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\begin{document}
\pagestyle{plain}
\title{Introduction to Algebraic Geometry, Class 15}
\author{Ravi Vakil}
\date{Thursday, October 26, 1999.}
\maketitle
\tableofcontents

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

Hand in problem sets at end.

Scheme aside:  All varieties can be considered as schemes.


\section{Non-singularity}

Last day:

{\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 $n-d$.  

{\em Remark.}  Derivatives are just ``formal'',  i.e. $dx^n/dx = n x^{n-1}$, even
in characteristic $p$; there are no limits here.

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

{\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 $\dim_{\kb} \fm/\fm^2 = \dim A$.
{\em I should have made clear that the dimension here is just the dimension of 
$\fm/ \fm^2$ as a $\kb$-vector space.}

The reason this definition 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.  {\em Leave on board until
proof is complete.}

I promised you a proof, and I'll delay the proof for a few minutes after I summarize last day some more.

Hence 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 look at an affine cover and use the Jacobian definition.  

{\bf Theorem (**).} Let $A$ be the localization of $\R$ at the origin, so
$A$ has dimension $n$.  Then $\fm / \fm^2$ is 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.

{\em Remark.}  $\fm / \fm^2$ is clearly basis-free.

Recap of proof of Theorem (**):  The proof identified the vector space $V$ of
``hyperplanes through the origin'' $\al_1 x_1 + \dots + \al_n x_n=0$
with $\fm / \fm^2$.  The map from left to right is easy; the map from
right to left 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).  We checked that this is well-defined,
i.e. if $f \in \fm^2$, then this is the 0-map; this was the chain rule.

{\em Proof of Theorem (*).}
You'll notice the similarity to the above proof.

Let $p$ be a point $(a_1, \dots, a_n)$ in $\A^n$, and let
$\fa = (x_1-a_1, \dots, x_n-a_n)$ be the maximal ideal corresponding
to $p$ in $A = \kb[x_1, \dots, x_n]$.  We define a linear map $\theta: A \rightarrow \kb^n$
by 
$$ \theta(f) = \left( 
\frac {\partial f} {\partial x_1} (p), \dots, 
\frac {\partial f} {\partial x_n} (p) \right)$$
for any $f \in A$.  Notice that this is exactly the same map as in
the proof about $\A^n$.  (You should really think of the $\kb^n$ as
being the space of linear forms vanishing at the origin.)
As we said before, $\theta$ induces an isomorphism $\theta':  \fa/ \fa^2 \rightarrow \kb^n$.

Now let's bring $Y$ into the picture.  Let $\fb$ be the ideal of $Y$
in $A$, and let $f_1$, \dots, $f_t$ be a set of generators of $\fb$.
Then the rank of the Jacobian matrix $J = ( \partial f_i /\partial x_j)
(p)_{i,j}$ is just the dimension of $\theta(\fb)$ as a subspace of
$\kb^n$.  Using the isomorphism $\theta'$, this is the same as the
dimension of the subspace $(\fb + \fa^2) / (\fa^2)$ of $\fa/\fa^2$.

(You should check: $R$ is a ring, $\fm$ is a maximal ideal, $R_\fm$
localization.  Then $R/\fm \cong R_\fm / \fm R_\fm \cong \kb$, $R /
\fm^2 \cong R_\fm / \fm^2 R_\fm$, and $\fm / \fm^2 \cong
\fm R_\fm / \fm^2 R_\fm$.  Exercise on Problem Set 8.)

Now the local ring $\oh_{Y,p}$ of $p$ on $Y$ is obtained by modding out
by $\fb$ and localizing at the maximal ideal $\fa$.  Thus if $\fm$ is
the maximal ideal of $\oh_{Y,p}$, then $\fm / \fm^2 \cong \fa / (\fb +
\fa^2)$.  Counting dimensions of vector spaces, $\dim \fm / \fm^2 + rk
J = n$.

Now the dimension of the local ring $\oh_{Y,p}$ (as a ring) is the dimension of $Y$ (as a variety), so $\oh_{Y,p}$ is regular
if and only if $\dim \fm/ \fm^2 = r$.  But this is equivalent to $rk J = n-r$, which says that $p$ is a nonsingular point of $Y$.
\epf

{\em Important observation from Theorem (**).} 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 we made the following definition:

{\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.} 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.


\subsection{Examples}

It isn't hard to check for singular points, especially on
hypersurfaces in $\A^n$.  For example, consider the plane curve $y^2 =
x^3- x^2$.  (Do it.)

{\em Exercise.} Hartshorne Ex. I.5.1.  Find the singular points, and
sketch the following plane curves: (a) $x^2 = x^4 + y^4$, (b) $xy =
x^6 + y^6$, (c) $x^3= y^2+ x^4 + y^4$, (d) $x^2 y + x y^2 = x^4 +
y^4$.  Hartshorne Ex. I.5.2, some two-dimensional examples. 

Hypersurfaces in projective space are also easy, because there is a
trick.

{\em Exercise.} Suppose the characteristic of $\kb$ is 0.  Suppose a
hypersurface $Y \subset \proj^n$ is given by $f(x_0, \dots, x_n)=0$.
Show that the locus of points $p \in \proj^n$ where $\partial f /
\partial x_i(p) = 0$ for all $i$ are precisely the singular points of
$Y$.  (In particular, if $\partial f / \partial x_i(p) = 0$ for all
$i$, then $f(p) = 0$, i.e. $p \in Y$!  To see why, calculate $\sum_i
\partial f / \partial x_i$.)

As an example, consider $y^2 z - x^3 = 0$.  The only singular point is
$(x;y;z) = (0;0;1)$.

{\em Scheme examples.}

{\em Exercise.}  (a)  Show that both $\Spec \zed$ and $\Spec \zed[i]$ are nonsingular curves.
(b)  Let $\fm = (1+i)$ in $\zed[i]$.  Then under the map $f:  \Spec \zed[i] \rightarrow \zed$, $f(\fm) = (2)$. 
Check that the map on cotangent spaces (or equivalently, that the dual
map on tangent spaces) is the zero-map.  For all other primes of
$\zed[i]$, calculate the map on cotangent spaces.

\subsection{The singular points form a closed subset}

{\bf Theorem.} Let $Y$ be a prevariety.  Then the set of singular
points $\Sing Y$ is a closed subset of $Y$.

In the proof, we'll use the following (which will come up in the
Commutative Algebra course).

{\bf Commutative Algebra Fact.}  If $A$ is a noetherian local ring with maximal ideal
$\fm$ and residue field $\kb$, then $\dim_{\kb} \fm/\fm^2 \geq \dim A$.  

{\em Remark.} It is also true that $\Sing Y \neq Y$, but I won't prove
this fact.

{\em Proof of Theorem.} First step: reduce to the affine case.  It
suffices to show that for some open covering $Y = \cup Y_i$ of $Y$, 
that $\Sing Y_i$ is closed for each $i$.  So assume that $Y$ is affine.

By the above C.A. fact, we know that the rank of the Jacobian matrix
is at most $n-r$, so the set of singular points is the set of points
where the rank is {\em less than} $n-r$.  Thus $\Sing Y$ is the set
defined by the ideal $I(Y)$ together with all determinants of $(n-r)
\times (n-r)$ submatrices of $( \partial f_i / \partial x_j )_{i,j}$, which
is a closed set.  \qed


\section{Curves:  Valuation rings and nonsingular points, take 1}

Dimension 1 varieties, or curves, are particularly simple, and most of
the rest of the course will concentrate on them.

We saw that nonsingularity has to do with local rings, so we'll discuss
{\em one-dimensional local rings}.

First we'll recall some facts about discrete valuation rings and Dedekind domains.

{\bf Definition.} Let $K$ be a field.  A {\em discrete valuation} of
$K$ is a map $v: K \setminus \{ 0 \} \rightarrow \zed$ such that for
all $x$, $y$ non-zero in $K$, we have: $v(xy) = v(x)+v(y)$, $v(x+y)
\geq \min (v(x), v(y))$.  Notice that the set $R = \{x \in K | v(x)
\geq 0 \} \cup \{ 0 \}$ is a subring of $K$; call this the {\em
discrete valuation ring}, or {\em DVR}, of $K$.  The subset $\fm = \{ x \in K | v(x)
> 0 \} \cup \{ 0 \}$ is an ideal in $R$, and $(R, \fm)$ is a local
ring.  A {\em discrete valuation ring} is an integral domain which is
the discrete valuation ring of some valuation of its quotient field.  If $\kb$
is a subfield of $K$ such that $v(x) = 0$ for all $x \in \kb \setminus
\{ 0 \}$, then we say $v$ is a {\em discrete valuation of $K/\kb$},
and $R$ is a {\em discrete valuation ring of $K / \kb$}.

{\em I'll need to patch this to prevent pathologies such as the ``zero
valuation''.  More on this next day.}

{\em Example.} Let $K = \kb(t)$, and for $f \in K$, let $v(f)$ be the
order of the zero of $f$ at $t=0$ (negative if $f$ has a pole).  Check
all properties.  Notice that {\em discrete valuation ring} of of $v$
are those quotients of polynomials whose denominator doesn't vanish at
0, i.e. $\kb[t]_{(t)}$.  In geometric language, it is the stalk of the 
structure sheaf of $\A^1$ at the origin.

Also, $\kb[t]_{(t)}$ is a discrete valuation ring:  it is indeed an integral domain,
and it is the valuation ring of some valuation in its quotient field $\kb(t)$.

Similarly, we could get other valuations by replacing 0 with any other element of $\kb$.
Have we found all the valuations?  No:

{\em Example.} Let $K =\kb(t)$ as before.  For $f \in K$, write $f$ in
terms of $u = 1/t$, and let $v(f)$ be the order of zero of $f$ at
$u=0$.  Again, it is indeed a valuation, and it has geometric meaning.
(Ask them.)  It corresponds to the point of $\proj^1$ ``at $\infty$''
(when looking at it with respect to the $t$-coordinate).

{\em Fact (that we'll later prove).}  These are all the valuations of $\kb(t)$, the
function field of $\proj^1$.  They naturally correspond to the points of $\proj^1$.

{\em Example.} Let $K = \Q$.  (Ask for valuations.)  If $f \in \Q$,
let $v(x)$ be the highest power of 2 dividing $x$, so $v(14) = 2$,
$v(3)=0$, $v(13/12) = -2$.  Check all properties.  What's the discrete
valuation ring?  Those fractions with no 2's in the denominators.
Geometrically, $\Q$ is the function field of $\Spec \zed$, and the
valuations turn out to correspond to the maximal prime ideals of $\zed$,
i.e.  the ``closed points'' of $\Spec \zed$.

{\bf Coming soon.} Describing discrete valuation rings in many ways.
Integral closure.  Dedekind domains.

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