}
\begin{theorem}
Let $f:X\rightarrow Y$ be a map of schemes. Then the following are equivalent:
\begin{enumerate}
\item For all quasi-compact open $U\subseteq Y$ we have $f^{-1}(U)$
quasi-compact.
\item There exists an open covering $\Spec B_i$ of $Y$ such that $f^{-1}(\Spec B_i)$
is covered by finitely many open affines $\Spec A_{ij}$.
\end{enumerate}
\end{theorem}
\begin{definition}
Any morphism satisfying these equivalent conditions will be called {\em quasi-compact}.
\end{definition}
\begin{proof}
Assuming (2), let $U\subset Y$ be quasi-compact and cover $U$ by finitely many open affines
$\Spec C_{\alpha}$. Then since $f^{-1}(U)$ is covered by the $f^{-1}(\Spec C_{\alpha})$,
it suffices to show that each $f^{-1}(\Spec C_{\alpha})$ is quasi-compact, that is, we reduce to the
case of affine $U=\Spec C$. Now we can cover $\Spec C$ by finitely many sets of the form
$\Spec (B_i)_{b_i}$ since $\Spec B_i$ cover $Y$. We must show that $f^{-1}(\Spec (B_i)_{b_i})$
is quasi-compact given that $f^{-1}(\Spec B_i)$ is. But we have
$$f^{-1}(\Spec (B_i)_{b_i})=\bigcup \Spec (A_{ij})_{b_i}$$ (where the union is finite),
where $f^{-1}(\Spec B_i)=\cup \Spec A_{ij}$. Thus $f^{-1}(\Spec (B_i)_{b_i})$
is quasi-compact.
\end{proof}
\begin{definition}
A morphism of schemes $f:X\rightarrow Y$ is {\em finite type} if it is quasi-compact and locally of finite type,
or equivalently, if every open affine $\Spec B\subseteq Y$ has $f^{-1}(\Spec B)$
covered by finitely many open affines $\Spec A_{ij}$ with each $A_{ij}$ a finitely generated $B_i$-algebra.
\end{definition}
\begin{corollary}
If $X\stackrel{f}{\rightarrow}Y\stackrel{g}{\rightarrow}Z$ with $f,g$ locally of finite type (resp. quasi-compact) then $g\circ f:X\rightarrow Z$
is locally of finite type (resp. quasi-compact).
\end{corollary}
\begin{proof}
We prove the assertion for locally of finite type as the proof for quasi-compact is similar.
Let $\Spec C\subseteq Z$ be open affine. Then $g^{-1}(\Spec C)$ is covered by $\Spec B_i$
with each $B_i$ a finitely generated $C$ algebra. Similarly, since $f$ is locally of finite type, each $f^{-1}(\Spec B_i)$
is covered by $\Spec A_{ij}$, with $A_{ij}$ a finitely generated $B_i$-algebra. Thus $f^{-1}g^{-1}(\Spec C)$
is covered by $\Spec A_{ij}$ with $A_{ij}$ a finitely generated $C$-algebra for all $i,j$.
\end{proof}
\begin{lemma}
Let $X$ be a scheme. Then the following are equivalent:
\begin{enumerate}
\item $\O_X(U)$ is reduced for all $U\subseteq X$.
\item There exists an open affine cover $\Spec A_i$ of $X$
with each $A_i$ reduced.
\item For all $x\in X$, the ring $\O_{X,x}$ is reduced.
\end{enumerate}
\end{lemma}
\begin{proof}
Clearly (1) implies (2). Since $\O_{X,x}$ is a localization of $\O_X(U)$
for any $U\ni x$, and since the localization of the nilradical of a ring
is the nilradical of the localization, we see that $\O_{X,x}$ is reduced if $\O_X(U)$
is. Finally, because of the injection
$$\O_X(U)\hookrightarrow \prod_{x\in U} \O_{X,x},$$
we see that if each $\O_{X,x}$ is reduced, so is their product, and hence $\O_X(U)$
is reduced as well.
\end{proof}
\end{document}