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\begin{document}
\smallskip
\centerline{\sc Math 632, Lecture 17\\ February 16, 2004}


\section{Base change}

Let $f:X\rightarrow S$ and $\pi:S'\rightarrow S$ be schemes.  Then we have the cartesian diagram
\begin{diagram}
 X'=X\times_S S' & \rTo & X\\
 \dTo 		&          & \dTo^f\\
 S'			&\rTo^{\pi} & S
\end{diagram}
and we say that $X'$ is the {\em base change} of $X$ by $\pi$.  This construction generalizes extension of scalars,
which is just the case where $X,S,S'$ are affine.
As an example, consider 
\begin{diagram}
 X_s=X\times_S \Spec\kappa(s) & \rTo & X\\
 \dTo^{f_s} 		&          & \dTo^f\\
 \Spec \kappa(s)			&\rTo & S
\end{diagram}
and suppose that $f$ is locally of finite type.  Then one can show that $X_s$ is a locally finite type $\kappa(s)$-scheme.

Recall that for any ring $A$ we defined $\P_A^n=\Proj(A[T_0,\ldots,T_n])$.  Then we have $\P_A^n\simeq \P_{\Z}^n\times_{\Spec \Z} \Spec A$.
More generally, if $S$ is any $\N$-graded ring and $A$ is an $S_0$-algebra then $S\otimes_{S_0} A$
has a natural $\N$-grading given by $(S\otimes_{S_0}A )_d=S_d\otimes_{S_0}A$ and we have,
by the isomorphism $S_(f)\otimes_{S_0} A\simeq (S\otimes_{S_0} A)_{f\otimes 1}$,
\begin{diagram}
 \Proj(S\otimes_{S_0}A) & \rTo & \Proj(S)\\
 \dTo		&          & \dTo\\
 \Spec A			&\rTo & \Spec S_0
\end{diagram}

As an example, let $g_j\in A[T_0,\ldots,T_n]$ be homogenous polynomials and consider the closed immersion
$$\Proj(A[T_0,\ldots,T_n]/(g_1,\ldots,g_r))\hookrightarrow \P_A^n.$$
The fiber over a point $\Spec \kappa(x)\rightarrow \Spec A$ is
$$\Proj(\kappa(x)[T_0,\ldots,T_n]/(g_1(x),\ldots,g_r(x))),$$
where $g_j(x)$ denotes the image of $g_j$ in $k(x)[T_0,\ldots,T_n]$.  

\begin{enumerate}
	\item Consider
	\begin{diagram}
	\Spec \C[\lambda,x,y]/(y^2-x(x-1)(x-\lambda)) &\rTo &  \A^2_{\C[\lambda]}\\
	&\rdTo & \dTo\\
	&        & \Spec \C[\lambda]=\A_{\C}^1
	\end{diagram}
	The fiber over $\lambda_0\in \C$ is $\Spec \C[x,y]/(y^2-x(x-1)(x-\lambda_0))$.
	
	\item Now consider $\Spec k[x,y]/(y^2-x)\rightarrow \Spec k[x]$ and let $x_0\in k$.
	The fiber over $x_0$ is $\Spec k[y]/(y^2-x_0),$ which consists of 2 points if $\Char k\neq 2$
	and $x_0\neq 0$ and which is the nonreduced scheme $\Spec k[y]/y^2$ if $x_0=0$.
	
	\item Suppose that $K/\Q$ is a number field and $\O$ its ring of integers.  Let $x=\{p\}\in\Spec\Z$
	and consider the diagram
	\begin{diagram}
	\Spec \O & \lTo & \Spec \O/p\O \\
	\dTo          &      & \dTo \\
	\Spec \Z   & \lTo     &\Spec \kappa(x)\\
	\end{diagram}
	Here $\O/p\O$ is a finite dimensional $\kappa(x)$-algebra
	
	\item From the point of view of base change, we can reduce a scheme over $\Spec A$ modulo an ideal $I\subseteq A$
	by taking the fiber product with $\Spec A/I$ over $\Spec A$.  One might be interested in understanding a scheme
	$A$ over $\Q$ ``modulo $p$.''  Concretely, this amounts to the existence of a ``nice model''
	$\mathcal{A}$ over $\Spec \Z$ with the commutative diagram
	\begin{diagram}
		A & \rTo & \mathcal{A} \\
		\dTo  &   &   \dTo \\
		\Spec\Q & \rTo & \Spec \Z
	\end{diagram}
\end{enumerate}

\section{Behavior of properties of morphisms under base change}

\begin{definition}
	A map of schemes $f:X\rightarrow Y$ is {\em flat} if for all $x\in X$
	the map of local rings $\O_{Y,f(x)}\rightarrow \O_{X,x}$ is flat.
\end{definition}

\begin{lemma}
	If $X=\Spec A$ and $Y=\Spec B$ are affine then $f:X\rightarrow Y$ is flat
	if and only if $A$ is a flat $B$-algebra.
\end{lemma}

\begin{proof}
	By commutative algebra, the map $\varphi:B\rightarrow A$ is flat if and only if the map $B\rightarrow A_\p$
	is flat for all $\p\in X$.  But this map factors through $B_{\varphi^{-1}(\p)}$ and is flat if and only if the
	induced map $\O_{Y,f(x)}=B_{\varphi^{-1}(\p)}\rightarrow A_{\p}=\O_{X,x}$ is flat.
\end{proof}

\begin{definition}
	A scheme is {\em regular} if it is locally Noetherian and all local rings are regular.
\end{definition}

By a famous Theorem of Serre, if $A$ is a regular local ring then so is $A_{\p}$ for any prime $\p\subseteq A$.


As a warning, we remark that base change does not necessarily preserve all ``nice'' properties of a scheme.  For 
example, it can happen that for a regular scheme $X$ of finite type over a field $k$, the base change to
$k'/k$ is {\em not} regular.  For example, one might take $k'/k$ a purely inseparable extension of
degree greater than 1.  Then $k'\otimes_k k'$ is not reduced, even though $k'$ is.

However, many nice properties of morphisms are preserved by base change.
Let $X,S'$ be schemes over a scheme $S$ and let $X'=X\times_S S'$.  Then we have the
cartesian diagram
\begin{diagram}
	X & \lTo & X'\\
	\dTo^f & & \dTo_{f'}\\
	S & \lTo_{\pi} & S'\\
\end{diagram}
and we claim that for the following list of properties $P$, the morphism $f$ has property $P$
implies that the morphism $f'$ has the property $P$:
Open immersion, closed immersion, affine, finite type and locally finite type, quasi-compact, flatness, surjective.
All of these facts are easy to show, with perhaps the exception of surjectivity.  Let $s'\in S'$
and put $s=\pi(s')$.  Since the map $X\twoheadrightarrow S$ is surjective, we can find $x\in X$ mapping to 
$s$.  This gives the diagram:
\begin{diagram}
&&	\Spec \kappa(x) &      & & \lTo & \Spec(\kappa(x)\otimes_{\kappa(s)}\kappa(s')) \\
&&	\dTo		         &      &   &\ldTo &	\dTo	\\
&&	X			& \lTo&  X'&      &          \\
&&	\dTo^{f}		&       &\dTo_{f'} &  &     \\
&&	S			&\lTo_{\pi} &  S' &  \lTo & \Spec\kappa(s')\\    
& \ruTo &      &               &      &         & \\
\Spec\kappa(s) & & & & & &\\
\end{diagram}
and since $\kappa(x)\otimes_{\kappa(s)}\kappa(s')$ is {\em nonzero}, this gives at least one point of $X'$ mapping to $s'$,
as required.  The same argument also shows that $s'\in S'$ is in $f'(X')$ if and only if $\pi(s')=s\in S$ lies in $f(X)$.
We remark that injectivity is usually destroyed by base change.  Indeed, let $L/K$ be an extension of fields.
Then $\Spec L\rightarrow \Spec K$ remains injective after arbitrary base change if and only if $L/K$
is purely inseparable.  


\begin{definition}
	A map $f:X\rightarrow S$ of schemes is {\em universally injective} (also called radical) if it is injective
	after arbitrary base change.  
\end{definition}

One can show that a map $f$ is radical if and only if $f$ is injective and for all $x\in X$ the extension $\kappa(x)/\kappa(f(x))$
is purely inseparable.  This is a very restrictive condition indeed.





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