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\begin{document}
\smallskip
\centerline{\sc Math 632, Lecture 15\\ February 11, 2004}
\section{Codiminsion}
\begin{definition}
Let $Y\subset X$ be an irreducible closed subset. We define the {\em codimension} of $Y$ in $X$
$$\codim(Y,X)=\sup_{n}(Y=Y_0\subsetneq Y_1\subsetneq\ldots \subsetneq Y_n=X),$$
with each $Y_i$ closed and irreducible.
For arbitrary closed $Y\subset X$ we define
$$\codim(Y,X)=\inf_{Y_i}\codim (Y_i,X),$$
where the infimum is over all irreducible components $Y_i$ of $Y$.
\end{definition}
\begin{definition}
For $X\neq\emptyset$, we define
$$\dim X=\sup_n (Y_0\subsetneq Y_1 \subsetneq \ldots\subsetneq Y_n\subseteq X)$$
with each $Y_i$ irreducible and closed.
\end{definition}
When $X=\Spec A$ with $A\neq 0$ then $\dim X$ is the Krull dimension of $A$ since there is an inclusion
reversion bijection between irreducible closed sets in $X$ and prime ideals of $A$.
When $A$ is a finitely generated domain over a field $k$ then we have
$$\dim A_\p+\dim A/\p=\dim A,$$ so that if $Y=\overline{\{\p\}}$ and $X=\Spec A$
we get $\dim Y+\codim(Y,X)=\dim X$. This is not true in general, and we only have
$\dim Y+\codim(Y,X)\le \dim X$.
\begin{definition}
We define the dimension of $X$ at the point $x\in X$ to be
$$\dim_x X=\sup_{Y}\dim Y,$$
where the supemum is over all irreducible components of $X$ passing through $x$.
\end{definition}
It is not difficult to see that we have a bijection between $\Spec \O_x$ and irreducible closed subsets of $X$ passing
through $x$, and moreover that $\dim X=\sup_{x\in X} \dim\O_x$.
\section{Closed subschemes}
Given a closed subset $Y$ of a scheme $X$ we would like to give $Y$ the structure of a closed subscheme, that
is, we want to find a sheaf of rings $\O$ on $Y$ such that the topological inclusion map $i:Y\hookrightarrow X$
induces $i_*\O \simeq \O_X/ \I$ for some ideal sheaf $\I$, and such that $(Y,\O)$ is a scheme.
In other words, we seek an ideal sheaf $\I\subseteq \O_X$ such that
\begin{enumerate}
\item $\supp (\O_X/\I)=Y$, and when this holds,
\item $\O_X/\I\simeq i_* i^{-1}(\O_X/\I)$,
\end{enumerate}
and such that $(Y,i^{-1}(\O_X/\I))$ is a scheme.
Observe that condition (1) is $Y=\{x\in X\ |\ f(x)=0\ \text{for all}\ f\in \I_x\}$.
\begin{definition}
We say that $\I\subseteq \O_X$ is {\em radical} if equivalently $\I_x\subseteq \O_x$ is radical for every $x\in X$
or $\I(U)\subseteq \O_X(U)$ is radical for all open $U$.
\end{definition}
\begin{lemma}
If $X$ is a scheme and $Y\subseteq X$ is a closed subset then there existst a unique radical ideal sheaf
$\I\subset \O_X$ with zero locus $Y$ such that $(Y,\O_X/\I)$ is a scheme.
\end{lemma}
\begin{proof}
Let $X=\Spec A$. Then $Y=\Spec A/I$ for a unique radical ideal $I\subseteq A$. For any $a\in A$
we have $X_a\cap Y=\Spec A_a/I_a$ and $I_a\subseteq A_a$ is again radical. Thus, for every open affine
$U\subseteq X$ we get a unique radical ideal $I_U\subseteq \O_X(U)$ such that $Y\cap U$ is the zero locus
of $I_U$ on $U$. When $U_a=V\subseteq U$ is a basic open then $(I_U)_a=I_V$ in $\O_X(U_a)=\O_X(U)_a$.
Now we imitate the construction of $\O_X$ on an affine scheme (i.e. the $\B$-sheaf construction)
to enhance $\{(I_U)_a\}_{a\in\O_X(U)}$
to an ideal sheaf $\I_{Y\cap U}\subseteq \O_U$ for each affine open $U\subseteq X$.
If $U,U'\subseteq X$ are two open affines then
$$\I_{U\cap Y}\big|_{U\cap U'}=\I_{U'\cap Y}\big|_{U\cap U'}$$ inside $\O_X\big|_{U\cap U'}$
(which can be deduced using Nike's trick locally on $U\cap U'$). Thus that $\I_{U\cap Y}$
glue to give $\I_Y\subseteq \O_X$ such that $\I_Y\big|_{U}=\I_{Y\cap U}$.
\end{proof}
We call $(Y,\O_X/\I_Y)$ the {\em induced reduced scheme structure} on $Y$. When $Y=X$, the ideal sheaf $\I_Y$
is just the sheaf of nilpotent elements, so we obtain $X_{\rm red}$ in this way.
Given any scheme structure on $Y$ making it a closed subscheme of $X$, say $\I=\ker(\O_X\rightarrow i_*\O_Y)$
we can look at $(Y,\O_X/\I^{n+1})$ for any $n\ge 0$. For example, giving $Y$ the reduced structure, we can contemplate
$(Y,\O_X/\I_Y^{n+1})$. On any affine $U=\Spec A\subseteq X$, the sheaf $\I_Y\big|_{U}$ comes from
$I\subseteq A$ so that $(Y,\O_X/\I_Y^{n+1})\big|_{U}\simeq \Spec A/I^{n+1}$. This is called
the $n$ th {\em infintessimal neighborhood of $Y$ in $X$}.
\begin{theorem}
If $X=\Spec A$ and $Y\hookrightarrow X$ is a closed subscheme then there is a unique ideal $\a\subseteq A$
and a unique isomorphism $Y\simeq \Spec A/\a$ such that the diagram
\begin{diagram}
Y & \rTo^{\sim} & \Spec A/\a\\
\dTo & \ldTo & \\
X & &\\
\end{diagram}
commutes.
Moreover, a map $\Spec A/\a\rightarrow \Spec A/\a'$ exists if and only if $\a\supseteq \a'$.
\end{theorem}
\begin{proof}
This is on the homework. The key point is to show that such a $Y$ as int he statement of the Theorem is
affine, for which one uses the criterion for affineness as in Hartshorne Ex. 2.17 (b).
\end{proof}
\end{document}