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\title{Algebraic Geometry}
\author{Zvezdelina Stankova-Frenkel\\ Mills College/UC Berkeley}
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\begin{document}
\vspace*{-10mm} \begin{center}
{\large Algebraic Geometry}
\smallskip
\hrule
\end{center}

\bigskip
%\maketitle

\medskip
\noindent{\bf Question:} What regular polygons can be inscribed in an
ellipse?  

\bigskip
\section{Varieties, Ideals, Nullstellensatz}

\bigskip 
Let $K$ be a field. We shall work {\it over} $K$, meaning,
our coefficients of polynomials and other scalars will lie in $K$.

\medskip \noindent{\bf Definitions.}

\begin{itemize}
\item[1.]  An {\it affine variety} $X$ in $\mathbb{A}^n$ is the zero 
locus of a collection of polynomials $\{f(x_1,...,x_n)\}$ in 
$K[x_1,...,x_n]$.  A {\it projective variety} X in $\mathbb{P}^n$ 
is the zero locus of a collection of {\it homogeneous} polynomials 
$\{F(Z_0,...,Z_n)\}$ in $K[Z_0,...,Z_n]$.

\medskip \item[2.]  For a given variety $X$, the set of {\it all} 
polynomials vanishing on $X$ is an ideal, called the {\it ideal} of 
$X$ and denoted by $I(X)$.  In other words, 
\[I(X)=(...,f_{\alpha},...)\,\,\text{or}\,\,I(X)=(...,F_{\alpha},...),\] 
depending on whether the variety is affine or projective.  In the 
second case, the ideal is called {\it homogeneous}, i.e.  generated by 
homogeneous polynomials.  Conversely, for a given ideal $I\subset 
K[x_1,...,x_n]$ (or homogeneous $I\subset K[Z_0,...,Z_n]$), the 
zero locus of $I$ is denoted by $Z(I)$.
\end{itemize}


\bigskip Projective varieties can be thought of as ``completions'', 
``compactifications'', or ``closures'' of affine varieties. Their
global properties are usually easier to describe than those  
of affine varieties. Conversely, affine varieties can be thought of as
building blocks of projective varieties (indeed, they constitute
an open cover), and hence local properties are easier to describe
using affine varieties. However, projective varieties vary ``nicely''
in families and hence parametrizing and moduli spaces are usually
constructed for projective varieties with certain defining common 
properties.

\begin{itemize}
\bigskip \item[3.]  A ring $R$ is called {\it Noetherian} if any 
inscreasing sequence of ideals terminates, i.e.  whenever $I_j$'s are 
ideals in $R$ such that \[I_1\subset I_2\subset \cdots \subset I_j 
\subset \cdots\] then for some $k\geq 1$: $I_k=I_{k+1}=I_{k+2}=\cdots$

\medskip
\item[4.] An ideal $I$ in a ring $R$ is called {\it radical}
if whenever $f^m\in I$ ($f\in R$, $m\in\mathbb{N}$), then $f\in I$.
In other words, $I$ contains all (positive integer) roots of
its elements. 
\end{itemize}

\bigskip \begin{thm} The polynomial ring $R=K[x_1,x_2,...,x_n]$ is 
Noetherian.  Consequently, any ideal $I$ of $R$ is finitely generated.  
In particular, for any affine (or projective) variety $X$, the ideal 
$I$ of $X$ is generated by finitely many polynomials.
\end{thm}

\begin{lem} The ideal of any algebraic variety $X$ is radical.  If $I$ 
is an arbitrary ideal, the set of all radicals of its elements: 
\[\sqrt{I}:=\{f\in R\,\,|\,\,f^n\in I\,\,\text{for 
some}\,\,n\in\mathbb{N}\}\] is also an ideal, called the {\it radical} 
of $I$.  If $I$ is a radical ideal, then its radical is itself, i.e.  
the operation of taking radicals stabilizes after one step.
\end{lem}

We construct maps bewteen the set $\mathcal{X}$ of all varieties $X$ 
over $K$ and the set $\mathcal{J}$ of all ideals $J\subset 
K[x_1,...,x_n]$ by sending \[i:X\mapsto I(X)\,\,\, 
\text{and}\,\,\,j:J\mapsto Z(J).\] It is immediate from definition of an 
ideal of variety that $Z(I(X))=X$, i.e.  $j\circ i=\text{id}_{\mathcal 
X}$.  Also, the image of $i$ is inside $\mathcal{R}$, the subset of 
radical ideals in $R$.  Thus, we have an injection 
$i:\mathcal{X}\hookrightarrow \mathcal{R}$ with a one--way inverse 
$j:\mathcal{R}\rightarrow \mathcal{X}$.  It is natural to ask whether 
$i$ and $j$ are inverses of each other, i.e.  whether $i\circ 
j=\text{id}_{\mathcal R}$.

\medskip For any ideal $J\subset \mathcal{J}$ (not necessarily 
radical), we consider $X:=Z(J)=j(J)$ -- a variety, and then take 
$i(X)=I(X)=I(Z(J))$.  It is evident that $I(X)$ will be a radical 
ideal containing $J$, but is it going to be $\sqrt{J}$?  To 
paraphrase the problem, start with $J$ being a radical ideal and take 
$I(Z(J))$.  Is this equal to $J$?

\medskip The answer in general is {\it no}.  For example, if 
$K=\mathbb{R}$ is the ground field, and $J=(x^2+y^2)$ is the ideal 
generated by the single polynomial $f(x,y)=x^2+y^2$ in the affine 
plane, then $J$ is obviously radical ($f$ is irreducible), and the 
zero locus of $J$ is $Z(J)=(0,0)$ -- just one point.  However, the 
ideal of $(0,0)$ is definitely much larger than $J$ -- it consists of 
all polynomials vanishing at $(0,0)$, i.e.  having no free terms: 
$I((0,0))=(x,y)\supset (x^2+y^2)$.  Thus, we end up with a (radical) 
ideal bigger than the original.

\medskip The above situation is possible because $\mathbb{R}$ is not 
an algebraically closed field.  This leads to the famous {\it 
Nullstelensatz}, a basic theorem in commutative algebra, on which much 
of algebraic geometry over algebraically closed fields is based.

\begin{thm} (Nullstelensatz) If $K$ is an algebraically closed field, 
then for any ideal $J\subset K[x_1,...,x_n]$:
\[i\circ j(J)= I(Z(J))=\sqrt{J}.\]
In particular, there is a one--to--one correspondence between
the set $\mathcal{X}$ of affine varieties $X$ in $\mathbb{A}^n$ and 
radical ideals $\mathcal{R}$ given by
\[i:\mathcal{X}\rightarrow \mathcal{R},\,\,\text{and}\,\,j:\mathcal{R}
\rightarrow 
\mathcal{X}.\]
\end{thm}

Note that the radical of the unit ideal is again the unit ideal:
$\sqrt{(1)}=(1)$. This implies the following corollary:

\begin{cor} If $f_1,...,f_k$ are polynomials in several variables over 
an algebraically closed field $K$, then they have no common zeros in 
$K$ iff 
\[1=g_1f_1+g_2f_2+\cdots+g_kf_k\]
for some polynomials $g_1,g_2,...,g_k$.
\end{cor}

Some other ``strange'' things happen over fields, which are not 
algebraically closed.  For example, we would like to call
a ``planar curve'' any variety $X$ in $\mathbb{A}^2$ which is
given by 1 polynomial. However, over $\mathbb{R}$, the ``curve''
defined by $x^2+y^2=0$ is really just a point, while over $\mathbb{C}$
(or any algebraically closed fields) it is a pair of intersecting
lines. Thus, many interesting and intuitive properties
of algebraic varieties hold only over algebraically closed fields.

\medskip There is an analog of Nullstelensatz for projective varieties 
(for $K=\overline{K}$, of course.)  There is one subtle point, though.  
We call $Ir=(Z_0,...,Z_n)$ the {\it irrelevant ideal} in 
$K[Z_0,Z_1,...,Z_n]$.  Note that $Ir$ is radical, and that 
$Z(Ir)=\emptyset$.  Yet, $Ir$ is {\it not} the whole ideal of 
$\emptyset$: $(1)=K[Z_0,Z_1,...,Z_n]$, the unit ideal, is the ideal of 
$\emptyset$.  Thus, we have two radical ideals competing for the 
$\emptyset$: $Z(Ir)=Z((1))=\emptyset$.  The bigger one ``wins'', 
because $I(\emptyset)=(1)$, and we state the Nullstelensatz as 
follows:

\begin{thm} There is a one--to--one correspondence between the set
$\mathcal{X}$ of projective
varieties $X\subset \mathbb{P}^n$ and the set $\mathcal{R}$
of radical homogeneous ideals minus $Ir$ given by $i$ and $j$ from
above. In particular, for $J\in\mathcal{J}$:
\[I(Z(J))=\left\{\begin{array}{ll}
\sqrt{J} & \text{if}\,\,Z(J)\not = \emptyset\\
(1)      & \text{if}\,\,Z(J)=\emptyset.
\end{array}\right.\]
\end{thm}

Note further that for a (homogeneous) ideal $J$, $Z(J)=\emptyset$ iff 
$\sqrt{J}=(1)$ or $\sqrt{J}=(Z_0,Z_1,...,Z_n)$.  In both cases, 
$\sqrt{J}\supset (Z_0,Z_1,...,Z_n)$, which can be shown to imply 
$J\supset (Z_0,Z_1,...,Z_n)^d$ for some $d>0$.

\begin{prop} Let $J$ be a homogeneous ideal.  Then $Z(J)=\emptyset$ 
iff $J$ contains a power of the irrelevant ideal.  In other words, a 
collection of homogeneous polynomials $\{F_{\alpha}\}$ will have no 
common zeros iff the ideal generated by the $F_{\alpha}$'s contains 
all (homogeneous) polynomials of a certain degree $d>0$.
\end{prop}

\newpage
\section*{Examples of Varieties}

\medskip
\noindent\underline{\bf Example 1.} Let $Q$ be the (smooth)
quadric surface in $\mathbb{P}^3$ given as the zero locus
of one homogeneous quadratic equation:
\[Q=Z(Z_0Z_3-Z_1Z_2).\]
\vspace*{30mm}

The quadric $Q$ consists of two families of lines, each of which sweeps
$Q$ on its own:
\[\{Z_1=\lambda Z_0,\,\,Z_3=\lambda Z_2\}\,\,\text{and}\,\,\{Z_2=\mu 
Z_0,\,\,Z_3=\mu Z_1\}.\] In terms of the matrix 
\[M=\left(\begin{array}{ll} Z_0& Z_1\\ Z_2& Z_3 \end{array}\right),\] 
$Q$ is the locus where $\text{det}M=0$, one family of lines consists 
of lines where the two column satisfy a given linear relation, the 
other family -- where the two rows satisfy a given linear relation.  Note 
that two lines in a family do not intersect (they are skew lines in 
$\mathbb{P}^3$), while any two lines from different families intersect 
in exactly one point.  The latter hints at an alternative description 
of $Q$ -- namely as the ``Cartesian product'' $\mathbb{P}^1\times 
\mathbb{P}^1$.

\vspace*{50mm}

More generally, for any two varieties $X$ and $Y$, there is a (unique) 
variety $X\times Y$ with projection maps $\pi_1:X\times Y\rightarrow 
X$ and $\pi_2:X\times Y\rightarrow Y$, the fibers of which are 
correspondingly copies of $Y$ and of $X$.  The uniqueness of such a 
variety is ensured by some extra (natural) properties coming from 
category theory -- such properties are exactly what we would expect 
from a variety deserving to be called the ``product'' of two 
varieties.

\bigskip The actual construction of this product is given by a 
(seemingly random) map, called the {\it Segre embedding}.  For starters, to 
construct the product $\mathbb{P}^n\times \mathbb{P}^m$, we define the 
Segre map by \[\sigma_{n,m}:\mathbb{P}^n_{[X]}\times 
\mathbb{P}^m_{[Y]}\rightarrow \mathbb{P}^{(n+1)(m+1)-1}_{[Z]}\]
\[\sigma_{n,m}([X_0,...,X_n],[Y_0,...,Y_m])=[X_0Y_0,...,X_iY_j,...,X_nY_m].\] 
If we label the coordinates on 
$\mathbb{P}^{(n+1)(m+1)-1}_{[Z]}=\mathbb{P}^N$ by $Z_{ij}$ for $0\leq 
i\leq n$ and $0\leq j\leq m$, then the image of $\sigma_{n,m}$ is the 
zero locus of all possible quadratic equations 
$Z_{ij}Z_{kl}=Z_{il}Z_{kj}$.  Thus, the image of the Segre map is a 
variety.  Factoring into this the injectivity of $\sigma_{n,m}$, 
allows us to ``identify'' the {\it set}--product $\mathbb{P}^n\times 
\mathbb{P}^m$ with a subvariety of $\mathbb{P}^N$, which is {\it 
define} as the {\it product} of $\mathbb{P}^n$ and $\mathbb{P}^m$ in 
the {\it category of varieties}.  One can show that this particular 
variety satisfies all the required properties of a product, and by 
virtue of the {\it uniqueness} of the product, it is {\it The product 
variety} $\mathbb{P}^n\times\mathbb{P}^m$.

\medskip Now it is not hard to construct the product of any two 
varieties $X$ and $Y$: if $X\subset \mathbb{P}^n$ and $Y\subset 
\mathbb{P}^m$, then $X\times Y$ is the subvariety of 
$\mathbb{P}^n\times \mathbb{P}^m$, satisfying the extra equations 
coming from $X$ and $Y$.

\bigskip
\noindent\underline{\bf Example 2.} The {\it twisted cubic} curve
$C$ is the zero locus of three polynomials:
\[C=Z(Z_0Z_3-Z_1Z_2, Z_0Z_2-Z_1^2, Z_1Z_3-Z_2^2).\]

\vspace*{40mm}

\noindent The twisted cubic $C$ lies on the quadric surface $Q$,
and is a curve of type $(2,1)$ on $Q$, i.e. $C$ meets every line in
one family in 2 points, and every line in the other family - in 1 
point (prove this!) Prove also that the zero locus of any two of the
three quadratic polynomials defining $C$ is the union of $C$ and a line
on $Q$ meeting $C$ in two points (or being tangent to $C$). How many
such tangent lines to $C$ are there in the family consisting 
generically of lines meeting $C$ in two points?

\medskip An alternative description of $C$ is the image of the {\it 
Veronese embedding} of $\mathbb{P}^1$ in $\mathbb{P}^3$: 
\[C=\{[X^3,X^2Y,XY^2,Y^3]\,\,|\,\,[X,Y]\in\mathbb{P}^1\}=\nu_3(\mathbb{P}^1).\] 
Show that this is indeed an embedding, and that its image indeed
coincides with the twisted cubic curve $C$.

\bigskip
\noindent\underline{\bf Example 3.} What are the hyperplane sections
of $Q$, i.e. $\mathbb{P}^2\cap Q$, as the hyperplane $\mathbb{P}^2$
varies in $\mathbb{P}^3$?

\vspace*{60mm}

The strategy here is to restrict the equation of $Q$ to the hyperplane
$\mathbb{P}^2$, and to realize that a homogeneous quadratic
polynomial in 3 variables is either irreducible (smooth plane
conic hyperplane section of $Q$),  or factors as the product of
two homogeneous linear factors (the hyperplane section here is the
union of two intersecting lines in $\mathbb{P}^2$). Prove that we will 
never get the quadratic polynomial to factor as a perfect square
of a linear form (i.e. no hyperplane in $\mathbb{P}^3$ intesects
$Q$ in a ``double line'').

\medskip We push the above considerations an inch further to ask the 
following question: can we construct a variety $\mathcal{H}_{Q}$ which 
in some reasonable way will be the family of {\it all} hyperplane 
sections of $Q$?  In other words, can we separate all hyperplane 
sections of $Q$, so that they do not intersect anymore, but stay as 
fibers of some map?  The answer is ``YES'', yet we have to work a bit 
to construct this variety.

\medskip
For starters, we can construct the {\it universal hyperplane} $\mathcal{H}$
in $\mathbb{P}^3$ -- apriori, we want this to be a variety, 
representing {\it all} hyperplanes in $\mathbb{P}^3$, in other words,
$\mathcal{H}$ should be a family of all hyperplanes in $\mathbb{P}^3$.
The first step is to realize what variety {\it parametrizes} these 
hyperplanes -- this is the {\it dual} $(\mathbb{P}^3)^*$,
which is really $\mathbb{P}^3$ all over again, by with different 
coordinates. If $\mathbb{P}^3$ has coordinates $[Z_0,Z_1,Z_2,Z_3]$,
then a hyperplane in $\mathbb{P}^3$ is given by a linear form
$W_0Z_0+W_1Z_1+W_2Z_2+W_3Z_3=0$ for some fixed $[W_0,W_1,W_2,W_3]\in
(\mathbb{P}^3)^*$. Thus, the dual $(\mathbb{P}^3)^*$ has coordinates
$W_0,W_1,W_2,W_3$; points in $(\mathbb{P}^3)^*$ correspond to
hyperplanes in $\mathbb{P}^3$, and hyperlanes in $(\mathbb{P}^3)^*$
correspond to points in $\mathbb{P}^3$.

\medskip Now, the universal hyperplane $\mathcal{H}$ should be defined 
as a set by \[\mathcal{H}=\{[p,H]\,\,|\,\,p\in\mathbb{P}^3,\,\,H\in 
(\mathbb{P}^3)^*, \,\,p\in H\}.\] Therefore, we naturally consider 
$\mathcal{H}\in \mathbb{P}^3\times (\mathbb{P}^3)^*$ - it is the zero 
locus of a single {\it bihomogeneous} polynomial: 
\[W_0Z_0+W_1Z_1+W_2Z_2+W_3Z_3=0.\] In terms of the coordinates 
$T_{ij}$ on $\mathbb{P}^{15}$, in which $\mathbb{P}^3\times 
(\mathbb{P}^3)^*$ is embedded, this polynomial reads: 
\[T_{00}+T_{11}+T_{22}+T_{33}=0.\] We conclude that $\mathcal{H}$ is a 
subvariety of the product $\mathbb{P}^3\times (\mathbb{P}^3)^*$; more 
precisely, it is a hyperplane section of $\mathcal{H}$ inside 
$\mathbb{P}^{15}$.

\vspace*{60mm}

\medskip Finally, to construct the {\it universal hyperplane section} 
$\mathcal{H}_Q$ of the quadric $Q$, we only have to intersect with 
$\mathcal{H}$: \[\mathcal{H}_Q=Q\cap \mathcal{H}\subset 
\mathbb{P}^3\times (\mathbb{P}^3)^*.\] Since the product on the right 
has two natural projections $\pi_1$ and $\pi_2$ onto the two factors 
$\mathbb{P}^3$ and $(\mathbb{P}^3)^*$, we can restrict these maps to 
$\mathcal{H}_Q$, and ask what the fibers of $\pi_1$\and $\pi_2$ are.  
Prove that the fibers of $\pi_1$ are all (isomorphic to) 
$\mathbb{P}^3$, while the fibers of $\pi_2$ are the hyperplane 
sections of $Q$ -- what we wanted in the first place.

\smallskip
Maps of varieties $\phi:X\rightarrow Y$ whose fibers are all 
isomorphic to some $\mathbb{P}^k$ are called $\mathbb{P}^k${\it -- 
bundles over} $Y$.  Sometimes it is important to classify all such 
bundles over a fixed variety $Y$ -- this describes additional 
invariants of $Y$, which may be used for instance to identify two
non--isomorphic varieties.

\bigskip \noindent\underline{\bf Example 4.} As we saw above, all 
hyperplanes in $\mathbb{P}^3$ can be parametrized by the variety 
$(\mathbb{P}^3)^*$ (which is isomorphic to $\mathbb{P}^3$.)  In this 
case, the points of $(\mathbb{P}^3)^*$ are in 1--1 correspondence with 
the hyperplanes in question, and $(\mathbb{P}^3)^*$ reflects (in a 
certain sense) how the hyperplanes vary in $\mathbb{P}^3$ - that is, 
for any ``nice'' family $\mathcal{F}$ of hyperplanes in $\mathbb{P}^3$ 
the subset of $(\mathbb{P}^3)^*$ corresponding to $\mathcal{F}$ is a 
{\it subvariety} of $(\mathbb{P}^3)^*$.  (The word ``nice'' has a very 
technical meaning, usually called ``flatness'' of families.  We shall 
not discuss this here since it will take us too far afield.)

\smallskip One can easily generalize the above construction to 
parametrize all hyperplanes in $\mathbb{P}^n$ by the dual projective 
space $(\mathbb{P}^n)^*$. A natural question arises: Can we find
varieties parametrizing other objects, say, conics in $\mathbb{P}^2$?
Such varieties are called {\it parameter spaces}.

\smallskip Let $\mathcal{P}$ be the set of all conics in 
$\mathbb{P}^2$.  If we fix the coordinates of $\mathbb{P}^2$ to be 
$X,Y,Z$, then a conic $C$ is determined upto a scalar by a quadratic 
equation: \[a_0X^2+a_1Y^2+a_2Z^2+a_3XY+a_4YZ+a_5ZX=0.\] Such an 
equation determines a projective point $[a_0,a_1,...,a_5]\in 
\mathbb{P}^5$, and conversely, any point of $\mathbb{P}^5$ determines 
a unique quadratic polynomial (upto a scalar).  Thus, the parameter 
space for all conics in $\mathbb{P}^2$ is $\mathcal{P}=\mathbb{P}^5$.

\smallskip
Similarly, the parameter space of all hypersurfaces of degree $d$ in
$\mathbb{P}^n$ (i.e. subvarieties given by single degree $d$ homogeneous
polynomials on $\mathbb{P}^n$) is $\mathcal{P}_{d,n}=\mathbb{P}^{N}$
where $N=\binom{n+d}{d}-1$. Note a slight technicality here: we have 
included as points in $\mathcal{P}_{d,n}$ ``hypersurfaces'' 
corresponding to polynomials with multiple factors. For example,
in the case of conics in $\mathbb{P}^2$, we included as points in
$\mathcal{P}_{2,2}$ all ``double'' lines. One can show that the
set of such ``multiple'' (or more precisely, {\it non-reduced})
hypersurfaces is in fact a subvariety of the corresponding parameter
space $\mathcal{P}_{d,n}$. 

\bigskip \noindent\underline{\bf Example 5.} Parameter spaces 
parametrize usually not just objects $X$ sharing some common 
properties, but also the embeddings of $X$ in projective space.
For example, there are really only three types of conics in 
$\mathbb{P}^2$ -- the irreducible (smooth) conics, the joins of
two different lines, and the double lines. Every irreducible conic
can be transformed into any other irreducible conic after a suitable
change of variables (coordinate change) on $\mathbb{P}^2$, etc.
Thus, in constructing $\mathcal{P}_{2,2}$, we grossly ``overcounted''
the irreducible conics (well, we were parametrizing, therefore,
not just the conics, but the pairs $(C,\phi)$ where $C$ is a plane
conic and $\phi:C\hookrightarrow \mathbb{P}^2$ is an embedding.)

\medskip The philosophy of viewing a variety as an object with a given 
embedding in some $\mathbb{P}^n$ is inherent to XIX century algebraic 
geometry, especially to the Italian school.  XX century changed this 
view by considering varieties as objects on their own, disregarding 
particular embeddings in projective space.  For example, any 
irreducible conic in $\mathbb{P}^2$ is really a $\mathbb{P}^1$ 
embedded in a certain way in $\mathbb{P}^2$: 
\[\nu_2:\mathbb{P}^1\hookrightarrow \mathbb{P}^2,\,\, 
\nu_2([X,Y])=[X^2,XY,Y^2].\] Similarly, the twisted cubic $C$ in 
$\mathbb{P}^3$ is isomorphic to $\mathbb{P}^1$: 
$C=\nu_3(\mathbb{P}^3)$.  We say that these curves are isomorphic to 
each other because there exist nicely defined maps via polynomials 
going back and forth between these varieties, whose compositions are 
identities.  Thus, the {\it intrinsic} properties of $\mathbb{P}^1$ 
are preserved under these isomorphism, and therefore the embeddings 
do not change the actual variety.

\medskip Some {\it extrinsic} properties, however, change, and these 
cause the different embeddings of $\mathbb{P}^1$ to look different.  
For example, define the {\it degree} of $X\cong\mathbb{P}^1\subset 
\mathbb{P}^n$ to be the number of points in the intersection of a 
general hyperplane in $\mathbb{P}^n$ with $X$.  Thus, the conics in 
$\mathbb{P}^2$ have degree $2$, and will keep their degree if we embed 
now $\mathbb{P}^2$ as a linear subspace of a bigger $\mathbb{P}^n$.  
However, the twisted cubic $C$ in $\mathbb{P}^3$ has degree $3$ (one 
way to see this is to recall that a line in one ruling meets $C$ in 1 
point, while a line in the other ruling meets $C$ in 2 points.)

\medskip While parameter spaces may take into account such extrinsic 
properties as {\it degrees} of varieties, {\it moduli spaces} usually 
parametrize objects according to only their intrinsic properties, and 
hence are much harder to be constructed.  To even state what common 
instrinsic properties can be characterized will take too much ink on 
this handout.  But let us mention one very famous example -- the 
moduli space $\mathfrak{M}_g$ of {\it smooth curves of genus $g$}.  
\footnote{Be forwarned that what follows in this section is littered 
with too many unexplained terms in order for the text to be 
self--contained.} These curves do not lie (and cannot be embedded in 
general) in the same projective space $\mathbb{P}^n$.  The best we can 
say is that each such (non--hyperelliptic) curve can be embedded in 
$\mathbb{P}^{2g-3}$, but we don't care about these embeddings anyways.  
Yet, $\mathfrak{M}_g$ can be constructed, and it is a variety of 
dimension $3g-3$ for $g\geq 2$.  For $g=1$ there is only one such 
curve -- $\mathbb{P}^1$, so $\mathfrak{M}_1$ is really just one point; 
for $g=1$ -- the elliptic curves can be effectively parametrized by a 
certain cross ratio, and hence $\mathfrak{M}_1\cong \mathbb{P}^1$.

\vspace*{50mm}

\medskip A further development of this theory is the {\it 
Deligne-Mumford} compactification of $\mathfrak{M_g}$.  Since 
$\mathfrak{M}_g$ is {\it not} a projective variety, one can have a 
nice family of {\it smooth curves} degenerating to a singular curve, 
but $\mathfrak{M}_g$ does not have any points to reflect the limiting 
singular member of the family.  The question arises -- what is the 
``minimal'' set of {\it singular curves} must be added to the set of 
smooth curves in order to obtain a ``nice'' moduli space 
$\overline{\mathfrak{M}}_g$, compactifying $\mathfrak{M}_g$?  Deligne 
and Mumford chose (and for very good reasons) the set of the 
so--called {\it stable} curves $C$ of genus $g$.  These are connected 
curves with at most {\it nodal} type of singularities (e.g.  take two 
lines intersecting in $\mathbb{P}^2$), and such that if they contain a 
$\mathbb{P}^1$--component, then the latter must meet at least 3 other 
components of the curve.  The last condition is added to ensure that 
the curves have finite groups of automorphisms.  With this said, 
$\overline{\mathfrak{M}}_g$ is {\it the moduli space of all stable 
curves of genus $g$}.  It is a projective variety which contains 
$\mathfrak{M}_g$ as an open dense set, and it reflects naturally the 
variation of ``nice'' families of stable curves.  Moreover, any 
``nice'' family whose general members are smooth curves, but whose 
special members can be as nasty as you wish, can be brought in an 
essentially one way to a family with only stable members.  This 
process is called {\it semistable reduction} and it is the basis for 
many related constructions in algebraic geometry.




\medskip



  
\bigskip \section*{Zariski Topology}

\medskip \noindent{\bf Definitions.} 

\begin{itemize}
\item[(a)] An ideal $P$ of a ring $R$ is called {\it prime} if whenever 
$ab\in P$ for $a,b\in R$, then $a\in P$ or $b\in P$ (or both).

\item[(b)] A variety $X$ is called {\it irreducible} if for any 
decomposition $X=X_1\cup X_2$ of $X$ into a union of two subvarieties, 
either $X_1=X$ or $X_2=X$.  In other words, there are no non--trivial 
decompositions of $X$ into smaller varieties.
\end{itemize}

\begin{prop} A variety $X$ is irreducible iff its ideal
$I(X)$ is prime.
\end{prop}

Thus, there exists a one--to--one correspondence between the set of 
varieties $\mathcal{X}$ and the set of prime (homogeneous if 
projective $X$) ideals $\mathcal{P}$ in the corresponding polynomial
ring.

\begin{thm} Any radical ideal $I\subset K[x_1,...,x_n]$ is uniquely
expressible as a finite intersection of prime ideals $P_i$
with $P_i\not\subset P_j$ for $i\not = j$. Equivalently, any
variety $X$ can be uniquely expressed as a finite union of irreducible
subvarieties $X_i$ with $X_i\not\subset X_j$ for
$i\not = j$.
\end{thm}

The varieties $X_i$ appearing in this unique decomposition
are called the {\it irreducible components} of $X$. 

\bigskip \noindent{\bf Definition.} Let $X$ be a set of points in some 
space.  A {\it topology} on $X$ is a set $\mathcal{T}$ of designated 
subsets of $X$, called the {\it open} sets of $X$, so that the 
following axioms are satisfied:

\begin{itemize}
\item[(a)] The union of any collection of open sets is open.

\item[(b)] The intersection of any finite collection of open sets
is open.

\item[(c)] $X$ and $\emptyset$ are open.
\end{itemize}

\noindent The {\it closed} sets in $X$ are the complements of the
open sets.

\bigskip We define below the so--called {\it Zariski topology} on 
algebraic varieties.  If we work over $\mathbb{C}$, every variety can 
be roughly viewed as a {\it complex manifold} $X$ (with the exception 
of a proper subset of {\it its singular points}).  Through its 
embedding in, say, $\mathbb{C}^n$, $X$ will inherit the usual {\it 
complex analytic} topology from $\mathbb{C}^n$ -- a basis for the open 
sets on $X$ will consist of the intersections of $X$ with any finite 
balls in $\mathbb{C}^n$.

\smallskip
The {\it Zariski topology} is a different kind of topology. A basis 
for the open sets in $X$ is given by the sets
\[U_f=\{p\in X\,\,|\,\,f(p)\not = 0\}\]
where $f$ ranges over polynomials (homogeneous if projective $X$).

\begin{lem} The Zariski topology is indeed a topology on $X$.
\end{lem}

\noindent{\bf Exercise.} Show that the Zariski topology on the 
projective line $\mathbb{P}^1_{\mathbb{C}}$ is different from the
analytic topology of $\mathbb{P}^1_{\mathbb{C}}$.

\smallskip Many statements in algebraic geometry are true for {\it 
general} points on varieties, i.e.  if $X$ is a variety and $U$ is an 
open dense set of $X$, then any point $p\in U$ is called a {\it 
general} point on $U$.  (If $X$ is irreducible, then any nonempty open 
set will be dense.  This, in particular, makes Zariski topology a {\it 
non-Housdorff} topology -- in the latter, one needs for any two points 
of $X$ to have two nonintersecting open sets containing each one of 
the points.  This confirms once again that the Zariski topology is 
much coarser than the analytic topology.)

\smallskip
\section{Bezout's Theorem}

\noindent\smallskip
{\bf Definition.} Suppose that $X$ and $Y\subset \mathbb{P}^n$ are 
two irreducible varieties and that their intersection has irreducible 
components $Z_i$. We say that $X$ and $Y$ intersect {\it generically 
transversally} if, for each $i$, $X$ and $Y$ intersect transversally at
a general point $p_i\in Z_i$, i.e., are smooth at $p_i$
with tangent spaces spanning $\mathbb{T}_{p_i}({\mathbb P}^n)$
(the tangent space to $\mathbb{P}^n$ at $p_i$.)

\begin{thm} (Bezout) Let $X$ and $Y\subset \mathbb{P}^n$ be 
subvarieties of pure dimensions $k$ and $l$ with $k+l\geq n$, and 
suppose they intersect generically transversely.  Then
\[\text{deg}\,(X\cap Y)=\text{deg}(X)\cdot \text{deg}(Y).\]
In particular, if $k+l=n$, this says that $X\cap Y$ will consist of 
$\text{deg}(X)\cdot \text{deg}(Y)$ points. 
\end{thm}

A pair of pure--dimensional varieties $X$ and $Y\subset\mathbb{P}^n$
{\it intersect properly} if their intersection has the expected 
dimension, i.e.,
\[\text{dim}(X\cap Y)=\text{dim}(X)+\text{dim}(Y)-n.\]

\begin{thm} If $X$ and $Y$ intersect properly, \[\text{deg}(X)\cdot 
\text{deg}(Y)=\sum m_Z(X,Y)\cdot \text{deg}(Z)\] where the sum is over 
all irreducible subvarieties $Z$ of the appropriate dimension (in 
effect, over all irreducible components $Z$ of $X\cap Y$).  Here 
$m_Z(X,Y)$ is the {\it intersection multiplicity} of $X$ and $Y$ along 
$Z$:
\begin{itemize}
\item[1.] $m_Z(X,Y)\geq 1$ for all $Z\subset X\cap Y$ ($m_Z(X,Y)=0$ otherwise.)

\smallskip
\item[2.] $m_Z(X,Y)=1$ if $X$ and $Y$ intersect transversely at a general
point of $Z$.

\smallskip
\item[3.]  $m_Z(X,Y)$ is additive, i.e.  $m_Z(X\cup X^{\prime},Y)= 
m_Z(X,Y)+m_Z(X^{\prime},Y)$ for any $X$ and $X^{\prime}$ as long as 
all three numbers are defined and $X$ and $X^{\prime}$ have no common 
components.
\end{itemize}
\end{thm}

\medskip \noindent In particular, for any subvarieties $X$ and $Y$ of 
pure dimension in $\mathbb{P}^n$ intersecting properly: 
\[\text{deg}(X\cap Y)\leq \text{deg}(X)\cdot \text{deg}(Y).\]

\end{document}
