\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}
\DeclareMathOperator{\codim}{codim}
\renewcommand{\Im}{{\rm Im}}
\newcommand{\B}{\mathscr{B}}
\newcommand{\I}{\mathscr{I}}
\newcommand{\calC}{\mathscr{C}}
\newcommand{\calV}{\mathscr{V}}
\renewcommand{\a}{\mathfrak{a}}
\newcommand{\G}{\mathbf G} 
 
\begin{document}
\smallskip
\centerline{\sc Math 632, Lecture 18\\ February 18, 2004}

\section{Scheme theoretic intersection}

If $i:Z\rightarrow X$ and $i':Z'\rightarrow X$ are closed immersions of ringed spaces and there
exists a compatible map $f:Z\rightarrow Z'$ then $f$ is also a closed immersion.  If in addition
there exists a compatible $f':Z'\rightarrow Z$ then necessarily  $f\circ f'=\id_{Z'}$ and $f'\circ f=\id_{Z}$.

We say the two closed
immersions $Z,Z'\rightarrow X$ are {\em equivalent} if there exists a compatible isomorphism $Z\simeq Z'$.
Such a map is unique if it exists.

\begin{definition}
	Let $Z,Z'\rightarrow X$ be closed immersions of schemes.  The {\em scheme theoretic intersection}
	of $Z,Z'$ is
	\begin{diagram}
		Z\cap Z'=Z\times_X Z' & \rTo & Z'\\
		\dTo & & \dTo \\
		Z & \rTo & X\\
	\end{diagram}
	and $Z\cap Z'$ is realized, via either projection, as a closed subscheme of $X$ whose image is contained
	in $Z$ and $Z'$.
\end{definition}

Observe that if $Y\rightarrow X$ is a closed subscheme contained in $Z$ and $Z'$ then we get the diagram
	\begin{diagram}
Y	&	 &			&		&\\
	&\rdTo(2,4)\rdTo\rdTo(4,2)	&				&            &    \\
	&	&Z\times_X Z' & \rTo & Z'\\
	&	&\dTo & & \dTo \\
	&	&Z & \rTo & X\\
	\end{diagram}
by the universal property of the fiber product $Z\times_X Z'$, so that $Y$ is contained in $Z\cap Z'$.  

\begin{proposition}
The topological space $|Z\cap Z'|$ of $Z\cap Z'$ is $|Z|\cap |Z'|$.
\end{proposition}

\begin{proof}
	Let $U=X-Z\times_X Z'$ and observe that $U$ is open.  Observe that $U=(X-Z)\cup (X-Z')$.
	Indeed, $U,(X-Z),(X-Z')$ are open subschemes of $X$ and $f:Y\rightarrow X$ factors through $U$
	if and only if for all $y\in Y$ we have $f(y)\not\in Z\times_X Z'$, that is, if and only if the map
	$\Spec\kappa(f(y))\rightarrow Y$ fails to factor through $Z$ and $Z'$ (using the universal
	property of the fiber product).  Since $\Spec \kappa(f(y))$ is reduced, such factorization fails if and only if $f(y)\not\in Z$ or $f(y)\not\in Z'$,
	{\em i.e.} if and only if $f(y)\in (X-Z)\cup (X-Z')$.  Thus, $f:Y\rightarrow X$ factors through $U$ if and only if
	it factors through $(X-Z)\cup (X-Z')$, so the two opens are uniquely isomorphic as they have the same universal
	property.
\end{proof}

We now ask: what is the ideal sheaf $\calI_{Z\cap Z'}\subseteq \O_X$?

\begin{proposition}
	We have $\calI_{Z\cap Z'}=\calI_{Z}+\calI_{Z'}$ inside $\O_X$, where for any two sheaves $\calF,\calG$
	on a space $Y$ the sum $\calF+\calG$ is the sheaf image of $\calF\oplus\calG$ under the addition map.
\end{proposition}

\begin{proof}
	Everything is local on $X$, so we may assume $X=\Spec A$ and $Z=\Spec A/I$ and $Z'=\Spec A/I'$.  Let us compute
	stalks at $x\in X$.  We have
	$$Z\times_X Z'=\Spec (A/I\otimes_A A/I')=\Spec A/(I+I')$$
	and 
	$$\calI_{Z\cap Z'}=(I+I')_{\p}=I_{\p}+I'_{\p}$$ in $A_{\p}=\O_{X,x}$.
	Thus we have $\calI_{Z\cap Z',x}=\calI_{Z,x}+\calI_{Z',x}=(\calI_{Z}+\calI_{Z'})_x$.
\end{proof}

As an example, suppose that $\Char k\neq 2$ and put
$C_i=\Spec k[x,y]/(y+(-1)^i x^n)$ for $i=1,2$.  Then 
$$C_1\cap C_2=\Spec k[x,y]/(y-x^n,y+x^n)=\Spec k[x]/x^n\hookrightarrow \Spec k[x].$$

\section{Remarks on base change}

If $k$ is a field and $X$ a finite type $k$-scheme, it is important to study behavior of properties 
of morphisms and of $X$ with respect to extension of $k$.  So for example, if $k'/k$ is an extension
and $X'=X\otimes_k k'=X\times_{\Spec k}\Spec k'$, then $X$ being irreducible does {\em not}
imply that $X'$ is, and similarly for reduced.

We would like to descend properties.  That is, we wish to know that if a property holds after base change, then it held before
base change.  It will turn out that if $\pi:S'\rightarrow S$ is any base change that is {\em faithfully flat}
(i.e. flat and surjective) then most ``nice'' properties will hold after base change if and only if they held before base change.
A useful principle for finite type schemes over a field $k$
is that we can check many properties over any algebraically closed extension of a field $k$.  We will see this later.

Flat maps will turn out to have especially nice properties.  For example, let $f:X\rightarrow Y$ be a flat map 
of locally noetherian schemes and suppose $f(x)=y$.  Then $\dim \O_{X,x}=\dim \O_{Y,y}+\dim \O_{X_y,x}$.
The key point here is that $\O_{X_y,x}\simeq \O_{X,x}/\m_y\O_{X,x}$.

\section{Group schemes}

Let $\calC$ be the category of sets and $\{\cdot\}$ the final object.  A {\em group object} $G$ is
an object $G$ together with a triple of morphisms:
$$\left(G,G\times G\stackrel{m}{\rightarrow} G,G\stackrel{i}{\rightarrow} G,\{\cdot\}\stackrel{e}{\rightarrow} G\right)$$
such that the following diagrams commute:
\begin{enumerate}
\item Associativity:
\begin{diagram}
	G\times G\times G & \rTo^{\id\times m} & G\times G \\
	\dTo^{m\times \id}	&			& \dTo_m\\
	G\times G & \rTo_{m}		& G\\
\end{diagram}

\item Identity:
\begin{diagram}
	G\times\{\cdot\}=\{\cdot\}\times G& \rTo^{e\times \id} & G\times G \\
	\dTo^{\id\times e}	&\rdTo_{\id}			& \dTo_m\\
	G\times G & \rTo_{m}		& G\\
\end{diagram}

\item Inverse:
\begin{diagram}
	G 				& \rTo^{\id\times i}     &	        & 	&G\times G \\
	\dTo^{i\times \id}	&\rdTo 			& 		& 	&\dTo_m	\\
					&				&\{\cdot\}   &	&		\\
					&				&		&\rdTo^{e} &	\\
	G\times G 		& \rTo_{m}			&	         & 		&G	\\
\end{diagram}
\end{enumerate}

It is clear that in any category with finite products and a final object we have a notion of a group 
object.  For example, let $\calC$ be the category of $S$-schemes.  Here the final object is $S$ and
products are fiber products over $S$.  So an $S$-group $G$ is a scheme $\pi:G\rightarrow S$ together
with a section $e:S\rightarrow G$ such that $\pi\circ e=e\circ \pi=\id$ and maps $m:G\times_S G\rightarrow G$
and $i:G\rightarrow G$ making the above three diagrams commute.

As an example, consider $\G_a=\A^1_{\Z}$.  We have
$\G_a(Y)=\Hom(Y,\A_{\Z}^1)=\Gamma(Y,\O_Y)$, which is an additive group,
and this is natural in $Y$, {\em i.e.} a map $Y'\rightarrow Y$ gives a map of groups
$\G_a(Y)\rightarrow \G_a(Y')$.  It will follow by Yoneda's lemma, which we will see later,
that $\G_a$ is a group scheme.

Another important example is $\GL_n=\Spec(\Z[X_i]_(\det))$, and $\GL_n(Y)$
is the set of $n\times n$ matrices having entries in $\Gamma(Y,\O_Y)$ with determinant
in $\Gamma(Y,\O_Y^{\times})$, functorially in $Y$ as a group via multiplication.
The simplest case is $\G_m=\GL_1=\Spec \A^1_{\Z}-\{0\}=\Spec \Z[X,X^{-1}]$.
Here, $\G_m(Y)=\Gamma(Y,\O_Y^{\times})$.
The map $\G_m(Y)\rightarrow \G_m(Y)$ given by $u\mapsto u^n$
has kernel $\mu_n(Y)$, the $Y$-points of $\mu_n=\Spec \Z[X]/(X^n-1)$.  This
is a closed subscheme of $\G_m$.  We can even consider the $\mu_p/\F_p$,
that is, the fiber of $\mu_p\rightarrow \Spec \Z$ over $p$.  This is naturally a group
scheme, $\Spec \F_p[T]/(T^p-1)$ whose $Y$ points are
$\mu_p(Y)=\{u\in \Gamma(Y,\O_Y^{\times}): u^p=1\}$.














\end{document}