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\begin{document}
\pagestyle{plain}
\title{Introduction to Algebraic Geometry, Class 17}
\author{Ravi Vakil}
\date{Thursday, November 4,  1999.}
\maketitle
\tableofcontents

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

Thanks for responses to diophantine e-mail.  I'll respond soon en masse.

Hand in problem sets at end.

\section{A little more on completions:  changing coordinates}

A few quick notes:

Note that the completion of $\kb[x,y]$ at $\fm=(x,y)$ is $\kb[[x,y]]$.

Note that you can make sense of the infinite sum $x_0 + x_1 + \dots$
in $\hat{R}$, the completion of a ring with respect to $\fm$ if
$x_i \subset \fm^i$ (in the case of a DVR, $v(x_i) \geq i$).

Note that if $f(x) \in \kb[[x]]$ and $u \in \fm \subset \hat{R}$, then $f(u)$ makes
sense as an element of $\hat{R}$.  Reason:  $u^n \in \fm^n$.

In $\kb[[x]]$, let $u = x + g_{>1}(x)$.  Then this invertible, i.e.
you can write $x=h(u)$.  Here's how, informally.
$x = u - g_{>1}(x)$; this will tell us the ``degree 1'' term 
of $h$. Then $x= u- g_{>1}(u- g_{>1}(x))$; this will tell us the
``degree 2'' term.  And so on.

This result can be expressed in another way: Suppose you have an
element $f \in \kb[[x]]$, which I prefer to think of as a ``function
in a formal neighbourhood of a point on a nonsingular curve''.
Suppose $v(f) = 1$, i.e. $f = ax + g_{>1}(x)$, where $a \neq 0$.  Then
there is a change of coordinates so that $f(u)=u$, which is given
by $u=ax+g_{>1}(x)$. (Note that changes of coordinates are by definition
invertible.)

In $\kb[[x,y]]$, suppose $f(x,y) = f_1 + f_2 + \dots$, and $f_1 \neq 0$, or equivalently,
$$
\left( \frac {\partial f } {\partial x} (0,0),
\frac {\partial f } {\partial y} (0,0) \right) \neq (0,0)
$$
(the Jacobian condition for nonsingularity!).  Let's change coordinates to put this function in nice form.

First, make a linear change of coordinates so that $f_1 = y$.  Hence 
$f(x,y) = y + f_{>1}$, where $f_{>1}$ has elements of degree greater than 1.  We can change
coordinates (invertibly, by definition of change of coordinates) again so that $f(u,v) = v$, by
taking $u=x$, $v = f(x,y) = y + f_{>1}$.  We need to check that this is invertible.  I sketched why;
in short, $y = v- f_{>1} (u, v - f_{>1}( u, \dots))$.  More precisely, we can read off the first $n$ degrees of $y$
(in terms of $u$ and $v$).

Here's the beginning of the second-last exercise on the problem set I just gave out.  Suppose the characteristic of $\kb$ is not 2, and
$f(x,y) = f_2 + f_{>2} \in \kb[[x,y]]$, with $f_2 \neq 0$.  Then  $f_2 = \al x^2 + \be xy + \ga y^2$.
Suppose further that $\be^2 -4\al \ga \neq 0$, i.e. $f_2$ factors into $(\de_1 x + \eps_1 y) (\de_2 x + \eps_2 y)$,
where
$$
\det \left( \begin{array} {cc} \de_1 & \eps_1 \\ \de_2 & \eps2 \end{array} \right) \neq 0.
$$
For convenience, make a linear change of coordinates $a=\de_1 x + \eps_1 y$, $b = \de_2 x + \eps_2 y$ (which is
invertible by linear algebra).
Then you can further change coordinates to $u$, $v$, so that
$u = a + g_{>1}(a,b)$, 
$v = b +  h_{>1}(a,b)$
and $uv = f$.  (In the problem set, you will justify this step, using {\em Hensel's Lemma}.)

This is indeed invertible, by the same trick:
$$ a = u - g_{>1} (  u - g_{>1}(...), v - h_{>1}(...))$$
$$ b = v - h_{>1} (  u - g_{>1}(...), v - h_{>1}(...)).$$
As a consequence, the change of coordinates is
$$
u = \de_1 x + \eps_1 y + \text{higher}
$$
$$
v = \de_2 x + \eps_2 y + \text{higher}
$$
so long as it is ``invertible to first order'', i.e.
$$
\det \left( \begin{array} {cc} \de_1 & \eps_1 \\ \de_2 & \eps2 \end{array} \right) \neq 0.
$$
This generalizes to more variables.

\section{Integral closure and Dedekind domains}

Reminder: $R$ is integrally closed means that it is integrally closed
in its function field $K$.  In other words, the only solutions in $K$
to $x^n + a_{n-1} x^{n-1} + \dots + a_0=0$ where $a_i \in R$, lie in
$R$. 

There are two ``canonical'' examples.  

{\em Example.} We already know $\kb[t]$ has dimension one, and it's
obviously a domain, and it is Noetherian by the Hilbert Basis Theorem.
$\kb[t]$ is integrally closed in its function field, by explicit
calculation: If $f \in \kb(t)$ is in the integral closure, so $f^n +
a_{n-1}(t) f^{n-1} + \dots + a_0(t)=0$, where $a_i(t)$ are
polynomials.  Write $f$ in lowest terms as $p(t)/q(t)$, where $p$ and
$q$ are relatively prime polynomials.  Then $p(t)^n + a_{n-1}(t)
p(t)^{n-1} q(t) + \dots + a_0(t) q(t)^n=0$.  Then $q(t)$ must divide
$p(t)^n$, so as $p$ and $q$ are relatively prime, $q(t)$ must be a
constant, so $f$ was indeed a polynomial.

Indeed its localizations at the primes $(t-a)$ are DVR's, as we
observed earlier.

{\em Example.} We already know that $\zed$ has dimension one, and it is
obviously a Noetherian domain.  Let's check that it is integrally closed in its function field.
(Mimic the above proof.)

{\em For more on integral closure, see next day's notes.}


We'll need two more facts about integral closures.

{\bf Theorem (Integral closure is a local property).} 
If $R$ is integrally closed if and only if the localization of $R$ at
each of its maximal ideals is integrally closed.

(One direction is easy; the other is harder.)

{\bf Definition.} 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 $A'$.

{\bf Theorem (Finiteness of integral closure).} Then $A'$ is a finite
$A$-module, and is also a finitely generated $\kb$-algebra.

{\bf Corollary.}  Every prime in $A$ has a preimage in $A'$.  {\em (Appears in class 19 or 20; should be here.)} 

{\bf Definition.}  A {\em Dedekind domain} is an integrally closed noetherian
domain of dimension one.  

{\em Quick scheme-theoretic remark.} It's basically immediate that a
ring $R$ is a Dedekind domain if and only if $\Spec R$ is a
nonsingular dimension one scheme --- just use that commutative algebra
fact from last day (the 4 equivalent definitions of DVR).  (That's not
quite right: we need to include {\em Noetherian} in the adjectives in
front of {\em scheme}.  I haven't defined what that means, but you
might be able to guess.)

{\em Remark.} Because a Dedekind domain $R$ has dimension one, it
means that its prime ideals are: the ideal (0), and the maximal
ideals.  If you localize $R$ at a maximal ideal $\fm$, you get a
discrete valuation ring, by the commutative algebra facts above.

{\em Example, generalizing $\kb[t]$.} Let $X$ be a nonsingular affine
curve, with ring of functions $R$.  Then $R$ is a Noetherian domain of
dimension one.  It is also integrally closed: by the commutative
algebra fact and the Theorem on ``locality of integral closure'' above, we need to check that for
each maximal ideal of $\fm$ of $R$, the localization $R_\fm$ is a
regular local ring.  But this is the stalk of the structure sheaf at a
point, so by our algebraic definition of nonsingularity, this is true.

To show you how nontrivial it is, we've shown for example that the
ring $\kb[x,y]/(y^2-x^3+x)$ is integrally closed in its function field
(in characteristic different from 2).  Just check that $y^2-x^3+x$ cuts
out a nonsingular curve in $\A^2$, using the Jacobian condition,
and you're done!

{\em Exercise.}  The function field of this curve is not isomorphic to $\kb(t)$.  
Hence this really is a different example.
(Hints given.)

{\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 $\kb$.  

(Perhaps keep this on the board.)

{\em Proof.} By the Theorem on finiteness of integral closure, $S$ is
a finitely generated algebra over $\kb$.  It is a domain, as it is a
subring of a domain $L$.  It has dimension 1, as its function field
$L$ has the same transcendence degree over $\kb$ as $K$ (as $L$ is
finite over $K$).  And it is integrally closed.  \epf

Something more general is true:  you can excise the ``finitely generated
algebra over $\kb$''.  I'll give the result because it will be of interest
to those of you with arithmetic interests, but as we won't use it in the course,
I won't prove it.  Those of you learning schemes will find it useful, though.

{\bf Theorem that we won't use.} Take any Dedekind domain $R$, 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.

{\em Example, generalizing $\zed$.}  Let $R = \zed$, so $K= \Q$.  Let $L$ be a finite
extension of $\Q$, called a {\em number field}.  Then $S$ in 
this case is called the {\em ring of integers} in the number field.
For example, if $L = \Q(i)$, then $S = \zed[i]$, the example 
that has come up in the problem sets.

\section{Extending rational maps of nonsingular curves}

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

Discuss reasons why you can't extend $\proj^1 \dashrightarrow \A^1$,
$\A^2 \dashrightarrow \proj^1$.  Perhaps mention same thing with cusp
and node.

{\em Discussed extending map from $\proj^1$ to $\proj^4$.  Then
analytically.  This is l'Hopital in a vague sense: can take limits.}

{\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 $P=Q$.

Recall that if any two points are contained in an affine open, then
$Y$ is separated, i.e. a variety.  Recall that all quasiprojective
varieties and affine varieties have this property; this was even how
we showed that quasiprojective varieties are in fact varieties.

{\em Proof.} We can replace $Y$ be $U$, so we may as well assume $Y$
is affine.  Let $A(Y)$ be the affine coordinate ring.  Then there are
maximal ideals $\fm, \fn \subset A(Y)$ such that $\oh_{Y,P} =
A(Y)_\fm$ and $\oh_{Y,Q} = A(Y)_\fn$.  If $\oh_{Y,Q} \subset
\oh_{Y,P}$, we must have $\fm \subset \fn$.  But $\fm$ is a maximal
ideal, so $\fm=\fn$, so $P=Q$ (as maximal ideals are in 1-1
correspondence with points, by the Nullstellensatz).  \qed

Also used in Class 19 or 20.  Compare also Lemma I used in proof of
Key Technical Theorem.

{\bf Key technical Theorem (tricky).} Let $K$ be a 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}.

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.

{\bf Coming up.} I'll prove the ``key technical theorem''.  Then I'll
prove that if $X$ is 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 morhpism $\bar{\phi} : X \rightarrow Y$
extending $\phi$.  From there, we will prove that every nonsingular
curve is quasiprojective, so every curve is birationally equivalent to
a nonsingular projective curve.

} % end of parskip; it started just before the introduction
\end{document}
