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\begin{document}
\pagestyle{plain}
\title{Introduction to Algebraic Geometry, Class 14}
\author{Ravi Vakil}
\date{Tuesday, October 26, 1999.}
\maketitle
\tableofcontents

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

Problem sets back at end.

If pages 26--27 of Eisenbud-Harris are missing in your copy, let me
know; Alex Ghitza has kindly made a bunch of copies.

Come by and ask me questions!

Last time I said that we would call a dominant rational map $Y
\dashrightarrow X$ {\em finite} if the induced morphism of function
fields $k(X) \dashrightarrow k(Y)$ is a finite extension.  That's not
really in keeping with standard usage, so let me redefine it as {\em
generically finite}.

\section{Dimension}

I'll start by reviewing what I mentioned last time on dimension.

But first, let me make a few algebraic remarks, on things such as
transcendence degrees of finitely-generated field extensions.  You
should think about the statements I make; I think you'll find them all
believable.  Of course, one needs to properly prove everything.
If you've already seen these remarks, great; if you're taking Commutative
Algebra, that's great too; and otherwise, you should convince yourself that
the statements are reasonable, and treat them as black boxes.

\subsection{Last time}


{\bf Definition.} If $X$ is a prevariety defined over $\kb$, define
$\dim X = tr. d._{\kb} k(X)$.  If $Z$ is a closed subset of $X$, then
$Z$ has {\em pure dimension} $r$ if each of its components has
dimension $r$.  A variety of dimension 1 is a {\em curve}, a
variety of dimension 2 is a {\em surface}, a variety of dimension $n$
is an $n$-fold.

{\em Observation.} If $U \subset X$ is a nonempty open subset, then $k(U) =
k(X)$, so $\dim U = \dim X$.

I mentioned an algebraic fact, which I proved (albeit imperfectly).

{\bf Lemma.} Let $R$ be an integral domain over $\kb$, $\fp \subset
R$ a prime ideal.  Then $tr.d._{\kb} R \geq tr.d._{\kb} R/\fp$, with
equality iff $P=\{ 0 \}$ or both sides are infinite.

Geometrically, this translates into:

{\bf Proposition.}  If $Y$ is a closed subprevariety of $X$, then
$\dim Y < \dim X$.  

Summary of proof: can reduce to an affine open meeting $Y$ by earlier
observation, then use lemma.

{\bf Definition.}  Call the difference the {\em codimension} of $Y$ in $X$.

So for example, if the codimension is 1, there are
no other subvarieties in between $Y$ and $X$.

I then quoted a result from commutative algebra:

{\bf Theorem (Krull's Hauptidealsatz, or Principal Ideal Theorem).}
Suppose $R$ is a finitely generated integral domain over $\kb$, $f \in
R$, $\fp$ a minimal prime of $(f)$ (i.e. minimal among the prime ideals
containing it).  Then if $f \neq 0$, $tr. d. R/\fp = tr. d. R- 1$.
(Proof omitted.)

Let me repeat why, in geometrical situations, this is very reasonable.
For example, let $R = \kb[x,y,z]$, and let $f$ be some polynomial, say
$xy (x-y^3-z^4)$.  Notice that the vanishing set $V(f) = \{ f=0 \}$
consists of two planes and this weird surface.  The minimal primes
containing $(f)$ correspond to maximum subvarieties of $\A^3$
contained in the vanishing set of $f$, so there are 3 of them.
Intuitively, it is reasonable that all 3 have dimension 2.

The immediate geometric consequence of this is: 

{\bf Theorem.} Let $X$ be a variety, $U \subset X$ open, $g \in
\oh_X(U)$ a regular function on $U$, $Z$ an irreducible component of
$V(g) \cap U$.  Then if $g \neq 0$, $\dim Z = \dim X - 1$.

($U$ is a red herring, and doesn't add any complexity to the proof.
Essentially: $V(g)$ has pure codimension 1 for any non-zero $g \in
\oh_X(X)$.)

{\em Proof.}  Take $U_0 \subset U$ to be any open
affine meeting $Z$.  Let $R = \oh_X(U_0)$ be its coordinate ring, and
$f = \res_{U,U_0} g \in R$ the restriction of our function $g$.  Then
$Z \cap U_0$ (being irreducible) corresponds to some prime ideal $\fp
\subset R$.  $Z$ is a maximal irreducible subset of $V(g) \subset U$,
so $Z \cap U_0$ is a maximal irreducible subset of $V(f) \subset U_0$,
so $\fp$ is a minimal prime containing $f$, and we're in the situation
of the Hauptidealsatz.
\qed

Conversely, if $Z$ is an irreducible closed subset of $X$ of
codimension 1, then for any open $U$ meeting $Z$ and for all non-zero
functions $f$ on $U$ vanishing on $Z$, $Z \cap U$ is a component of
$f=0$.



{\bf Corollary.}  If $X$ is a variety, with subvariety $Z$ of codimension at least 2.
Then there is a subvariety of $W$ of codimension 1 containing $Z$.

{\em Proof.} We can restrict to an affine open meeting $Z$, so without
loss of generality assume $X$ is affine.  Then $Z$ corresponds to a
prime ideal $\fp \neq (0)$.  Let $f$ be any nonzero element of $\fp$.
Then all the components of $V(f)$ have codimension 1 (by the Principal
Ideal Theorem).  Hence $Z$ isn't a component, so it must be contained
in one.

{\bf Corollary.}  If $X$ is a variety, and $Z$ is a maximal closed
irreducible subset, smaller than $X$.  Then $\dim Z = \dim X - 1$.

{\bf Corollary.} Suppose $\emptyset \neq Z_0 \subset Z_1 \subset Z_2 \subset \dots
\subset Z_r = X$ (where no $\subset$ is an inclusion) is a
maximal chain of closed irreducible subsets of $X$.  Then $\dim X =
r$.

{\em Proof.}  By induction on $X$. \qed

(Discuss a bit.)

{\em Remark.} This is a good initial definition to make for schemes;
our original variety definition involves the field $\kb$.  {\em
Exercise (possibly on the next problem set).} Check that $\Spec \zed$
is a curve.

{\bf Corollary.}  Let $X$ be a variety and let $Z$ be a component of $V( (f_1, \dots, f_r) )$, where
$f_i \in \oh_X(X)$.  Then $\codim Z \leq r$.

(Explain.)

{\bf Corollary.}  
Let $U$ be an affine variety, $Z$ a closed irreducible subset.  Let $r=\codim Z$. Then there
are $f_1$, \dots, $f_r$ in $A(U) = \oh_U(U)$ such that $Z$ is a component of
$V(( f_1, \dots, f_r))$.

{\em Proof.}  We did the case $r=1$ earlier, and the proof is the same.  \qed

\subsection{An algebraic definition of dimension.}

Given a Noetherian local ring $\oh$, you can attach an integer called the {\em Krull dimension}.
(This will come up towards the end of this semester's commutative algebra class.)

It's defined as the length $r$ of the longest chain of prime ideals
$P_0 \subset P_1 \subset \dots \subset P_n=\fm \subset \oh$.
(Here $\fm$ is the maximal ideal.)  

{\bf Corollary.} The dimension of a variety $X$ is the Krull dimension
of any of the stalks of the structure sheaf.

{\em Proof.} Fix a point $p$. Translating the Krull definition into
geometry, we're asking about the longest chain of subvarieties of $X$
containing $p$.  But we already know that this is $\dim X$.  \qed

\subsection{Other facts that are not hard to prove} (Proofs are omitted for the first 2.)


{\bf Proposition.}  If $X$ is an affine variety with coordinate ring $R$, where
$R$ is a unique factorization domain.  Then every closed codimension 1 subset
equals $V((f))$ for some $f \in R$.

{\bf Proposition.}  $\dim X \times Y = \dim X + \dim Y$.

{\bf Proposition.}  The Zariski topology on a dimension 1 prevariety is the cofinite topology.

(Explain.)  We'll use this later when we study curves.

\section{Non-singularity:  a beginning} 

For a reference, see Hartshorne I.5 or Shafarevich, the start of Ch. II.

Some intuition, in the classical topology.  Consider the plane curve $y = x + x^2$ in $\com^2$.
Why is it smooth?  What is the tangent line?  (Discuss.)

What about $y=x+z+y^2$, $x=y+z+x^3$?  How about $y=x+z+y^2$, $y=x+z+y^4$?

There are no constant terms.  All we care about are the linear terms.  In the ring $\kb[x,y,z]$
with maximal ideal $\fm$, we care about $\fm/\fm^2$.

Classically, something is smooth of dimension $n$ if there is a local
isomorphism with $\com^n$.  I'll let you check that this is the same as the following definition.

{\bf Definition.} Let $Y$ be a dimension $d$ affine variety in $\A^n$
(with coordinates $x_1$, \dots, $x_n$.  Suppose $Y$ is defined by
equations $f_1$, \dots, $f_t$ (i.e. $I(Y)$ is generated by the $f_i$;
recall that any ideal in $\R$ is finitely generated!).  Warning: We
know that $t$ is at least the codimension $n-d$, but they two aren't
necessarily equal! Then $Y \subset \A^n$ is {\em nonsingular at a point $p \in Y$}
if the rank of the {\em Jacobian matrix} $(
\partial f_i / \partial x_j (p) )_{i,j}$ is $n-d$.  

I still must convince you that this is a reasonable definition, and in
particular, that this agrees with the old definition.  I'll let you
check that in the classical topology, if $Y$ is nonsingular at $a$ in
this sense, then there is a (classical) neighbourhood of $a$
isomorphic to an open set in $\com^n$.  I'll do an example first.

Consider our earlier example, $x-y+z+y^2=0$, $-x+y+z+x^3 = 0$.  So our
point is the origin, $f_1 = x-y+z+y^2$, $f_2 = -x+y+z +x^3$.  The two intersect in 
a curve, of dimension 1.  The Jacobian
matrix is
$$J= \left(
\begin{array}{ccc}
1 & -1 & 1 \\
-1 & 1 & 1 \end{array} \right).
$$
Indeed the rank of the matrix $J$ is 3-1=2.

I argued (with too much hand-waving) that the implicit function
theorem shows (in the case $\kb=\com$) that the Jacobian condition
implies that $X$ is a manifold at $p$; in the case above, consider the
inverse of the projection $(x,y,z) \mapsto t=x+y+z$.

\subsection{A more algebraic definition of nonsingularity; hence nonsingularity is intrinsic}

You'd think that the nonsingularity of a point of $Y$ wouldn't depend
on how you stuck it in an affine space, and you'd be right; but the
above definition {\em does} depend on that, so it isn't clear that
nonsingularity really is intrinsic.
We'll show this now.

{\bf Algebraic Definition.} Let $A$ be a noetherian local ring with
maximal ideal $\fm$ and algebraically residue field $\kb$.  Then $A$
is a {\em regular local ring} if $\dim_k \fm/\fm^2 = \dim A$.

The reason this will be relevant is:  

{\bf Theorem.} Let $Y \subset \A^n$ be an affine variety.  Let $p \in
Y$ be a point.  Then $Y$ is nonsingular at $p$ if and only if the
local ring $\oh_{Y,p}$ is a regular local ring.

We'll prove this soon.

Thus the concept of nonsingularity is intrinsic, so we can make the following definitions:

{\bf Definition.}  Let $Y$ be any prevariety.  Then $Y$ is {\em nonsingular at a
point $p \in Y$} if the local ring $\oh_{Y,p}$ is a regular local
ring; otherwise it is {\em singular at $p$}.  $Y$ is {\em nonsingular}
if it is nonsingular at any point.  Otherwise it is {\em singular}.

{\em Remark.} To check that something is singular, it is still easier
to use the Jacobian definition.  We make this more general definition
because it is, well, more general.

{\bf Theorem.} Let $A$ be the localization of $\R$ at the origin, so
$A$ has dimension $n$.  Then $\fm / \fm^2$ are naturally isomorphic to
the vector space $(\al_1, \dots, \al_n) \in \kb^n$ (call it $V$),
where points of the vector space can be associated with linear forms
$\al_1 x_1 + \dots \al_n x_n$.  Hence $A$ is a regular local ring.

The proof will be enlightening (hopefully) for several reasons.

{\em Proof.}  The morphism $V \rightarrow \fm / \fm^2$ is just given by 
$$(\al_1, \dots, \al_n) \mapsto  \al_1 x_1 + \dots \al_n x_n.$$
The morphism $\fm / \fm^2$ is given by
$$
f \in A \mapsto (\partial f / \partial x_1 (0,...,0) , \dots, \partial f / \partial x_n (0,...,0))
$$
(where $f$ vanishes at the origin). 
To show that this is well-defined, we need to check that if $f \in \fm^2$, then
$f \mapsto (0,...,0)$.  But if $f \in \fm^2$, then $f = \sum_i g_i h_i$, where $g_i$, $h_i$ are 
in $\fm$.    Then by the chain rule, $\partial (g_i h_i)  / \partial x_1 = g_i \partial h_i / \partial x_1 + 
h_i \partial g_i / \partial x_1$, so indeed if $f \in \fm^2$ then $f \mapsto (0,...,0)$

Finally, we need to show that they compositions are the identity.  The map $V \mapsto \fm/ \fm^2 \mapsto V$
is the identity (show it), so what's left is to show that $\fm / \fm^2 \mapsto V \mapsto \fm / \fm^2$ is
also the identity; this comes down to the fact that if $f \mapsto 0$, then $f \in \fm^2$.
\qed

{\em Important observation.} Notice that the elements of $\fm / \fm^2$
are naturally identified with linear functions on $\A^n$.  Now $\A^n$
can canonically be identified with the tangent space of $\A^n$ at the
origin.  So we've made an identification of $\fm / \fm^2$ with the
{\em cotangent space} of $\A^n$ at the origin.

Based on this observation:

{\bf Definition.} Let $(A,\fm)$ be the local ring of a point $p \in
Y$.  Call $\fm/\fm^2$ the {\em Zariski co-tangent space} to $Y$ at $p$,
and $(\fm/\fm^2)^*$ the {\em Zariski tangent space}.

{\em Exercise (that I will give on Thursday).} Suppose $f: X \rightarrow Y$ is a morphism of
varieties, with $f(p) = q$.  Show that there are natural morphisms
$f^*: \fm_q / \fm_q^2 \rightarrow \fm_p / \fm_p^2$ (the induced map on cotangent spaces)  and $f^*: (\fm_p /
\fm_p^2)^* \rightarrow (\fm_q / \fm_q^2)^*$ (the induced map on tangent spaces).  (If you imagine what is
happening on the level of tangent spaces and cotangent spaces of
smooth manifolds, this is quite reasonable.)  If $\phi$ is the
vertical projection of the parabola $x=y^2$ onto the $x$-axis, show
that the induced map of tangent spaces at the origin is the zero map.

Coming next:  Examples.  Checking for nonsingularity in projective space.  The singular points form a closed subset.


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