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\begin{document}
\smallskip
\centerline{\sc Math 632, Lecture 19\\ February 20, 2004}


\section{The diagonal and immersions}

\begin{definition}
	A subset $T\subseteq X$ of a topological space is {\em locally closed}
	if there exists an open $U\subseteq X$ containing $T$ such that $T$ is
	closed in $U$.
\end{definition}

\begin{definition}
	A morphism $f:Y\rightarrow X$ of locally ringed spaces is an {\em immersion}
	if $f$ is a homeomorphism onto a locally closed subset of $X$ and 
	$f^{\#}:\O_{X,f(y)}\rightarrow \O_{Y,y}$ is surjective for all $y$.
\end{definition}

Thus we see immediately that open immersions and closed immersions are immersions.
Similarly, any quasi-projective variety $X$ over $k$ comes equipped with an immersion
$X\rightarrow \P_k^n$.

\begin{theorem}\label{list}
	\begin{enumerate}
		\item Let $f:Y\rightarrow X$ be a morphism of locally ringed spaces and $\{U_i\}$ an open
		cover of $X$ (equivalently of $\im f$).  Then  $f$ is an immersion if and only if $f:f^{-1}(U_i)\rightarrow U_i$
		are immersions for all $i$.
		
		\item A composite of immersions is an immersion.
		
		\item If $f:Y\rightarrow X$ is an immersion and $X'\rightarrow X$
		any base change, then $f':Y\times_X X'\rightarrow X'$ is an immersion as well.
		
		\item Suppose that $f:Y\rightarrow X$ is an immersion.  Then $f$ is a closed immersion
		if and only if $f(Y)$ is closed.  SImilarly, $f$ is an open immersion if and only if
		$f(Y)$ is open and $\O_{X,f(y)}\rightarrow \O_{Y,y}$ are isomorphisms
		(equivalently injective) for all $y\in Y$
	\end{enumerate}
\end{theorem}

\begin{proof}
Left as an exercise.
\end{proof}

\begin{definition}
	A {\em subscheme} of $X$ is an immersion $f:Y\rightarrow X$ up to the equivalence relation
	$Y\sim Y'$ if and only if there exists a commutative diagram
	\begin{diagram}
		Y &  & \rTo^{\sim}&  & Y' \\
		  & \rdTo_f &      & \ldTo^{f'} & \\
		  & & X & & \\
	\end{diagram}
\end{definition}

\begin{definition}
	Let $X\rightarrow S$ be a morphism.  The {\em diagonal} morphism $\Delta_{X|S}$, or simply $\Delta:X\rightarrow X\times_S X$,
	is the morphism induced by the diagram
	\begin{diagram}
		X 		&  									&   				&  		 &  \\
		 		&  \rdTo(2,4)_{\id}\rdTo^{\Delta}\rdTo(4,2)^{\id}    &   			&   		& \\
				&  		     							& X\times_S X	& \rTo	& X\\
				&		     							&  \dTo			&		& \dTo\\
				&		    							&   X			&\rTo		& S\\
	\end{diagram}
\end{definition}

When $X=\Spec A$ and $S=\Spec B$ are both affine, we obtain the map $\Delta^*:A\otimes_B A\rightarrow A$.
Since we must have $\Delta^*(1\otimes a)=a=\Delta^*(a\otimes 1)$ (by virtue of the fact
that the maps $X\rightarrow X$ inducing the diagonal are the identity maps), and hence that
$\Delta^*(a_1\otimes a_2) = a_1a_2$.

\begin{lemma}
		The diagonal $\Delta_{X|S}:X|rightarrow X\times_S X$ is an immersion.
\end{lemma}

\begin{proof}
	First of all, the property of being an immersion is a local property,
	and since an open cover of $S$ pulls back to an open cover of $X\times_S X$,
	it suffices to treat the case of affine $S$.  Similarly, as in a previous lecture,
	we may assume that $X$ is affine.  But in the case of $X,S$ affine, 
	the lemma is clear.
\end{proof}

\begin{corollary}
	The diagonal $\Delta_{X|S}$ is a closed immersion if and only if $\im\Delta$ is closed.
\end{corollary}

\begin{definition}
	A morphism $X\rightarrow S$ is {\em separated} if $\im\Delta_{X|S}$ is closed.  
\end{definition}

We observe that this condition is the right scheme-theoretic analog of the Hausdorff
condition as for a topological space $X$, the diagonal $X\rightarrow X\times X$
has closed image if and only if $X$ is Hausdorff.

\begin{remark}
	Observe that a map from $T$ to the diagonal ({\em i.e.} the image of $\Delta_{X|S}:X\rightarrow X\times_S X$
	is a map to $X\times_S X$ that factors through $X$.  Such maps are just the image of
	the diagonal set map $X(T)\rightarrow X(T)\times X(T)$.
\end{remark}

Suppose we have a commutative diagram
\begin{diagram}
	X & & \rTo^f & & Y\\
	& \rdTo & & \ldTo & \\
	&& S &&\\
\end{diagram}
We can form the {\em graph} morphism $\Gamma_f:X\rightarrow X\times_S Y$: it is the morphism induced
by the following diagram:
	\begin{diagram}
		X 		&  									&   				&  		 &  \\
		 		&  \rdTo(2,4)_{\id}\rdTo^{\Gamma_f}\rdTo(4,2)^{f}    &   			&   		& \\
				&  		     							& X\times_S Y	& \rTo	& Y\\
				&		     							&  \dTo			&		& \dTo\\
				&		    							&   X			&\rTo		& S\\
	\end{diagram}
We claim that $\Gamma_f$ is an immersion.  Indeed, by Theorem \ref{list}, it suffices to
show that the diagram 
\begin{diagram}
	X & \rTo^{\Gamma_f} & X\times_S Y \\
	\dTo^f &			& \dTo_{f\times 1}\\
	Y     &  \rTo^{\Delta_{Y|S}} & Y\times_S Y\\
\end{diagram}
is cartesian.  To check that it is indeed cartesian, we observe that to give maps $T\stackrel{(\phi_1,\phi_1')}{\longrightarrow} X\times_S Y$
and $T\stackrel{\phi_2}{\longrightarrow} Y$ such that the diagram
\begin{diagram}
	T & \rTo^{(\phi_1,\phi_1')} & X\times_S Y \\
	\dTo^{\phi_2} &			& \dTo_{f\times 1}\\
	Y     &  \rTo^{\Delta_{Y|S}}_{1\times 1} & Y\times_S Y\\
\end{diagram}
commutes requires that we have $(\phi_2,\phi_2)=(f\phi_1,\phi_1')$, i.e. that $\phi_1'=\phi_2$
and $f\phi_1=\phi_2$, so that the map $\phi_1:T\rightarrow X$ determines $\phi_1'$ and $\phi_2$.
Thus, to give such maps is to give a map $T\rightarrow X$, whence the cartesian claim follows.


Now put $U_i=\A^1$ for $i=1,2$ and let $X$ be the scheme obtained by glueing $U_1,U_2$
along $\A^1-\{0\}$, so that $U_1\cap U_2=\A^1-\{0\}$.  
We obtain the diagonal $X\rightarrow X\times_S X$ whose image is covered by $U_i\times U_j$,
and $\Delta$ is a closed immersion if and only if each of the induced morphisms $\Delta:{\id}^{-1}(U_i\times U_j)\rightarrow U_i\times U_j$
is a closed immersion.  But as $\id^{-1} (U_1\times U_2)=U_1\cap U_2,$ this entails that
the map $U_1\cap U_2\stackrel{(\id,\id)}{\rightarrow} U_1\times U_2$ be a closed immersion.
But this is a morphism
$\A^1-\{0\}\rightarrow \A^1\times \A^1=\A^2$, whose image is the line $t_1=t_2$ (where
$t_i$ are coordinates on the two copies of $\A^1$) with the origin deleted, and hence is not closed.
Thus $X\rightarrow S$ is not separated.














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