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\begin{document}
\pagestyle{plain}
\title{Introduction to Algebraic Geometry, Class 18}
\author{Ravi Vakil}
\date{Tuesday, November 9,  1999.}
\maketitle
\tableofcontents

{\parskip=12pt % closing bracket is just before the bibliography 

No class Thursday.
Problem sets back at end.
New problem set handed out.

\section{Extending rational maps of nonsingular curves}

\subsection{More on integral closure in a field extension}

I neglected to say a few small facts about integral closure in a field extension.  Recall:

Let $A$ be an integral domain, which is
a finitely generated algebra over $\kb$.  Let $K$ be the quotient field
of $A$, and let $L$ be a finite algebraic extensions of $K$.
Then the {\em integral closure of $A$ in $L$}
consists of those elements of $L$ satisfying $x^n + a_{n-1} x^{n-1} +
\dots + a_0=0$ where $a_i \in A$.  

For example, the {\em integral closure of $A$ in its quotient field}
is {\em the integral closure of $A$}.

It isn't hard to show that integral closure $A'$ of $A$ in $L$ must be
integrally closed, i.e. the only solutions to such equations in $L$,
where now the $a_i$ are supposed to lie in $A'$, lie in $A'$.

{\em Remark 1.}  $R \subset S$.  Indeed, if $r \in R$, then it is a solution
to the equation $x-r=0$.

{\em Remark 2.}  The quotient field of $A'$ is $L$.  This follows from:

{\em Remark 3.}  If $l \in L$, then some multiple of it $al$ ($a \in A$) is
in $A'$.  {\em Proof.}  $l$ satisfies some $l^n   + a_{n-1} l^{n-1} + \dots + a_0 = 0$ for some $a_i \in K$.
Clear denominators, to write $b_n l^n   + b_{n-1} l^{n-1} + \dots + b_0 = 0$ for some $b_i \in A$. Then rewrite as
$(b_n l)^n   + b_{n-1} b_n (b_n l)^{n-1} + \dots + b_0 b_n^n = 0$.  Thus $b_n l \in A'$.  \qed

{\em Remark 4.} A reminder: A theorem I proved (using a couple of
theorems I {\em didn't} prove), that we'll use today:

{\bf Theorem.} Take any Dedekind domain $R$ that is a finitely
generated algebra over $\kb$, and let $K$ be its field of fractions.
Let $L$ be a finite extension of $K$, and let $S$ be the integral
closure of $R$ in $L$.  Then $S$ is also a Dedekind domain,
and is also a finitely generated algebra over $\kb$.  

\subsection{Last time}

Goal:  Rational maps of nonsingular curves to projective varieties can
be extended to morphisms.

We discussed reasons why you can't extend $\proj^1 \dashrightarrow \A^1$,
$\A^2 \dashrightarrow \proj^1$. 

{\bf Lemma.} Let $Y$ be a prevariety, and suppose $P$ and $Q$ are two
points contained in a single affine open $U$, and $\oh_{Y,Q} \subset
\oh_{Y,P}$ (as subrings of $k(Y)$).  Then $P=Q$.

{\bf Key Technical Theorem (tricky).} Let $K$ be a finitely generated function field of
dimension one over $\kb$, and let $x \in K$.  Then the set of discrete
valuations of $K / \kb$ where $v(x)<0$ is {\em finite}.

{\em Warning:  We're going to get a lot of mileage out of the proof this, so pay attention to it,
even though it will get heavy!}

Here's the geometric idea.  Suppose $K= \kb(t)$, the function field of $\proj^1$.
$x \in K$, so $x$ is a rational function of $\proj^1$.  Then the discrete valuations
of $K/\kb$ correspond to points of $\proj^1$ (exercise).  The points/valuations $v_p$ where
$v_p(x)<0$ are precisely the points where $x$ has poles; there are only a finite number
of such points/valuations.

\subsection{New material starts here}

The proof will use the following:

{\bf Commutative algebra lemma.}  Suppose $(S, \fn)$, $(R,\fm)$ are two discrete valuation rings
of $K/\kb$, with $S \subset R$ and $\fn = S \cap \fm$.  Then $S=R$ and $\fn = \fm$:  they
are the {\em same} valuation ring.

(Compare to lemma just above.)

{\em Proof.} The two valuation rings give two discrete valuations
$v_\fn$ and $v_\fm$ on $K$ (so $S$ are the elements $s$ of $K$ with
$v_\fn(s) \geq 0$, and $R$ are the elements $r$ of $K$ with $v_\fm(r)
\geq 0$; we can recover the DVRs from the valuations.)  Let $u$ be a
uniformizaer of $S$ (i.e. a generator of $\fn$), so $v_\fn(u) = 1$ (and
$v_\fm(u) > 0$).  Suppose $r \in K$, $r \notin S$; we'll see that $r \notin R$.
As $v \notin S$, $v_\fn(r) <0$, so $r=u^{-n}w$ with $n>0$, $v_\fn(w) =0$, so $w \in S \setminus \fn$.
Then $w \in R \setminus \fm$, hence $v_\fm(w) = 0$.  But then $v_\fm(r)  = -n v_\fm(u) + v_\fm(w) < 0$,
so $r \notin R$ as desired.  \qed


{\em Proof of ``Key Technical Theorem''.}
{\em i) The set-up.}
Note that if $(R, \fm_R)$ is a valuation ring of $K/\kb$, then if $x$
is not in $R$ then it has negative valuation, so $y:=1/x$, having
positive valuation, is in $\fm_R$.  So we have to show that if $y \in
K$, $y \neq 0$, then the set of discrete valuations where $y \in
\fm_R$ is a finite set.  (Geometric picture: show that a non-zero
rational function on $\proj^1$ has only a finite number of zeroes.)
If $y \in \kb$, then $v(y)=0$ for all discrete valuations of $K/\kb$,
so we can assume $y$ is not in $\kb$.  (Geometric picture:  
can ignore constant functions.)

{\em ii) Part of proof involving $y$.} Consider $\kb[y]$, the subring of
$K$ generated by $y$.  Since $\kb$ is algebraically closed, $y$ is
transcendental over $\kb$, so $\kb[y]$ is a polynomial ring, i.e. $y$
doesn't satisfy any relations.  Hence $K$ is a finite field extension
of $\kb(y)$.  Let $B$ be the integral closure of $\kb[y]$ in $K$.
Then by an earlier theorem proved last day (and repeated as Remark 4
above), $B$ is a Dedekind domain, and is also a finitely generated
algebra over $\kb$.

Hence $B$ corresponds to an affine variety $Y$, which maps to $\A^1$
with coordinate $y$; draw it!  It is dimension 1 and nonsingular.
Also, $y$ is a function on $Y$.

{\em iii) Now bring in the valuation.} If $y$ is contained in a
discrete valuation ring $(R, \fm_R)$ of $K/\kb$, then $\kb[y] \subset
R$, and since $R$ is integrally closed in $K$, we have $B \subset R$
as well.  Let $\fn = \fm_R \cap B$.  Then $\fn$ is a maximal ideal of
$B$.  (Hence it corresponds to a point of $Y$ mapping to zero.)  Also,
$B_\fn \subset R$.  And also $\fn B_\fn \subset \fm_R$.  Now $B_\fn$
is also a DVR of $K/\kb$, so $B_\fn = R$ by the Commutative Algebra
Lemma.

Hence if $y$ is in $\fm_R$, then $y$ is in $\fn$.
To say that $y$ is in $\fn$,
means that $y$, as a regular function on $Y$, vanishes at the corresponding
point.  As non-zero functions vanish at only finitely many points (closed zero-dimensional
subsets are only finite numbers of points), we're done.  \qed 

From the proof, we can also extract another result, which we can use later. 

{\bf Corollary.} Let $v$ be a discrete valuation of the field $K/\kb$,
a finitely-genenerated function field of transcendence degree 1.  Then
there is a nonsingular curve $C$, with function field $K$, and a point
$p \in C$, such that $v$ is the valuation induced by $p$.

\bpf Take any $y \in K$ of positive valuation.  Then construct
$Y$ as in the previous proof.  \epf

{\bf Corollary.} Given $y \in K \setminus \kb$, all discrete
valuations of $K/\kb$ such that $v(y) \geq 0$ (i.e. that $y$ is in the
corresponding DVR) are accounted for by points on the curve $Y$ above.

{\em (Remark:  just proof above.)}

The only ``missing'' valuations are those for which
$v(y)<0$, so by the Key Technical Theorem, we're only missing a finite number.

\bpf  Let $(R,\fm)$ be the corresponding DVR.  
As $v_\fm(y) \geq 0$, $y \in R$.  Then let $a \in \kb$ be the residue of $y$
modulo $\fm$, i.e. $y \equiv a \pmod \fm$.  If $a=0$, then
the proof above shows that $(R,\fm)$ is the local ring of one of the points of $Y$
mapping to 0, so we're done.

If $a \neq 0$, then replace $y$ with $y-a$, and we're done again. \epf

{\em This confused them.}

\subsection{Extension of morphisms to projective varieties, over nonsingular points of curves}
Now we're ready to prove the main result of this section.


{\bf Theorem.} Let $X$ be a nonsingular curve, $p$ a point of $X$, $Y$
a projective variety, and $\phi: X-p \rightarrow Y$ a morphism.  Then
there exists a unique morphism $\overline{\phi }: X \rightarrow Y$
extending $\phi$.

By separatedness, as $Y$ is a variety, we have uniqueness; all that
is necessary is existence.

Last day, we discussed an example of a map from $\proj^1$ to
$\proj^n$: $[x;y] \mapsto [x^3; x^2 y ; xy^2]$ is defined away from
$[0;1]$; it's clear how to extend.  You divide by $x$.   We'll carry
this example along as we do the proof. 

Hence this is l'Hopital's rule in a vague sense.

\bpf 
It is sufficient to show that $f$ extends to a morphism of $X$ into $\proj^n$ (with coordinates
$x_0$, \dots, $x_n$.  Let $U$ be the open set where all $x_i$ are non-zero.
By changing coordinates if necessary, assume $f(X-p)$ meets $U$.

For each $i$, $j$, $x_i/x_j$ is a regular function on $U$; pulling back by $f$, we
have regular function $f_{ij}$ on an open subset of $X$, which 
we view as a rational function on $X$, so $f_{ij}$ is in $k(X)$.

Let $v$ be the valuation associated with $p$.  Let $r_i= v(f_{i0}) $
for $i = 0, 1, \dots, n$.  Then $v(f_{ij}) = r_i - r_j$.  
Choose $r_k$ minimal, so $v(f_{ik}) \geq 0.$ 

Consider the map of sets $\overline{\phi}: X \rightarrow \proj^n$,
given by $\overline{\phi}$ is the same as $\phi$ on $X-p$, and
$\overline{\phi}(p) = ( f_{0k}(p); \dots; f_{nk}(p))$ (observe that
not all coordinates are 0).  I claim this is a morphism.  To show
this, I need only show an affine neighbourhood $X_{\aff}$ of $X$ that
maps pointwise to an affine $U$ of $\proj^n$, such that the pullback
of every regular function on $U$ is a regular function on $X_{\aff}$.

Let $U=U_k$ be the open set where $x_k \neq 0$ (so $\overline{\phi}(p) \in
U_k$, since $f_{kk}(p)=1$).  The coordinate ring of the affine $U_k$ is
$\kb[x_0/x_k, \dots, \widehat{x_k/x_k},\dots,   x_n/x_k]$.  These functions pull back to
$f_{0k}$, \dots, $f_{nk}$ which are regular at $p$ by construction.
Let $X_{\aff}$ be any affine neighbourhood of $p$ where the rational functions
$f_{ik}$ are defined.

So we're done.
\epf

As consequences, we will get (next day):

{\bf Theorem.}  Every nonsingular separated curve $C$ is quasiprojective.  

{\bf Corollary.} Hence every nonsingular separated curve is birational to
a projective curve.

{\bf Proposition.}  Let $C_1$ and $C_2$ be two separated nonsingular curves, and let $\al:  C_1 \dashrightarrow C_2$
be a {\em birational map} between them.   Then they can be glued together via $\al$, i.e. there is
another nonsingular curvve $C = U_1 \cup U_2$, with $C_i$ isomorphic to $U_i$, and the birational map $U_1 \dashrightarrow U_2$
induced by $\al$.

{\bf Proposition.} If two nonsingular projective curves $C_1$ and
$C_2$ are birational, then they are isomorphic.

Coming up: We'll discuss these, and begin talking about why
finitely-generated fields of transcendence dimension 1 correspond to
nonsingular projective curves (over $\kb$).  } % end of parskip; it
started just before the introduction
\end{document}
