\documentclass[10pt]{amsart} \usepackage[leqno]{amsmath} \usepackage{amssymb,mystyle,amscd,diagrams} \topmargin=-0.2in \oddsidemargin=-0.2in \evensidemargin=-0.2in \textwidth=6.9in \textheight=8.7in \DeclareMathOperator{\Mor}{Mor} \DeclareMathOperator{\Rad}{Rad} \DeclareMathOperator{\Frac}{Frac} \DeclareMathOperator{\supp}{Supp} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\End}{End} \DeclareMathOperator{\Ann}{Ann} \DeclareMathOperator{\id}{id} \DeclareMathOperator{\height}{ht} \DeclareMathOperator{\Char}{char} \renewcommand{\Im}{{\rm Im}} \newcommand{\B}{\mathscr{B}} \newcommand{\calC}{\mathscr{C}} \newcommand{\calV}{\mathscr{V}} \renewcommand{\a}{\mathfrak{a}} \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}