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\begin{document}
\smallskip
\centerline{\sc Math 632, Lecture 14\\ February 9, 2004}
\section{Types of maps}
Let $f:X\rightarrow Y$ be a map of schemes.
\begin{enumerate}
\item $f$ is {\bf quasi-compact} if equivalently
\begin{enumerate}
\item There exists an open affine cover $U_i$ of $Y$ such that $f^{-1}(U_i)$ is quasi-compact.
\item For every open quasi-compact $U\subseteq Y$ we have $f^{-1}(U)$ quasi-compact.
\end{enumerate}
\item $f$ is {\bf locally of finite type} if equivalently
\begin{enumerate}
\item There exists an open affine cover $U_i$ of $Y$ and an open affine cover $V_{ij}$ of $f^{-1}(U_i)$
such that $\O_X(V_{ij})$ is a finitely generated $\O_Y(U_i)$-algebra.
\item For every open affine $U\subseteq Y$ and any open affine $V\subseteq f^{-1}(U)$, $\O_X(V)$
is a finitely generated $\O_Y(U)$-algebra.
\end{enumerate}
\item $f$ is {\bf of finite type} if it is locally of finite type and quasi-compact.
\item $f$ is {\bf affine} if equivalently
\begin{enumerate}
\item There exists an open affine cover $U_i$ of $Y$ such that $f^{-1}(U_i)$ is affine.
\item For every open affine $U\subseteq Y$, $f^{-1}(U)$ is affine.
\end{enumerate}
\item $f$ is {\bf finite} if
\begin{enumerate}
\item There exists an open affine cover $U_i$ of $Y$ with $f^{-1}(U_i)=V_i$ affine and $\O_X(V_i)$
is a finite $\O_Y(U_i)$-module.
\item For any open affine $U\subseteq Y$, $f^{-1}(U)=V$ is affine and $\O_X(V)$ is a finite $\O_Y(U)$-module.
\end{enumerate}
\end{enumerate}
The equivalence of each part of these definitions follows easily from ``Nike's Trick."
\begin{definition}
Let $X$ be a scheme. We say $X$ is {\em integral} if $\O_X(U)$ is a domain for every open $U\subseteq X$.
\end{definition}
\begin{definition}
Let $X$ be a scheme. We say $X$ is {\em irreducible} if the underlying topological space $|X|$ of $X$ is irreducible
({\em i.e.} can not be written as the union of two proper closed subsets).
\end{definition}
\begin{definition}
Let $X$ be a scheme. Then $X$ is {\em reduced} if $\O_X(U)$ is a reduced ring for all open $U\subseteq X$,
or, equivalently, if $\O_{X,x}$ is reduced for all points $x\in X$.
\end{definition}
\begin{theorem}
Let $X$ be a scheme. Then $X$ is integral if and only if it is irreducible and reduced.
\end{theorem}
\begin{proof}
Clearly $X$ integral implies that $X$ is reduced. Suppose that $X$ is reducible. Then we can write
$X=Z_1\cup Z_2$ with $Z_i$ proper and closed subsets of $X$. Let $U_i=Z_i-X_i$. Then
the $U_i$ are open and disjoint, so by the sheaf axiom we have $\O_X(U_1\cup U_2)=\O_X(U_1)\times\O_X(U_2)$.
Now as $U_1,U_2$ are nonempty the rings $\O_X(U_i)$ are nonzero (as they map to the local ring $\O_{X,x}\neq 0$
for any $x\in U_i$). It follows that since $\O_X(U_1)\times \O_X(U_2)$ is not a domain, we must have had $X$
irreducible.
Conversely, suppose that $X$ is reduced and irreducible. Suppose that $a_1,a_2\in\O_X(U)$ satisfy $a_1a_2=0$.
Put $Y_i=\{x\in U\ |\ a_i(x)=0\ \text{in}\ k(x) \}$. Evidently $Y_1\cup Y_2=U$. We claim that $Y_i$ is closed.
Indeed, the complement $U-Y_i=\{x\in U\ |\ a_i(x)\neq 0\ \text{in}\ k(x)\}$. But for any $x\in U-Y_i$
the image of $a_i$ in $\O_{U,x}$ is not in the maximal ideal and is hence a unit, so there exists an open $V_x$
containing $x$ with $a_i$ a unit in $\O_U(V_x)$, so that for every $y\in V_x$ we have $a_i(y)\neq 0\in k(y)$,
that is $V_x\subseteq U-Y_i$. Hence $U-Y_i$ is open. Now as $U$ is irreducible (since $X$ is) we have $Y_1=U,$ say.
It follows that $a_1(x)=0$ for all $x\in U$. Thus for any affine $V=\Spec R\subseteq U$ we have $a_1\big|_{V}=0$
since $a_1\big|_{V}$ is contained in every prime ideal of $R$ is then nilpotent, and $R$ is reduced. It follows
that $a_1=0$ and $\O_X(U)$ is reduced.
\end{proof}
\begin{definition}
A scheme $X$ is {\em locally Noetherian} if there exists an open affine cover $\{U_i\}$ of $X$
with each $\O_X(U_i)$ Noetherian.
\end{definition}
The equivalence of this definition with the corresponding ``string definition" (that is, for any open affine $U\subseteq X$
$\O_X(U)$ is Noetherian) follows from ``Nike's Trick" and
\begin{theorem}
If $A$ is a ring and $(f_1,\ldots f_n)A=A$ with $A_{f_i}$ Noetherian then $A$ is Noetherian.
\end{theorem}
\begin{proof}
Let $\a_1\subseteq \a_2\subseteq\ldots$
be an ascending chain of ideals in $A$. Then since each $A_{f_i}$ is Noetherian, we can omit some initial terms
of this sequence so that $\a_i A_{f_j}=\a_{i+1}A_{f_j}$ for all $i,j$. But as the $f_j$ generate the unit ideal, for any
prime ideal $\p$ of $A$ there exists some $f_j\not\in \p$, whence $A_{\p}$ is a localization of $A_{f_j}$
so that $\a_i A_{\p}=\a_{i+1}A_{\p}$ for all $i$. As this holds for all $\p$ we see that $\a_i=\a_{i+1}$
for all $i$ so that the chain stabilizes. Hence $A$ is Noetherian.
\end{proof}
\begin{definition}
A scheme $X$ is {\em Noetherian } if it is locally Noetherian and quasi-compact.
\end{definition}
\begin{example}
Let $R=\prod_{i=1}^{\infty} \F_2$. Then every element of $R$ is idempotent, so that
the localization of $R$ at any maximal ideal is $\F_2$ (since any local ring
in which every element is idempotent is a field) and is hence Noetherian.
But $\Spec R$ is decidedly not Noetherian.
\end{example}
\begin{definition}
A subspace $Y$ of $X$ is an {\em irreducible component} if it is a maximal closed and irreducible subset.
\end{definition}
A fact (which appears on the homework) is that and irreducible closed subset of a scheme $X$ lies
in some irreducible component. (This is equivalent to the fact that any prime of a ring contains a minimal prime,
which follows from Zorn's lemma). Moreover, we have a bijection
$$\{\text{Irreducible components of}\ X\}\leftrightarrow \{\xi\in X\ |\ \O_{X,\xi}\ \text{is 0-dimensional}\}$$
given by
$$\overline{\{\xi\}}\leftarrow \xi.$$
\end{document}