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\begin{document}
\smallskip
\centerline{\sc Math 632, Lecture 8\\ January 26, 2004}

\section{Closed immersions}

\begin{definition}
	A map of ringed spaces $(\iota,\iota^{\#}):(Y,\O_Y)\rightarrow (X,\O_X)$ is a {\em closed immersion} if
	\begin{enumerate}
		\item $\iota:Y\rightarrow X$ is a closed embedding, i.e. a homeomorphism onto a closed subset of $X$.
		\item $\iota^{\#}:\O_X\rightarrow \iota_* \O_Y$ is surjective.
	\end{enumerate}
\end{definition}

Condition (2) roughly says that functions on $Y$ locally lift to $X$.  

\begin{remark}
	If $\iota:Y\hookrightarrow X$ is an embedding and $\calF$ is a sheaf on $Y$ then
	$(\iota_*\calF)_x=\calF_x$ if $x\in \iota(Y)$.  If $x\not\in\iota(Y)$, the situation is more complicated,
	but if $\iota(Y)$ is closed then $U=X-\iota(Y)$ is open and $U\cap Y=\emptyset,$ so that $(\iota_*\calF)_x=0$
	when $x\not\in\iota(Y)$.
\end{remark}

As an example, suppose that $X,Y$ are $C^{\infty}$ manifolds.  Then $\iota:Y\rightarrow X$ is a closed immersion
if and only if $\iota(Y)$ is a closed submanifold.  This follows from the Immersion Theorem.  Indeed the difficult
direction is to show that surjectivity of $\O_{X,\iota(y)}\rightarrow \iota_*\O_{Y,\iota(y)}\rightarrow \O_{Y,y}$
implies that $Y$ is locally described as the zero-locus of some local functions.  That is, we need to show that
$T_y(Y)\stackrel{d\iota(y)}{\longrightarrow} T_{\iota(y)}(X)$ is injective, or dually, that
$T_y(Y)^*\longleftarrow T_{\iota(y)}(X)^*$ is surjective.  But we have the identification
$\O_{X,x}/\m_x^2\simeq \R\oplus \m_x/\m_x^2$ and $T_x(X)^*\simeq \m_x/\m_x^2$, so that surjectivity of
the map $\O_{X,\iota(y)}\rightarrow \O_{Y,y}$ implies the surjectivity of $T_y(Y)^*\longleftarrow T_{\iota(y)}(X)^*$.

\begin{remark}
	Given a closed immersion $\iota:Y\rightarrow X$ we let $\calI_Y=\ker(\O_X\rightarrow \iota_*\O_Y)$.
	Then $\calI_Y$ is a sheaf of ideals in $\O_X$ and $\iota_*\O_Y\simeq \O_X/\calI$, so pulling back we
	obtain $\O_Y\simeq \iota^{-1}(\O_X/\calI_Y)$.  We conclude that closed immersions are ``classified"
	by sheaves of ideals.  One should take this interpretation with a grain of salt, however, as many ideal sheaves
	will correspond to ``crazy" spaces $Y$ that one would not want to work with.
\end{remark}

\begin{definition}
	Let $X$ be a topological space and $\B$ a base for the topology.  A $\B${\em-presheaf}
	is an assignement $U\mapsto \calF(U)$ for all $U\in \B$ with restriction maps $\rho_{U,V}:\calF(U)\rightarrow \calF(V)$
	whenever $V\subseteq U$ for $U,V\in\B$.  A $\B$-presheaf is a $\B$-sheaf if
	\begin{enumerate}
		\item it is separated: for all $U\in \B$ and $\{U_i\}$ an open cover of $U$ by $U_i\in\B$ the map
		$$\calF(U)\longrightarrow \prod_i\calF(U_i)$$ is injective.
		
		\item glueing works: if $U\in\B$ and $\{U_i\}$ is a cover of $U$ by $U_i\in\B$ then for any covers
		$\{U_{ijk}\}_{k\in K_{ij}}$ of $U_i\cap U_j$ with $U_{ijk}\in \B$, the sequence
		\begin{diagram}
		\calF(U) & \rTo & \prod_i \calF(U_i) & \pile{\rTo\\ \rTo} & \prod_{\substack{(i,j)\\ k\in K_{ij}}} \calF(U_{ijk})
		\end{diagram}
		is exact.
	\end{enumerate}
\end{definition}

Observe that condition (2) says that for a collection of $s_i\in \calF(U_i)$ with $s_i\big|_{U_{ijk}}=s_j\big|_{U_{ijk}}$
for all $k$ there exists $s\in \calF(U)$ with $s\big|_{U_i}=s_i$.  This $s$ is unique by condition (1).
We remark that any sheaf on $X$ restricts to a $\B$-sheaf and that if $f:X'\rightarrow X$
is continuous, and $\B',\B$ are bases on $X',X$ respectively, such that $f^{-1}(U)\in \B'$
fir all $U\in\B$, then the pushforward under $f$ of any $\B'$-sheaf is a $\B$-sheaf.

\begin{lemma}
	If $X$ and $\B$ are as above and $\calF$ is a $\B$-sheaf on $X$, then there exists a sheaf $\calF^+$
	on $X$ together with an isomorphism $\calF^+\big|_{\B}\simeq \calF$ which is universal in the sense
	that given any sheaf $\calG$ on $X$ and any map of $\B$-sheaves $\varphi:\calF\rightarrow \calG\big|_{\B}$,
	there exists a unique sheaf map $\varphi^+:\calF^+\rightarrow \calG$ such that $\varphi^+\big|_{\B}=\varphi$.
\end{lemma}

\begin{proof}
	For any open $V\subset X$ let
	$$\calF^+(V)=\left\{(s_U)\in\prod_{\substack{U\in\B\\ U\subseteq V}} \calF(U)\ :\  s_U\big|_W = s_{U'}\big|_W\ 
	\text{for all}\ W\subseteq U\cap U'\ \text{with}\ W,U,U'\in\B\right\}.$$
	Since $\calF$ is a $\B$-sheaf and $\B$ is a base, it is not hard to check that $\calF^+$ is a sheaf
	with $\calF^+\big|_{\B}=\calF$.  Verification of the stated universal property is left as an exercise, or one
	can consult EGA I, \S 3.2.
\end{proof}

Let $A$ be a commutative ring and put
$X=\Spec A$ with the Zariski topology.  A base of opens for $X$ is given by 
$$X_a=\Spec A_a\simeq \{\p\in X: a\not\in \p\}=\{\p\in X : a\neq 0\ \text{in}\ \kappa(\p)=\Frac(A/\p)\}.$$

\begin{lemma}
	With this notation, the assignment $X_a\mapsto A_a $ is a $\B$-sheaf.
\end{lemma}

Observe that the ring $A_a$ depends only on the set $X_a$, and not on the particular element $a$, for we have the description
$A_a=S^{-1}A$ where $S=\{\alpha\in A\ :\ \alpha(x)\neq 0\ \text{for all}\ x\in X_a\}$.

\begin{proof}
	Notice that $X_{a_1}\cap X_{a_2}=X_{a_1a_2}$ and that $X_a$ is quasi-compact.  It therefore suffices to check the sheaf
	condition for a finite cover of $X_a$ by $X_{a_1},\ldots,X_{a_n}$.  Moreover, since $X_a\supseteq X_{a_i}$ for all $i$,
	we see that $a\in A$ is nonvanishing on $X_{a_i}$, so that the map $A\rightarrow A_{a_i}$ takes $a$ to a unit,
	that is $a$ is inverted in $A_{a_i}$ so $A_{a_i}=A_{aa_i}$ for all $i$.  Therefore, replacing $a_i$ by $aa_i$
	throughout, we reduce to the case $a=1$, so that $\{X_{a_i}\}$ cover $\Spec A$, and hence $(a_1,\ldots,a_n)=1$.
	We must show that 
	\begin{align}
	\begin{diagram}
		A & \rTo & \prod_{i=1}^n A_{a_i} & \pile{\rTo \\ \rTo} & \prod_{(i,j)} A_{a_i a_j}
	\end{diagram}\label{d1}
	\end{align}
	is exact.
	
	Suppose $\alpha,\alpha'\in A$ have the same image in $A_{a_i}$ for all $i$.  Since the $a_i$
	generate the unit ideal, any $\p\in \Spec A$ fails to contain some $a_i$, so that $A_{\p}$ is a localization of
	$A_{a_i}$ whence $\alpha,\alpha'$ have the same image in $A_{\p}$.  As this holds for all $\p$,
	we see that $\alpha=\alpha'$.
	
	Now given $(s_i)\in\prod_{i=1}^n A_{a_i}$ such that $s_i,s_j$ have the same image in $A_{a_i a_j}$,
	we must construct some $a\in A$ with $a\mapsto s_i$ in $A_{a_i}$.  For this, we use faithfully flat descent.
	
	Let $A'=\prod_{i=1}^n A_{a_i}$.  Since this is a finite product of localizations of $A$ it is faithfully flat over
	$A$.  Since $A_{a_i a_j}=A_{a_i}\otimes_A A_{a_j}$ we have 
	$$A'\otimes_A A'=\prod_{(i,j)} A_{a_i}\otimes_A A_{a_j}=\prod_{(i,j)} A_{a_i a_j}.$$
	Moreover, 

	The maps of diagram (\ref{d1}) are {\em precisely} the maps $A'\rightarrow A'\otimes_A A' $
	given by $a'\mapsto 1\otimes a'$ and $a'\mapsto a'\otimes 1$.  Since
	$A'$ is faithfully flat over $A$, we have an injection $A\hookrightarrow A'$.  We must then check that the diagram
	\begin{diagram}
		A & \rInto & A' & \pile{\rTo^{a'\mapsto a'\otimes 1} \\ \rTo_{a'\mapsto 1\otimes a'}} & A'\otimes_A A'
	\end{diagram}
	is exact.  We begin with this next lecture.
	
\end{proof}















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