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\begin{document}
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\title{Algebraic Geometry}


\author{Diane Maclagan\\
Notes by Florian Bouyer}

\maketitle
\begin{center}
Copyright (C) Bouyer 2011. 
\par\end{center}

\begin{center}
Permission is granted to copy, distribute and/or modify this document 
\par\end{center}

\begin{center}
under the terms of the GNU Free Documentation License, Version 1.3 
\par\end{center}

\begin{center}
or any later version published by the Free Software Foundation; 
\par\end{center}

\begin{center}
with no Invariant Sections, no Front-Cover Texts, and no Back-Cover
Texts. 
\par\end{center}

\begin{center}
A copy of the license can be found at \href{http://www.gnu.org/licenses/fdl.html}{http://www.gnu.org/licenses/fdl.html}
\par\end{center}

\tableofcontents{}

\newpage{}

Books: 
\begin{itemize}
\item Hasset: \emph{Introduction to Algebraic Geometry}
\item Cox, Little, O'shea \emph{Ideals, Varieties and Algorithm}
\end{itemize}

\section{Introduction and Basic Definitions}

Algebraic geometry starts with the study of solutions to polynomial
equations. \\
e.g.: $\{(x,y)\in\cc^{2}:y^{2}=x^{3}-2x+1\}$ (an elliptic curve)\\
e.g.: $\{(x,y,w,z)\in\cc^{4}:x+y+z+w=0,x+2y+3z=0\}$ (Subspace of
$\cc^{4}$)

The goals of this module is to understand solutions to polynomial
equations ``\emph{varieties}''. That is properties, maps between
them, how to compute them and examples of them. Why would we do that?
Because varieties occurs in many different parts of mathematics:\\
e.g.: A robot arm: any movement can be described by polynomial equations
(and inequalities)\\
e.g.: $\{(x,y)\in(\qq\setminus\{0\})^{2}:x^{4}+y^{4}=1\}=\emptyset$
(by Fermat's Last Theorem)

Algebraic geometry seeks to understand these spaces using (commutative)
algebra.
\begin{defn}
Let $S$ be the ring of polynomial with coefficients in a field $k$. \end{defn}
\begin{notation*}
$S=k[x_{1},\dots,x_{n}]$\end{notation*}
\begin{defn}
The \emph{affine space} is $\aa^{n}=\{(y_{!},\dots,y_{n}):y_{i}\in k\}$.
That is $k^{n}$ without the vector space structure. 
\end{defn}

\begin{defn}
Given polynomial $f_{1},\dots f_{r}\in S$ the \emph{affine variety}
defined by the $f_{i}$ is $V(f_{1},\dots,f_{r})=\{y=(y_{1},\dots,y_{n})\in\aa^{n}:f_{i}(y)=0\,\forall i\}$\end{defn}
\begin{example*}
$V(x^{2}+y^{2}-1)=$ circle of radius 1\end{example*}
\begin{note*}
Two different sets of polynomials can define the same varieties. 
\begin{example*}
$V(x+y+z,z+2y)=V(y-z,x+2z)=\{(2a,-a,-a):a\in k\}$
\end{example*}
\end{note*}
Recall: The ideal generated by $f_{1},\dots,f_{r}\in S$ is $I=\left\langle f_{1},\dots,f_{r}\right\rangle =\{\sum_{i=1}^{r}h_{i}f_{i}:h_{i}\in S\}$.
It is closed under addition and multiplication by elements of $S$.
\begin{lem}
$V(f_{1},\dots,f_{r})=\{y\in\aa^{n}:f(y)=0\forall f\in\left\langle f_{1},\dots,f_{r}\right\rangle \}$.
Thus if $\left\langle f_{1},\dots,f_{r}\right\rangle =\left\langle g_{1},\dots g_{s}\right\rangle $
then $V(f_{1},\dots,f_{r})=V(g_{1},\dots,g_{s})$.\end{lem}
\begin{proof}
We show the inclusion both ways:

$\subseteq$: Let $y\in V(f_{1},\dots,f_{r})$. Then $f_{i}(y)=0\forall i$,
so let $f=\sum_{i=1}^{r}h_{i}f_{i}\in\left\langle f_{1},\dots,f_{r}\right\rangle $,
then $f(y)=0$. 

$\supseteq$: Conversely if $f(y)=0\forall f\in\left\langle f_{1},\dots,f_{r}\right\rangle $
then $f_{i}(y)=0\,\forall i$. Hence $y\in V(f_{1},\dots,f_{r})$.\end{proof}
\begin{notation*}
If $I=\left\langle f_{1},\dots,f_{r}\right\rangle $ we write $V(I)$
for $V(f_{1},\dots,f_{r})$.\end{notation*}
\begin{defn*}
Let $X\subseteq\aa^{n}$ be a set. The \emph{ideal of function vanishing
on $X$ }is $I(X)=\{f\in S:f(y)=0\forall y\in X\}$\end{defn*}
\begin{example*}
$X=\{0\}\subseteq\aa^{1}$. Then $I(X)=\left\langle x\right\rangle $.
\end{example*}
Note that $I\subseteq I(V(I))$. To see this we have $f\in I\Rightarrow f(y)=0\,\forall y\in V(I)\Rightarrow f\in I(V(I))$.
On the other hand we don't have always equality.\\
e.g., $I=\left\langle x^{2}\right\rangle \in k[x]$, then $V(I)=\{0\}\subseteq\aa^{n}$,
so $I(V(I))=\left\langle x\right\rangle \neq\left\langle x^{2}\right\rangle $.\\
e.g., $k=\rr$ and $I=\left\langle x^{2}+1\right\rangle $. Then $V(I)=\emptyset$
so $I(V(I))=\left\langle 1\right\rangle =\rr[x]\neq I^{2}$.

\newpage{}


\section{Grobner Bases}

Question: Given $f_{1},\dots,f_{r},f\in S$, how can we decide if
$f\in\left\langle f_{1},\dots,f_{r}\right\rangle $? That is: given
generators for $I(X)$ how can we decide if $f$ vanishes on $X$?
\begin{example}
{}
\begin{itemize}
\item $n=1$, $k=\qq$. Is $\left\langle x^{2}-3x+2,x^{2}-4x+4\right\rangle =\left\langle x^{3}-6x^{2}+12x-8,x^{2}-5x+6\right\rangle =\left\langle x-2\right\rangle $?
Yes since we are in a PID so we can use Euler's algorithm to find
the generator. This is a solved problems
\item Any $n$ and $f_{1},\dots,f_{r}$ are linear

\begin{itemize}
\item Is $y-z\in\left\langle x+y+z,x+2y\right\rangle ?$ Yes.
\item Is $5x_{1}+3x_{2}-7x_{4}+8x_{5}\in\left\langle x_{1}+x_{2}+x_{3}+x_{4}+x_{5},3x_{1}-7x_{4}+9x_{5},2x_{1}+3x_{4}\right\rangle =\left\langle f_{1},f_{2},f_{3}\right\rangle $?
If $f\in\left\langle f_{1},f_{2},f_{3}\right\rangle $ then $f=af_{1}+bf_{2}+cf_{3}$
for $a,b,c\in k$. So the question now becomes: is 
\[
(5,3,0,7,8)\in\mbox{row}\begin{pmatrix}1 & 1 & 1 & 1 & 1\\
3 & 0 & 0 & -7 & 9\\
2 & 0 & 0 & 3 & 0
\end{pmatrix}?
\]
To solve this we use Gaussian elimination from Linear Algebra
\end{itemize}
\end{itemize}
\end{example}
As we seen from the above examples, we need a common generalization.
This is the Theory of Grobner bases.
\begin{defn}
A \emph{term order} (or \emph{monomial order}) is a total order on
the monomials (polynomial in one variable) is $S=k[x_{1},\dots,x_{n}]$
such that:
\begin{enumerate}
\item $1<x^{u}$ for all $u\neq0$
\item $x^{u}<x^{v}\Rightarrow x^{u+w}<x^{v+w}$ for all $w\in\nn^{n}$.
\end{enumerate}
\end{defn}
Several term orders:
\begin{description}
\item [{Lexicographic~order}] $X^{u}<X^{v}$ if the first non-zero element
of $v-u$ is positive.

\begin{example*}
$f=3x^{2}-8xz^{9}+9y^{10}$. If $x>y>z$, then $x^{2}>xz^{9}>y^{10}$
(since if $v=(2,0,0),u=(1,0,9)$ then $v-u=(1,0,-9)$.
\end{example*}
\item [{Degreelicographic~order}] $X^{u}<X^{v}$ if $\begin{cases}
\deg(X^{u})<\deg(X^{v})\\
X^{u}<_{\mathrm{lex}}X^{v} & \mbox{if }\deg(X^{u})=\deg(X^{v})
\end{cases}$.

\begin{example*}
$f=3x^{2}-8xz^{9}+9y^{10}$. Then $xz^{9}>y^{10}>x^{2}$.
\end{example*}
\item [{Reverse~lexicographic~order~(revlex)}] $X^{2}<X^{v}$ is $\begin{cases}
\deg(X^{u})<\deg(X^{v})\\
\mbox{the last non-zero entry of }v-u\mbox{ is negative} & \mbox{if }\deg(X^{u})=\deg(X^{v})
\end{cases}$

\begin{example*}
$f=3x^{2}-8xz^{9}+9y^{10}$. Then $y^{10}>xz^{9}>z^{2}$. 
\end{example*}
\end{description}
\begin{defn}
Given a polynomial $f=\sum c_{u}X^{u}\in S$ and a term order $<$,
the \emph{initial term} of $f$ is $c_{v}X^{v}$ with $X^{v}>X^{u}$
for all $u$ and $c_{v}\neq0$. This is denoted $\ld(f)$.
\end{defn}

\begin{defn}
The \emph{initial ideal} of $I$ with respect to $<$ is $\ld(I)=\left\langle \ld(f):f\in I\right\rangle $
\end{defn}
\textbf{Warning: }If $I=\left\langle f_{1},\dots,f_{r}\right\rangle $
then $\ld(I)$ is not necessarily generated by $\left\langle \ld(f_{1}),\dots\ld(f_{r})\right\rangle $.
\\
e.g., Let $I=\left\langle x+y+z,x+2y\right\rangle $ and let the term
ordering be $x>y>z$. Then $\ld(I)=\left\langle x,y\right\rangle $. 
\begin{defn}
A set $\{g_{1},\dots g_{s}\}$ is a \emph{Grobner basis} for $I$
if $\{g_{1},\dots,g_{s}\}\subseteq I$ and $\ld(I)=\left\langle \ld(g_{1}),\dots,\ld(g_{s})\right\rangle $.
\end{defn}
The point of this is that long division by a Grobner basis decides
the ideal membership problem, that is, is $f\in\left\langle f_{1},\dots,f_{r}\right\rangle $?
\begin{defn}
A \emph{monomial ideal} is an ideal $I\subseteq S$ generated by monomials
$X^{u}$.\end{defn}
\begin{lem}
Let $I$ be a monomial ideal, $I=\left\langle X^{u}:u\in A\right\rangle $
for some $A\subseteq\nn^{n}$. Then:
\begin{enumerate}
\item $X^{v}\in I$ if and only if $X^{u}|X^{v}$ for some $u\in A$.
\item If $f=\sum c_{v}X^{v}\in I$ then each $X^{v}$ with $c_{v}$ non-zero
is divisible by some $X^{u}$ for $U\in A$, hence they lies in $I$.
\end{enumerate}
\end{lem}
\begin{proof}
Note that part 1.~is a special case of part 2.

Since $f\in I$ we can write $f=\sum h_{u}X^{u}$ with $u\in A$,
$h_{u}\in S$ and all but finitely many are $0$. Let us expand the
RHS as a sum of monomials. Then each term is a multiple of some $X^{u}$
so lies in $I$, hence the same is true for the terms of $f$.\end{proof}
\begin{thm}[Dickson's Lemma]
 Let $I=\left\langle X^{u}:u\in A\right\rangle $ for some set $A\subseteq\nn^{n}$,
then there exists $a_{1},\dots a_{s}\in A$ with $I=\left\langle X^{a_{1}},\dots,X^{a_{s}}\right\rangle $.\end{thm}
\begin{proof}
The proof is by induction on $n$.
\begin{lyxlist}{00.00.0000}
\item [{$n=1$:}] We have $I=\left\langle X^{u}\right\rangle $ for $U=\min\{U:U\in A\}$,
this uses the fact that $\nn$ is well ordered
\item [{$n>1$:}] Name the variables of the polynomial ring $x_{1},\dots,x_{n-1},y$..
Let $J=\left\langle X^{u}:\exists j\geq0\mathrm{\, with\,}x^{u}y^{j}\in I\right\rangle \subseteq k[x_{1},\dots,x_{n-1}]$.
By induction hypothesis $J=\left\langle X^{a_{i_{1}}},\dots X^{a_{i_{s}}}\right\rangle $
where $(a_{i_{j}},m_{j})\in A$ for some $m_{j}\in\nn$. Let $m=\max(m_{j})$.
For $0\leq l\leq m-1$, let $J_{l}=\left\langle X^{u}:x^{u}y^{l}\in I\right\rangle \subseteq k[x_{1},\dots,x_{n-1}]$.
So again by induction we have that $J_{l}=\left\langle x^{b_{l1}},\dots,x^{b_{r(l)}}\right\rangle $
where $b_{ls}\in\nn^{n-1}$ and $x^{b_{ls}}y^{l}\in I$.\\
We now claim that $I=\left\langle x^{b_{ls}}y^{l}:0\leq l\leq m-1,1\leq s\leq r(l)\right\rangle +\left\langle x^{a_{ij}}y^{m_{j}}:1\leq j\leq s\right\rangle $.\\
Indeed if $x^{u}y^{j}\in I$, if $j<m$ then $x^{u}\in J_{j}$ so
$x^{b_{js}}|x^{u}$ for some $b_{js}$ so $x^{b_{js}}y^{j}|x^{u}y^{j}$.
If $j\geq m$ then $x^{u}\in J$, so there is $a_{i}$ with $X^{a_{i}}|X^{u}$
so $X^{a_{i}}y^{m_{i}}|X^{u}y^{j}$. In particular, every monomial
generator of $I$ lies in $\left\langle x^{b_{ls}}y^{l},x^{a_{ij}}y^{m_{j}}\right\rangle $
so the ideals are equal and $I$ is finitely generated.\\
For each of the finite number of generators we can find $a_{i}\in A$
with $X^{a_{i}}$ dividing the generator (using the previous lemma).
\end{lyxlist}
\end{proof}
\begin{cor}
A term order is well ordered (every set of monomials has a least element)\end{cor}
\begin{proof}
If not, there would be an infinite chain $X^{u_{1}}>X^{u_{2}}>\dots$.
Let $I=\left\langle X^{u_{i}}:i\geq1\right\rangle \subseteq k[x_{1},\dots,x_{n}]$,
then by Dickson's lemma $I=\left\langle X^{u_{i_{1}}},\dots,X^{u_{i_{s}}}\right\rangle $
for some $i_{1}<i_{2}<\dots<i_{s}$. In particular for $j\geq i_{s}$
there exists $l$ such that $X^{u_{i_{l}}}|X^{u_{j}}$. Thus $X^{u_{j}}=X^{u_{i_{l}}}X^{w}$,
but then $X^{u_{i_{l}}}<X^{u_{j}}$ because $1<X^{W}$. This is a
contradiction.\end{proof}
\begin{cor}
Let $I$ be an ideal in $k[x_{1},\dots,x_{n}]$ then there exists
$g_{1},\dots g_{s}\in I$ with $\ld(I)=\left\langle \ld(g_{1}),\dots,\ld(g_{s})\right\rangle $.
Hence a Grobner basis exists.\end{cor}
\begin{proof}
By definition $\ld(I)=\left\langle \ld(f):f\in I\right\rangle $.
By Disckson's lemma, there exists $g_{1},\dots,g_{s}\in I$ with $\left\langle \ld(g_{1}),\dots,\ld(g_{s})\right\rangle =\ld(I)$.
\end{proof}

\subsection{The Division Algorithm}
\begin{lyxcode}
Input:~$f_{1},\dots,f_{s},f\in S$,~$<$~the~term~order

Output:~Expression~of~the~form~$\sum_{i=1}^{s}h_{i}f_{i}+r$~where~$h_{i}\in S$~and~$r=\sum c_{u}X^{u}$~with~\{$c_{u}\neq0\Rightarrow X^{u}$~is~not~divisible~by~$\ld(f_{i})\,\forall i$\},~such~that~if~$\ld(f)=c_{u}X^{u},\ld(h_{i}f_{i})=c_{v_{i}}X^{v_{i}}$~then~$X^{u}\geq X^{v_{i}}\,\forall i$.



Step~1:~Initialize~$h_{1}=\dots=h_{s}=0,r=0,p=f$.

Step~2:~While~$p\neq0$~do:

~~~~~~~~~~~~$i=1$

~~~~~~~~~~~~Divisionoccured~=~false

~~~~~~~~~~~~While~$i\leq s$~and~Divisionoccured~=~false~do:

~~~~~~~~~~~~~~~~If~$\ld(f_{i})|\ld(p)$~then:

~~~~~~~~~~~~~~~~~~~~$h_{i}=h_{i}+\frac{\ld(p)}{\ld(f_{i})}$

~~~~~~~~~~~~~~~~~~~~$p=p-\frac{\ld(p)}{\ld(f_{i})}$$f_{i}$

~~~~~~~~~~~~~~~~~~~~Divisionoccured~=~true

~~~~~~~~~~~~~~~~Else:

~~~~~~~~~~~~~~~~~~~~$i=i+1$

~~~~~~~~~~~~~~~~If~Divisionoccured~=~false~then:

~~~~~~~~~~~~~~~~~~~~$r=r+\ld(p)$

~~~~~~~~~~~~~~~~~~~~$p=p-\ld(p)$

Step~3:~Output:~$h_{1},\dots,h_{s},r$.\end{lyxcode}
\begin{example}
{}\end{example}
\begin{lyxcode}
Input:~$f_{1}=x+y+z$,~$f_{2}=3x-2y$,~$f=5y+3z$,~$<$~lex~($x<y<z$)

Step~1:~$h_{1}=0$,~$h_{2}=0$,~$r=0$,~$p=5y+3z$.

Step~2:~$i=1$

~~~~~~~~Divisionoccured~=~false

~~~~~~~~does~$\ld(f_{1})|\ld(p)$?~Yes:

~~~~~~~~~~~~$h_{1}=0+3$

~~~~~~~~~~~~$p=5y+3z)-3\cdot(x+y+z)=-3x+2y$

~~~~~~~~~~~~Divisionoccured~=~true

Step~2:~$i=1$

~~~~~~~~Divisionoccured~=~false

~~~~~~~~does~$\ld(f_{1})|\ld(p)$?~No:

~~~~~~~~$i=2$

~~~~~~~~does~$\ld(f_{2})|\ld(p)$?~Yes:

~~~~~~~~~~~~$h_{2}=0+-1$

~~~~~~~~~~~~$p=-3x+2y+(-1)\cdot(3x-2y)=0$

~~~~~~~~~~~~Divisionoccured~=~true

Step~3:~Output:~$h_{1}=3,h_{2}=-1,r=0$
\end{lyxcode}
Note that the division algorithm depends on the ordering. (In the
above example if $x>y>z$ then the output is $h_{1}=h_{2}=0$ and
$r=5y+3z$)
\begin{prop}
The above algorithm terminates with the correct output .\end{prop}
\begin{proof}
As each stage the initial term $\ld(p)$ decreases with respect to
$<$. Since $<$ is a well-order, this cannot happen an infinite number
of times, hence the algorithm must terminate.

At each stage we have $f=p+\sum h_{i}f_{i}+r$, where $h_{i}f_{i}$
and $r$ satisfy the condition, so when it outputs with $p=0$, the
output has the desired correct form.\end{proof}
\begin{prop}
If $\{g_{1},\dots,g_{s}\}$ is a Grobner basis for $I$ with respect
to $<$, then $f\in I$ if and only if the division algorithm outputs
$r=0$.\end{prop}
\begin{proof}
The division algorithm writes $f=\sum h_{i}g_{i}+r$, where no monomial
in $r$ is divisible by $\ld(g_{i})$. Thus $f\in I$ if and only
if $r\in I$. Now if $r\neq0$  then $\ld(r)\notin\ld(I)=\left\langle \ld(g_{1}),\dots,\ld(g_{s})\right\rangle $,
so $r\notin I$. Hence $r=0$ if and only if $r\in I$.\end{proof}
\begin{cor}
If $\{g_{1},\dots,g_{s}\}$ is a Grobner basis for $I$ then $I=\left\langle g_{1},\dots,g_{s}\right\rangle $\end{cor}
\begin{proof}
We have $\left\langle g_{1},\dots,g_{s}\right\rangle \subseteq I$
by the definition of Grobner basis. If $f\in I$, then we divide $f$
by $g_{1},\dots,g_{s}$ to get $f=\sum h_{i}g_{i}+r$, but $r=0$.
So we have $f\in\left\langle g_{1},\dots,g_{s}\right\rangle $, hence
$I\subseteq\left\langle g_{1},\dots,g_{s}\right\rangle $.\end{proof}
\begin{cor}[Hilbert Basis Theorem]
 Let $I\subseteq S$ be an ideal. Then $I$ is finitely generated.\end{cor}
\begin{proof}
We know that $I$ has a finite Grobner basis (since monomial ideals
are finitely generated). By the previous corollary, this Grobner basis
generates $I$\end{proof}
\begin{defn}
A ring $R$ is \emph{Noetherian} if all its ideals are finitely generated.
\end{defn}
Hence the Hilbert basis theorem says $S$ is Noetherian. Note that
there is a standard algorithm (the Buchberger algorithm) to compute
Grobner bases.
\begin{defn}
A \emph{reduced Grobner basis} for $I$ with respect to $<$, is a
Grobner basis of $I$ which satisfies:
\begin{enumerate}
\item Coefficients of $\ld(g_{i})$ is 1
\item No $\ld(g_{i})$ divides any other-way
\item No $\ld(g_{i})$ divides any other term of $g_{j}$.
\end{enumerate}
\end{defn}
Such a reduced Grobner basis exists and is unique. With this we can
check whether two ideals are equal. To do this we fix a term order
and compute a reduced Grobner basis for $I$ and $J$. 

\newpage{}


\section{Zariski Topology}

Recall that a topological space is a set $X$ and a collection $\theta=\{U\}$
of subsets of $X$ called open sets, satisfying:
\begin{enumerate}
\item $\emptyset\in\theta$
\item $X\in\theta$
\item If $U,U'\in\theta$ then $U\cap U'\in\theta$
\item If $U_{\alpha}\in\theta$ for $\alpha\in A$, then $\cup_{\alpha}U_{\alpha}\in\theta$.
\end{enumerate}
A set $Z$ is closed if its compliment is open.
\begin{defn}
The \emph{Zariski Topology} on $\aa^{n}$ has close set $V(I)$ for
$I\subseteq S$ an ideal.\end{defn}
\begin{example*}
In $\aa^{1}$, under the Zariski Topology, the closed sets are finite
set, $\aa^{1}$ or $\emptyset$. ($\aa^{1}=V(0)$ and $\emptyset=V(S)$)
\end{example*}
Recall: If $I,J$ are ideals in $S$ then $I+J=\{i+j:i\in I,j\in J\}$,
while $IJ=\left\langle ij:i\in I,j\in J\right\rangle $. In terms
of generators, if $I=\left\langle f_{1},\dots,f_{s}\right\rangle $
and $J=\left\langle g_{1},\dots,g_{r}\right\rangle $ then $I+J=\left\langle f_{1},\dots,f_{s},g_{1},\dots,g_{s}\right\rangle $
and $IJ=\left\langle f_{i}g_{j}:1\leq i\leq s,1\leq j\leq r\right\rangle $. 
\begin{prop}
Let $X=V(I)$ and $Y=V(J)$ be two varieties in $\aa^{n}$ then: 
\begin{itemize}
\item $X\cap Y=V(I+J)$
\item $X\cup Y=V(I\cap J)=V(IJ)$
\end{itemize}
\end{prop}
\begin{proof}
{}
\begin{itemize}
\item Let $y\in X\cap Y$. Then $f(y)=0$ for all $f\in I$ and $g(y)=0$
for all $g\in J$. So $(f+g)(y)=0$ for all $f\in I$ and $g\in J$.
Hence by definition $y\in V(I+J)$.\\
Conversely: let $y\in V(I+J)$, then $h(y)=0$ for all $h=f+0$ with
$f\in I$, hence $y\in V(I)$. Similarly $h(y)=0$ for all $h=0+g$
with $g\in J$, hence $y\in V(J)$. So $y\in X\cap Y$.
\item Let $y\in X\cup Y$. Then $y\in X$ or $y\in Y$. If $y\in X$ then
$f(y)=0\forall f\in I$, so $f(y)=0\,\forall f\in I\cap J$, hence
$y\in V(I\cap J)$. Similarly if $y\in Y$ then $g(y)=0\forall g\in I$,
so $g(y)=0\,\forall g\in I\cap J$, hence $y\in V(I\cap J)$.\\
Let $y\in V(IJ)$. Then $h(y)=0\forall h=fg$ with $f\in I,g\in J$.
Thus $h(y)=f(y)g(y)\,\forall f\in I,g\in J$. Suppose $y\notin Y$,
that is there exists $g\in J$ with $g(y)\neq0$, then $f(y)=0\,\forall f\in I$,
hence $y\in V(I)=X$. Thus we have $y\in X\cup Y$. So $V(IJ)\subseteq X\cup Y$.\\
Note that $I\cap J\supseteq IJ$ so $V(I\cap J)\subseteq V(IJ)$ (This
follows from the general fact $I\subseteq J\Rightarrow V(J)\subseteq V(J)$)\\
We have shown $V(I\cap J)\subseteq V(IJ)\subseteq X\cup Y\subseteq V(I\cap J)$,
thus they are all equal.
\end{itemize}
\end{proof}
In fact, if $\{X_{\alpha}:\alpha\in A\}$ is a collection of varieties
in $\aa^{n}$ with $X_{\alpha}=V(I_{\alpha})$, then $\cap_{\alpha}X_{\alpha}=V(\left\langle \cup I_{\alpha}\right\rangle )$.
Challenge question: What goes wrong with arbitrary union.
\begin{cor}
The Zariski topology is a topology on $\aa^{n}$.
\end{cor}
Note: This topology is weird compare to the Euclidean topology, for
example it is not Haussdorf and open sets are dense.


\subsection{Morphism}
\begin{defn}
A \emph{morphism} is a map $\phi:\aa^{n}\to\aa^{m}$ with $\phi(y_{1},\dots,y_{n})=(\phi_{1}(y_{1},\dots,y_{n}),\dots,\phi_{m}(y_{1},\dots,y_{n}))$
where $\phi\in k[x_{1},\dots,x_{n}]$.\end{defn}
\begin{example*}
$\phi:\aa^{2}\to\aa^{2}$ defined by $\phi(x,y)=(x^{2}-y^{2},x^{2}+2xy+3y^{2})$.
\end{example*}
Morphism plays the role of continuous functions in topology. Questions:
are all continuous functions morphism? No.
\begin{example*}
$f(x)=\begin{cases}
x+1 & x\notin\qq\\
x & x\in\qq
\end{cases}$. This is a continuous function in the Zariski topology. We don't
want this, hence why we restrict to morphism.\end{example*}
\begin{defn}
For $f\in k[z_{1},\dots,z_{m}]$, $\phi:\aa^{n}\to\aa^{m}$, the function
$f\circ\phi\in k[x_{1},\dots,x_{n}]$ is called the \emph{pullback
}if $f$ by $\phi.$\end{defn}
\begin{note*}
$\phi^{*}f=f\circ\phi$.
\end{note*}
Recall: a $k$-algebra is a ring $R$ containing the field $k$. A
$k$-algebra homomorphism is a ring homomorphism $\phi$ with $\phi(a)=a\,\forall a\in k$.
\begin{lem}
The map $\phi^{*}:k[z_{1},\dots,z_{m}]\to k[x_{1},\dots,x_{n}]$ is
a $k$-algebra homomorphism
\begin{itemize}
\item $\phi^{*}(1)=1$
\item $\phi^{*}(0)=0$
\item $\phi^{*}(a)=a\,\forall a\in k$
\item $\phi^{*}(fg)=\phi^{*}(f)\phi^{*}(g)$
\item $\phi^{*}(f+g)=\phi^{*}(f)+\phi^{*}(g)$
\end{itemize}
\end{lem}
\begin{proof}
Exercise
\end{proof}
Note: The polynomial ring is the ring of morphism from $\aa^{n}$
to $\aa^{1}$.
\begin{defn}
The coordinate ring $k[X]$ of a variety $X=V(I)\subseteq\aa^{n}$
is the ring of polynomial functions from $X$ to $\aa^{1}$. 

Equivalently: $k[X]=\{f\in k[x_{1},\dots,x_{n}]\}/\sim$ where $f\sim g$
if $f(y)=g(y)$ for all $y\in X$.\end{defn}
\begin{note*}
$f(y)=g(y)\,\forall y\in X$ if and only if $(f-g)(y)=0\,\forall y\in X$,
that is, if and only if $f-g\in I(X)$. So $k[X]=k[x_{1},\dots,x_{n}]/I(X)$
and in particular $k[X]$ is a ring.\end{note*}
\begin{example*}
{}
\begin{itemize}
\item $X=V(x^{2}+y^{2}-1)$ then $k[X]=k[x,y]/\left\langle x^{2}+y^{2}-1\right\rangle $
\item $X=V(x^{3})\subseteq\aa^{1}$ then $k[X]=k[x]/\left\langle x\right\rangle \cong k$.
\end{itemize}
\end{example*}
\begin{defn}
Fix $X=V(I)\subseteq\aa^{n}$. Two morphism $\phi,\psi:\aa^{n}\to\aa^{m}$
\emph{are equal in $X$ }if the induced pullback $\phi^{*},\psi^{*}:k[z_{1},\dots,z_{m}]\to k[Z]=k[x_{1},\dots,x_{n}]/I(X)$
are equal.
\end{defn}

\begin{defn}
A morphism $\phi:X\to\aa^{n}$ is an equivalence class of such morphism.\end{defn}
\begin{example}
Let $X=V(x^{2}+y^{2}-1)$, $\psi:\aa^{2}\to\aa^{1}$ defined by $\psi(x,y)=x^{4}$
and $\phi:\aa^{2}\to\aa^{1}$ defined by $\phi(x,y)=(y^{2}-1)^{2}$.
We claim that $\phi=\psi$ on $X$ since $\psi^{*}:k[z]\to k[x,y]$
is defined by $z\mapsto x^{4}$ while $\phi^{*}:k[z]\to k[x,y]$ is
defined by $z\mapsto(y^{2}-1)^{2}$. But $k[X]=k[x,y]/(x^{2}+y^{2}-1)$,
and in there $x^{4}=(y^{2}-1)^{2}$, hence $\phi^{*}=\psi^{*}$.\end{example}
\begin{lem}
If $\phi,\psi:\aa^{n}\to\aa^{m}$ are equal on $X$ then $\phi(y)=\psi(y)$
for all $y\in X$.\end{lem}
\begin{proof}
If $\phi(y)\neq\psi(y)$ for some $y\in X$ then they differ in some
coordinate $i$. Then $z_{i}(\phi(y))\neq z_{i}(\psi(y))$, so $\phi^{*}z_{i}(y)\neq\psi^{*}z_{i}(y)$.
Hence $\phi^{*}z_{i}-\psi^{*}z_{i}\notin I(X)$, so the pullback homomorphism
$\phi^{*}$ and $\psi^{*}$ are different.\end{proof}
\begin{defn}
Let $X\subseteq\aa^{n}$ and $Y\subseteq\aa^{m}$ be varieties. A
morphism $\phi:X\to Y$ is a morphism $\phi:X\to\aa^{m}$ with $\phi(X)\subseteq Y$.\end{defn}
\begin{example*}
Let $X=\aa^{1}$ and $Y=V(cy-y^{2})\subseteq\aa^{3}$ and let $\phi:\aa^{1}\to\aa^{3}$
be defined by $\phi(t)=(t,t^{2},t^{3})$. Then $\phi^{*}:k[x,y,z]\to k[t]$
is defined by $x\mapsto t$, $y\mapsto t^{2}$ and $z\mapsto t^{3}$.
Since $tt^{3}-(t^{2})^{2}=0$, $\phi(\aa^{1})\subseteq Y$, so $\phi$
is a morphism from $\aa^{1}\to Y$.\end{example*}
\begin{prop*}
Let $X\subseteq\aa^{n}$, $Y\subseteq\aa^{m}$ be varieties. Any morphism
$\phi:X\to Y$ induces a $k$-algebra homomorphism $\phi^{*}:k[Y]\to k[X]$.
Conversely given a $k$-algebra homomorphism from $k[Y]\to k[X]$
is $\phi^{*}$ for some morphism $\phi:X\to Y$.\end{prop*}
\begin{proof}
Let $\phi:X\to Y$ be a morphism. Since $\phi(X)\subseteq Y$ we have
$f\circ\phi(x)=0\,\forall x\in X$ and $f\in I(Y)$. Hence $\phi^{*}f\in I(X)\,\forall f\in I(Y)$,
therefore the induced map $\phi^{*}:k[z_{1},\dots,z_{m}]\to k[X]=k[x_{1},\dots,x_{n}]/I(X)$
factors through $k[Y]$. So given a morphism $\phi:X\to Y$ we get
$\phi^{*}:k[Y]\to k[X]$.

Conversely given a $k$-algebra homomorphism $\alpha:k[Y]\to k[X]$
it suffices to find a $k$-algebra homomorphism $\widetilde{\alpha}^{*}:k[z_{1},\dots,z_{m}]\to k[x_{1},\dots,x_{n}]$
for which we have a commutating diagram
\[
\xymatrix{k[z_{1},\dots,z_{m}]\ar^{\widetilde{\alpha}^{*}}[r]\ar_{i_{Y}^{*}}[d] & k[x_{1},\dots,x_{n}]\ar^{i_{X}^{*}}[d]\\
k[Y]\ar^{\alpha}[r] & k[X]
}
\]
Then $\widetilde{\alpha}$ will be a morphism $\aa^{n}\to\aa^{m}$
with $\widetilde{\alpha}(X)\subseteq Y$. We construct such $\widetilde{\alpha}^{*}$
as follow. Let $g_{i}$ be any polynomial in $k[x_{1},\dots,x_{n}]$
with $i_{X}^{*}(g_{i})=\alpha(i_{Y}^{*}(z_{i}))$. Set $\widetilde{\alpha}^{*}=g_{i}$
and extend as a $k$-algebra homomorphism. ($g_{i}$ exists since
the map $i_{X}^{*}$ is surjective). This defines $\widetilde{\alpha}^{*}:k[z_{1},\dots,z_{m}]\to k[x_{1},\dots,x_{n}]$
and $i_{X}^{*}\circ\widetilde{\alpha}^{*}(z_{1})=\alpha\circ i_{Y}^{*}(z_{i})$
by construction, hence the diagram commutes.\end{proof}
\begin{example}
Let $\phi^{*}:k[t]\to k[x,y,z]/(x^{2}-y,x^{3}-z)$. Then $\phi^{*}(t)=x$
and $\phi^{*}(t)=x+x^{2}-y$ is the same. This is $\phi^{*}$ for
$\phi:V(x^{2}-y,x^{3}-z)\to\aa^{1}$ defined by $\phi(x,y,z)=x$ (or
$\phi(x,y,z)=x+x^{2}-y$ as while they are different morphism they
agree on $X$)
\end{example}
So to sum up: Morphism $\phi:X\to Y$ are the same as $k$-algebra
homomorphism of the coordinate rings $\phi^{*}:k[Y]\to k[X]$. note
that the homomorphism goes the other way! (contragradient).
\begin{xca}
If $X\overset{\phi}{\to}Y\overset{\psi}{\to}Z$ with $X\overset{\alpha}{\to}Z$.
Then $\alpha^{*}:k[Z]\to k[X]$ is $\phi^{*}\circ\psi^{*}$. \end{xca}
\begin{defn}
An \emph{isomorphism} of affine varieties is a morphism $\phi:X\to Y$
for which there is a morphism $\phi^{-1}:Y\to X$ with $\phi\circ\phi^{-1}=\id_{Y}$
and $\phi^{-1}\circ\phi=\id_{X}$.

An \emph{automorphism }of an affine variety is an isomorphism $\phi:X\to X$.
\end{defn}
WARNING: A morphism that is a bijection needs not be an isomorphism.


\subsection{Images of varieties under morphism}

That is, given $\phi:\aa^{n}\to\aa^{m}$ what is $\phi(X)$?

Warning: $\phi(X)$ needs not to be a variety. For example $X=V(xy-1)\subseteq\aa^{1}$
and $\phi:\aa^{2}\to\aa^{1}$ defined $(x,y)\to x$. Then $\phi(X)=\aa^{1}\setminus\{0\}$.
(REMEMBER THIS EXAMPLE!). Notice that the closure of $\phi(X)$, is
$\overline{\phi(X)}=\aa^{1}$.

Another question is: Given $X\subseteq\aa^{n}$ and $\phi:\aa^{n}\to\aa^{m}$,
how do we compute $\overline{\phi(X)}$. We use the following clever
trick: let $X\subseteq\aa^{n}$, first we send $x\mapsto(x,\phi(x))$,
then project unto the last $m$ coordinates, i.e., $\phi(X)$ is the
composition of the inclusion of $X$ into the graph of $\phi$ with
the projection onto the last $m$ coordinates.

This breaks the problem into two parts:
\begin{itemize}
\item Describe the image of $X\mapsto\aa^{n}\times\aa^{m}$
\item Describe $\overline{\pi(Y)}$ for $Y\subseteq\aa^{n}\times\aa^{m}$,
where $\pi$ is the projection onto the last $m$ coordinates.
\end{itemize}
For part 1, the image of $X=V(I)$ is $V(I)\cap V(z_{i}-\phi_{i}(x))\subseteq\aa^{n}\times\aa^{m}=(x_{1},\dots,x_{n},z_{1},\dots,z_{m})$
\begin{example*}
Let $\phi:\aa^{2}\to\aa^{2}$ defined by $\phi(x,y)=(x+y,x-y)$ and
let $X=V(x^{2}-y^{2})$. Then the graph of $X$ in $\aa^{2}\times\aa^{2}$
is $V(x^{2}-y^{2},z_{1}-z-y,z_{2}-x+y)\subseteq(x,y,z_{1},z_{2})$.
Then $\phi(x,y)=(z_{1},z_{2})$\end{example*}
\begin{thm}
Let $X\subseteq\aa^{n}$ be a variety and let $\phi:\aa^{n}\to\aa^{m}$
be the projection onto the last $m$ coordinates. Then $\overline{\pi(X)}=V(I(X)\cap k[x_{n-m+1},\dots,x_{n}])$\end{thm}
\begin{note*}
We'll soon show that if $k=\overline{k}$ then we can replace $I(X)$
by $I$. But it is not true otherwise, for example, consider $k=\rr$
and $X=V(x^{2}y^{2}+1)\subseteq\aa^{2}$ and $\pi:(x,y)\mapsto y$.
Then $X=\emptyset$, $\pi(X)=\emptyset$ and $I(X)=\left\langle 1\right\rangle $.
But $\left\langle x^{2}y^{2}+1\right\rangle \cap k[y]=\left\langle 0\right\rangle $\end{note*}
\begin{proof}
If $f\in I(X)\cap k[x_{n-m+1},\dots,x_{m}]$ then $f(y)=0\,\forall y\in X$,
so $f(y_{n-m+1},\dots,y_{n})=0\,\forall(y_{n-m+1},\dots,y_{n})$ with
$y\in X$, hence $f(\pi(y))=0\,\forall y\in X$ and thus $\pi(X)\subseteq V(I(X)\cap k[x_{n-m+1},\dots,x_{n}])$.

Conversely if $g\in I(\pi(X))$ then $g(y_{n-m+1},\dots,y_{n})=0\,\forall y=(y_{1},\dots,y_{n})\in X$.
So $g\in I(X)\cap k[x_{n-m+1},\dots,x_{n}]$ so $I(\pi(X))\subseteq I(X)\cap k[x_{n-m+1},\dots,x_{n}]$.
But since $\overline{\pi(X)}=V(I(\pi(X))$ this shows $V(I(X)\cap k[x_{n-m+1},\dots,x_{n}])\subseteq\overline{\pi(X)}$.
\end{proof}
This leaves the question: Given $I\subseteq k[x_{1},\dots,x_{n},z_{1},\dots,z_{m}]$
how can we compute $I\cap k[z_{1},\dots,z_{m}]$? The answer is to
use Grobner basis.

Recall: the lexicographic term order with $x_{1}>\dots>x_{n}>z_{1}>\dots>z_{m}$
has $x^{u}z^{v}>x^{u'}z^{v'}$ if $(u-u',v-v')$ has first non-zero
entry positive.
\begin{prop}
Let $I\subseteq k[x_{1},\dots,x_{n}]=S$ and let $G=\{g_{!},\dots,g_{s}\}$
be a lexicographic Grobner basis for $I$. Then a lexicographic Grobner
basis for $I\cap k[x_{n-m+1},\dots,x_{n}]$ is given by $G\cap k[x_{n-m+1},\dots,x_{n}]=S'$,
i.e., those elements of $G$ that are polynomials in $x_{n-m+1},\dots,x_{n}$.\end{prop}
\begin{proof}
$G\cap S'$ is a collection of polynomials in $I\cap S'$, so we just
need to show that $\left\langle \mathrm{in}_{<\mathrm{lex}}(g):g\in G\cap S'\right\rangle =\mathrm{in}_{<\mathrm{lex}}(I\cap S')\subseteq S'$.
Let $f\in I\cap S'$. Then $\mathrm{in}_{<\mathrm{lex}}(f)\in\mathrm{in}_{<\mathrm{lex}}(I)$,
so there is $g\in G$ with $\mathrm{in}_{<\mathrm{lex}}(g)|\mathrm{in}_{<\mathrm{lex}}(f)$.
Since $f\in S'$, $\mathrm{in}_{<\mathrm{lex}}(g)$ is not divisible
by $x_{1},\dots,x_{n-m}$ and thus $g\in S'$. Hence $\mathrm{in}_{<\mathrm{lex}}(f)\in\left\langle \mathrm{in}_{<\mathrm{lex}}(g):g\in G\cap S'\right\rangle $,
so $G\cap S'$ is a Grobner basis for $I\cap S'$.
\end{proof}
The next question is: Given $X=V(I)$, what is $I(X)$?
\begin{namedthm}[Hilbert's Nullstellensatz]
 If $k=\overline{k}$, then $I(V(I))=\sqrt{I}$, where $\sqrt{I}$
is the radical of $I$. (Denoted $r(I)$ in Commutative Algebra)\end{namedthm}
\begin{proof}
This proof will come later in the course.

\newpage{}
\end{proof}

\section{Sylvester Matrix}

Given $f,g\in k[x]$, how can we decide if they have a common factor? 
\begin{defn*}
$f=5x^{5}+6x^{4}-x^{3}+2x^{2}-1$ and $g=7x^{5}+8x^{3}-3x^{2}+1$.

Or $f=ax+b,g=cx+d$. In this case we have that $f,g$ has a common
factor if and only if $\begin{vmatrix}a & b\\
c & d
\end{vmatrix}=0$. Notice the analogy with $\zz$, that is, $n,m\in\zz$ have a common
factor when there is no $a,b$ such that $an+bm=1$. This naturally
leads to the next proposition.\end{defn*}
\begin{prop}
Let $f=\sum_{i=0}^{l}a_{i}x^{i}$ and $g=\sum_{j=0}^{m}b_{j}x^{j}$
be two polynomials in $k[x]$. Then the following are equivalent.
\begin{enumerate}
\item $f,g$ have a common root, i.e., there exists $\alpha\in\overline{k}$
such that $f(\alpha)=g(\alpha)=0$
\item $f,g$ have a non-constant common factor $h$
\item There does not exists $A,B\in k[x]$ with $Af+Bg=1$
\item $\left\langle f,g\right\rangle \neq k[x]$
\item There exists $\widetilde{A},\widetilde{B}\in k[x]$ with $\deg(\widetilde{A})\leq m-1,\deg(\widetilde{B})\leq l-1$
and $\widetilde{A}f+\widetilde{B}g=0$.
\end{enumerate}
\end{prop}
\begin{proof}
{}
\begin{lyxlist}{00.00.0000}
\item [{$1\Rightarrow3$:}] If $f(\alpha)=g(\alpha)=0$ and $Af+Bg=1$
then $A(\alpha)f(\alpha)+B(\alpha)g(\alpha)=1\Rightarrow0+0=1$ which
is a contradiction, hence no such $A,B$ exists.
\item [{$3\Rightarrow4$:}] Suppose $\left\langle f,g\right\rangle =1=k[x]$,
then $1\in\left\langle f,g\right\rangle $ so there exists $A,B\in k[x]$
with $Af+Bg=1$
\item [{$4\Rightarrow2$:}] If $\left\langle f,g\right\rangle \neq k[x]$
then, since $k[x]$ is a PID, the ideal $\left\langle f,g\right\rangle =\left\langle h\right\rangle $
for some $h\in k[x]$ non-constant. So $f,g\in\left\langle h\right\rangle ,$that
is, $f=\widetilde{f}h$, $g=\widetilde{g}h$ and thus $f,g$ have
a non-constant common factor.
\item [{$2\Rightarrow5$:}] We write $f=\widetilde{f}h$, $g=\widetilde{g}h$
and set $\widetilde{A}=\widetilde{g}$ and $\widetilde{B}=-\widetilde{f}$.
Then $\widetilde{A}f+\widetilde{B}g=0$ and $\widetilde{A},\widetilde{B}$satisfy
the degree bound.
\item [{$5\Rightarrow2$:}] If $\widetilde{A}f+\widetilde{B}g=0$, then
every irreducible factor of $g$ divides $\widetilde{A}f$, since
$k[x]$ is a UFD. Since $\deg(g)>\deg(\widetilde{A})$ at least one
irreducible factor must divide $f$. Hence $f$ and $g$ have a common
factor.
\item [{$2\Rightarrow1$:}] If $f,g$ have a non-constant common factor
$h$, let $\alpha$ be any root of $h$, then $f(\alpha)=g(\alpha)=0$.
So $f$ and $g$ have a common root.
\end{lyxlist}
\end{proof}
Part 5 is the key idea here. Given $f=\sum a_{i}x^{i}$ and $g=\sum b_{j}x^{j}$
with $0\leq i\leq l$ and $0\leq j\leq m$, write $\widetilde{A}=\sum_{i=0}^{m-1}c_{i}x^{i}$
and $\widetilde{B}=\sum_{j=0}^{l-1}d_{j}x^{j}$ where $c_{i},d_{j}$
are undeterminate coefficients. 
\begin{eqnarray*}
0 & = & (c_{m-1}x^{m-1}+\dots+c_{0})(a_{l}x^{l}+\dots+a_{0})+(d_{l-1}x^{l-1}+\dots+d_{0})(b_{m}x^{m}+\dots+b_{0})\\
 & = & (c_{m-1}a_{l}+d_{l-1}b_{m})x^{l+m-1}+(c_{m-1}a_{l-1}+c_{m-2}a_{l}+d_{l-1}b_{m-1}+d_{l-2}b_{m})x^{l+m-2}+\dots+(c_{0}a_{0}+d_{0}b_{0})
\end{eqnarray*}


Thus all the coefficients of $x^{j}$ are zero. Remember that $a_{i}$
and $b_{j}$ are given, so we have a set of linear equations in the
$c$ and $d$ variables. We can count that we have $l+m$ variables
and linear equations. This gives the following matrix
\[
\underset{\underset{m}{\underbrace{\qquad\qquad\qquad\qquad\qquad}}\underset{l}{\underbrace{\qquad\qquad\qquad\qquad\qquad}}}{\begin{pmatrix}a_{l} & 0 & \dots &  & b_{m} & 0 & \dots\\
a_{l-1} & a_{l} & \dots &  & b_{m-1} & b_{m} & \dots\\
a_{l-2} & a_{l-1} & \ddots &  & b_{m-2} & b_{m-1} & \ddots\\
\vdots & \vdots & \ddots &  & \vdots & \vdots & \ddots\\
\\
\\
\\
 &  &  & a_{0} &  &  &  & b_{0}
\end{pmatrix}}
\]
There exists non-zero $\widetilde{A},\widetilde{B}$ if the correct
degree with $\widetilde{A}f+\widetilde{B}g=0$ if and only if the
determinant of this matrix is zero.
\begin{defn}
Let $f=\sum_{i=0}^{l}a_{i}x^{i}$ and $g=\sum_{j=0}^{m}b_{j}x^{j}$
be polynomials in $k[x]$ with $a_{l},b_{m}\neq0$. The \emph{Sylvester
matrix} of $f,g$ with respect to $x$ is the $(l+m)\times(l+m)$
matrix 
\[
\mbox{Syl}(f,g,x)=\underset{\underset{m}{\underbrace{\qquad\qquad\qquad\qquad\qquad}}\underset{l}{\underbrace{\qquad\qquad\qquad\qquad\qquad}}}{\begin{pmatrix}a_{l} & 0 & \dots &  & b_{m} & 0 & \dots\\
a_{l-1} & a_{l} & \dots &  & b_{m-1} & b_{m} & \dots\\
a_{l-2} & a_{l-1} & \ddots &  & b_{m-2} & b_{m-1} & \ddots\\
\vdots & \vdots & \ddots &  & \vdots & \vdots & \ddots\\
\\
\\
\\
 &  &  & a_{0} &  &  &  & b_{0}
\end{pmatrix}}
\]
The determinant of $\mbox{Syl}(f,g,x)$ is a polynomial in $a_{i},b_{i}$
with integer coefficients. This is called the \emph{resultant} of
$f$ and $g$ and is denoted $\mbox{Res}(f,g,x)$.\end{defn}
\begin{example*}
Let $f=x^{2}+3x+a$ and $g=x+b$. Then 
\[
\mbox{Syl (f,g,x)}=\begin{pmatrix}1 & 1 & 0\\
3 & b & 1\\
a & 0 & b
\end{pmatrix}
\]
so $\mbox{Res}(f,g,x)=b^{2}-(3b-a)=b^{2}-3b+a$, so $f$ and $g$
have a common factor if and only if $a=3b-b^{2}$.\end{example*}
\begin{thm}
Fix $f,g\in k[x]$, then $f,g$ have a common factor if and only if
$\mbox{Res}(f,g,x)=0$\end{thm}
\begin{proof}
This is what the previous work has been about.\end{proof}
\begin{example*}
$f=x^{2}+2x+1$, $g=x^{2}+3x+2$
\[
\mbox{Syl}(f,g,x)=\begin{pmatrix}1 & 0 & 1 & 0\\
2 & 1 & 3 & 1\\
1 & 2 & 2 & 3\\
0 & 1 & 0 & 2
\end{pmatrix}
\]
We see that $(r_{3}-r_{1})-(r_{2}-r_{1})-r_{4}=0$, so $\mbox{Res}(f,g,x)=0$.
(In fact the common factor is $x+1$)

$f=ax^{2}+bx+c$, $g=f'=2ax+b$
\[
\mbox{Syl}(f,g,x)=\begin{pmatrix}a & 2a & 0\\
b & b & 2a\\
c & 0 & b
\end{pmatrix}
\]
So $\mbox{Res}(f,g,x)=ab^{2}-2a(b^{2}-2ac)=-ab^{2}+4a^{2}c=-a(b^{2}-4ac)$
\end{example*}
Notice how in the second example we nearly ended up with the discriminant
of a quadratic equations.
\begin{defn}
Let $f=\sum_{i=0}^{l}a_{i}x^{i}$. Then the \emph{discriminant }of
$f$, $\mbox{disc}(f)=\frac{(-1)^{l-1}}{a_{l}}\mbox{Res}(f,f',x)$\end{defn}
\begin{prop}
The polynomial $\mbox{disc}(f)$ lies in $\zz[a_{0},\dots,a_{l}]$.
The polynomial $f$ has a multiple root if and only if $\mbox{disc}(f)=0$.\end{prop}
\begin{proof}
Note that the first row of $\mbox{Syl}(f,f',x)$ is $(a_{l},0,\dots,0,la_{l},0,\dots,0)$
so $a_{l}|\mbox{Res}(f,f',x)$ and thus $\mbox{disc}(f)\in\zz[a_{0},\dots,a_{l}]$.

Since $\deg(f)=l$ and $a_{l}\neq0$, we have $\mbox{disc}(f)=0$
if and only if $f$ and $f'$ have a common root, so we just need
to check that this happens if and only if $f$ has a multiple root.
Fix a root $\alpha$ of $f$ and write $f=(x-\alpha)^{m}\widetilde{f}$
where $\widetilde{f}(\alpha)\neq0$. Then $f'=m(x-\alpha)^{m-1}\widetilde{f}+(x-\alpha)^{m}\widetilde{f}'$,
so $f'(\alpha)=0$ if $m>1$. If $m=1$ then $f'(\alpha)=\widetilde{f}(\alpha)\neq0$.
So $\alpha$ is a root of $f'$ if and only if $\alpha$ is a multiple
root of $f$.
\end{proof}
\textbf{Generalizations:}
\begin{enumerate}
\item More variables:


Given $f,g\in k[x_{1},\dots,x_{n}]$, write $f=\sum_{i=0}^{l}a_{i}x_{1}^{i}$
and $g=\sum_{j=0}^{m}b_{j}x_{1}^{j}$ where $a_{i},b_{j}\in k[x_{2},\dots,x_{n}]$
and $a_{l},b_{m}\neq0$. Then $\mbox{Res}(f,g,x_{1})=\det(\mbox{Syl}(f,g,x_{1}))\in k[x_{2},\dots,x_{n}]$.
\begin{note*}
We can think about $f,g$ as polynomials in $k(x_{2},\dots,x_{n})[x_{1}]$
(fields of rational functions). So this is a special case of the first
one. In particular, either $\mbox{Res}(f,g,x_{1})=0$ or there exists
$A,B\in k(x_{1},\dots,x_{n})[x_{1}]$ with $Af+Bg=1$.\end{note*}
\begin{example*}
$\widetilde{A}=A\mbox{Res}(f,g,x_{1}),\widetilde{B}=B\mbox{Res}(f,g,x_{1})$
are polynomials in $k[x_{1},x_{2},\dots,x_{n}]$ so $\widetilde{A}f+\widetilde{B}g=\mbox{Res}(f,g,x_{1})$.
$A$ and $B$ comes from solution to 
\[
\mbox{Syl}(f,g,x_{1})\begin{pmatrix}c_{m-1}\\
\vdots\\
c_{0}\\
d_{l-1}\\
\vdots\\
d_{0}
\end{pmatrix}=\begin{pmatrix}0\\
\\
\vdots\\
\\
1
\end{pmatrix}
\]
Cramer's rule states $Ax=b$, $x_{i}=\frac{(-1)|A_{i}|}{|A|}$ where
$A_{i}$ is $A$ with $i^{\mathrm{th}}$ column replaced by $b$.
By Cramer's rule, the $c_{i}$ and $d_{j}$ have the form polynomial
is $x_{2},\dots x_{n}/\mbox{Res}(f,g,x_{1})$. So $A\mbox{Res}(f,g,x_{1})$
is a polynomial in $k[x_{1},\dots,x_{n}]$
\end{example*}

As a corollary to all of this we have that $\mbox{Res}(f,g,x_{1})\in\left\langle f,g\right\rangle \cap k[x_{2},\dots,x_{n}]$.
This is a cheaper way to do elimination/projection.
\begin{prop}
Fix $f,g\in k[x_{1},\dots,x_{n}]$ for degrees $l,m$ in $x_{1}$
respectively. If $\mbox{Res}(f,g,x_{1})\in k[x_{2},\dots,x_{n}]$
is zero at $(c_{2},\dots,c_{n})\in k^{n-1}$ then either $a_{l}(c_{2},\cdots,c_{n})=0$
or $b_{m}(c_{2},\dots,c_{n})=0$ or $\exists c_{1}\in\overline{k}$
such that $f(c_{1},\dots,c_{n})=g(c_{1},\dots,c_{n})=0$.\end{prop}
\begin{proof}
Let $f(x_{1},\underline{c})=f(x_{1},c_{2},\dots,c_{n})\in k[x_{1}]$
and similarly let $g(x_{1},\underline{c})\in k[x_{1}]$. If neither
$a_{l}(\underline{c})$, $b_{m}(\underline{c})=0$ then $f(x_{1},\underline{c})$
had degree $l$ and $g(x_{1},\underline{c})$ has degree $m$. So
$\mbox{Syl}(f(x_{1},\underline{c}),g(x_{1},\underline{c},),x_{1})$
is $\mbox{Syl}(f,g,x_{1})$ with $c_{2},\dots,c_{n}$ substituted
for $x_{2},\dots,x_{n}$. Thus $\mbox{Res}(f(x_{1},\underline{c}),g(x_{1},\underline{c}),x_{1})=0$,
so $f(x_{1},\underline{c})$ and $g(x_{1},\underline{c})$ have a
common root $c_{i}\in\overline{k}$. Hence $f(c_{1},c_{2},\dots,c_{n})=g(c_{1},c_{2},\dots,c_{n})=0$.
\end{proof}
\item Resultants of several polynomials.


Given $f_{1},\dots,f_{s}\in k[x_{1},\dots,x_{n}]$ we introduce new
variables $u_{2},\dots,u_{s}$ and let $g=u_{2}f_{2}+\dots u_{s}f_{s}$.
Write $\mbox{Res}(f_{1},g,x_{1})=\sum h_{\alpha}(x_{2},\dots,x_{n})u^{\alpha}$
with $\alpha\in\nn^{s-1}$. We call $h_{\alpha}\in k[x_{2},\dots,x_{n}]$
the generalised resultant.
\begin{example*}
Let $f_{1}=x^{3}+3x+2$, $f_{2}=x+1$, $f_{3}=x+5$. Then $g=u_{2}(x+1)+u_{3}(x+5)$
and 
\[
\mbox{Syl}(f,g,x_{1})=\begin{pmatrix}1 & u_{2}+u_{3} & 0\\
3 & u_{2}+5u_{3} & u_{2}+u_{3}\\
2 & 0 & u_{2}+5u_{3}
\end{pmatrix}
\]
so $\mbox{Res}(f_{1},g,x_{1})=-4u_{2}u_{3}+12u_{3}^{2}$. Hence $h_{1,1}=-4$
and $h_{0,2}=12$.\end{example*}
\begin{lem}
The polynomial $h_{\alpha}$ lies in $\left\langle f_{1},\dots,f_{s}\right\rangle \cap k[x_{2},\dots,x_{n}]$\end{lem}
\begin{proof}
Write $\mbox{Res}(f_{1},g,x_{1})=Af_{1}+Bg$ for $A,B\in k[u_{2},\dots,u_{s},x_{1},\dots,x_{n}]$.
Write $A=\sum A_{\alpha}u^{\alpha}$ and $B=\sum B_{\beta}u^{\beta}$
for $A_{\alpha},B_{\beta}\in k[x_{1},\dots,x_{n}]$. Then $\mbox{Res}(f_{1},g,x_{1})=\sum h_{\alpha}u^{\alpha}=\sum_{\alpha}(A_{\alpha}f_{1}+\sum_{i=2}^{s}B_{\alpha-e_{i}}f_{i})u^{\alpha}$.
So $h_{\alpha}=A_{\alpha}f_{1}+\sum B_{\alpha-e_{i}}f_{i}\in\left\langle f_{1},\dots,f_{s}\right\rangle $.
Furthermore $h_{\alpha}\in k[x_{2},\dots,x_{n}]$ by construction.
\end{proof}
\end{enumerate}

\subsection{Hilbert's Nullstellensatz}

Consider $\pi:\aa^{n}\to\aa^{m}$ projection onto the last $m$ coordinates.
We saw $\overline{\pi(X)}=V(I(X)\cap k[x_{n-m+1},\dots,x_{n}])$.
The question is what do we add then we take the closure? Given $y\in\overline{\pi(X)}$
is $y\in\pi(X)$?
\begin{thm}[Extension Theorem. ]
Let $k=\overline{k}$. Let $X=V(I)\subseteq\aa^{n}$ and let $\pi:\aa^{n}\to\aa^{n-1}$
be projection onto the last $n-1$ coordinates. Write $I=\left\langle f_{1},\dots,f_{s}\right\rangle $
with $f_{i}=g_{i}(x_{2},\dots,x_{n})x_{1}^{N_{i}}+\mathrm{l.o.t.\, in\,}x_{i}$.
Let $(c_{2},\dots,c_{n})\in V(I\cap k[x_{2},\dots,x_{n}])$. If $(c_{2},\dots,c_{n})\notin V(g_{1},\dots,g_{s})\subseteq\aa^{n-1}$
then $\exists c_{1}\in k$ with $(c_{1},\dots,c_{n})\in X$.\end{thm}
\begin{example*}
$X=V(xy-1)$, $f_{1}=xy-1,g_{1}=x$. Then the theorem say if $c_{1}\in V(0)=\aa^{1}$
and $c_{1}\notin V(x)$ then there exists $c_{2}$ with $(c_{1},c_{2})\in V(xy-1).$
Note that $V(0)$ comes from $\left\langle xy-1\right\rangle \cap k[x]$.\end{example*}
\begin{note*}
$I\subseteq I(X)$ so $I\cap k[x_{2},\dots,x_{n}]\subseteq I(X)\cap k[x_{2},\dots,x_{n}]$
so $V(I\cap k[x_{2},\dots,x_{n}])\supseteq\overline{\pi(X)}.$ How
useful this is depends on the choice of the generators of $I$. The
theorem talks about $I$, not $I(X)$, so this brings us closer to
the Nullstellensatz.\end{note*}
\begin{proof}
{}
\begin{lyxlist}{00.00.0000}
\item [{$s=1$:}] In this case $f=g_{1}(x_{2},\dots,x_{n})x_{1}^{N}+\mbox{\ensuremath{\mathrm{l.o.t}}.}$We
have $\left\langle f\right\rangle \cap k[x_{2},\dots,x_{n}]=\left\langle 0\right\rangle ,$
and $(c_{2},\dots,c_{n})\in V(\left\langle f\right\rangle \cap k[x_{2},\dots,x_{n}]$

\begin{casenv}
\item $N\neq0$: $g_{1}(c_{1},\dots,c_{n})\neq0$, then $f(x_{1},c_{2},\dots c_{n})$
is a polynomial of degree $N$ in $x_{1}$ so has a root $c_{1}$
in $k$.
\item $N=0$ then $g_{1}=f_{1}$, so if $(c_{2},\dots,c_{n})\in V(\left\langle f\right\rangle \cap k[x_{2},\dots,x_{n}])=V(f)\subseteq\aa^{n-1}$
\end{casenv}
\item [{$s=2$:}] The (previous) proposition shows that if $g_{1}(c_{1},\dots,c_{n})\neq0$
and $g_{2}(c_{2},\dots,c_{n})\neq0$ then the desired $c_{1}$ exists.
Suppose $(c_{2},\dots,c_{n})\notin V(g_{1},g_{2})$ then without loss
of generality $g_{1}(c_{2},\dots,c_{n})\neq0$. If $g_{2}(c_{2},\dots,c_{n})\neq0$
then $c_{1}$ exists. Otherwise replace $f_{2}$ by $f_{2}+x_{1}^{N}f_{1}$
for $N\gg0$. This does not change the ideal $\left\langle f_{1},f_{2}\right\rangle $
and it does not change $(c_{2},\dots,c_{n})\notin V(g_{1},g_{2})=V(g_{1},g_{1})$.
Then the proposition implies there exists $c_{1}$ with $f_{1}(c_{1},\dots,c_{n})=f_{2}(c_{1},\dots c_{n})=0$.
\item [{$s\geq3$:}] Also assume $g_{1}(c_{2},\dots,c_{n})\neq0$. Replace
$f_{2}$ by $f_{2}+x_{1}^{N}f_{1}$ for $N\gg0$ if necessary to guarantee
$g_{2}(\underline{c})\neq0$ and $\deg_{x_{1}}(f_{2})>\deg_{x_{1}}(f_{i})$
for $i>2$. Write $\mbox{Res}(f,\sum_{i=2}^{s}u_{i}f_{i},x_{1})=\sum h_{\alpha}u^{\alpha}$.
Since $h_{\alpha}\in I\cap k[x_{2},\dots,x_{n}]$ we have $h_{\alpha}(c_{2},\dots,c_{n})=0\,\forall\alpha$.
Thus $\mbox{Res}(f,\sum u_{i}f_{i},x_{1})(c_{2},\dots,c_{n},u_{1},\dots,u_{s})$
is the zero polynomial.\\
By construction the coefficients of the maximal power of $x_{1}$
in $f_{1}$ and in $\sum u_{i}f_{i}$ are $g_{1}$ and $g_{1}u_{1}$,
so are non-zero are $(c_{2},\dots,c_{n})$. Thus $0=\mbox{Res}(f,\sum u_{i}f_{i},x_{1})(c_{2},\dots,c_{n},u_{1},\dots,u_{s})=\mbox{Res}(f_{1}(x_{1},c_{2},\dots,c_{n}),\sum u_{i}f_{i}(x_{1},c_{2},\dots,c_{n}),x_{1})$.
Thus there exists $F\in k(u_{2},\dots,u_{s})[x_{1}]$ with $\deg_{x_{1}}F>0,F|f_{1}(x_{1},c_{2},\dots,c_{n})$.
Write $F=\widetilde{F}/g$ where $\widetilde{F}=k[u_{2},\dots,u_{s},x_{1}]$,
$g\in k[u_{2},\dots,u_{s}]$. Then $\widetilde{F}$ divides $f_{1}(x_{1},c_{2},\dots,c_{n})g(u_{2},\dots,u_{s})$.
Let $F''$ be an irreducible factor of $\widetilde{F}$ with positive
degree in $x_{1}$. Then $F''|f_{1}(x_{1},c_{2},\dots,c_{n})$. Thus
it does not contain any $u_{i}$. So $F''|\sum u_{i}f_{i}(x_{i},c_{2},\dots,c_{n})$
but $F''\in k[x_{1}]$ thus $F''|f_{i}(x_{1},c_{2},\dots,c_{n})$
for $2\leq i\leq n$. Then $F''|f_{i}(x_{1},c_{2},\dots,c_{n})$ for
all $1\leq i\leq s$. Then choose a root $c_{1}$ of $F''$. Then
$F''(c_{1})=0$ so $f_{i}(c_{1},\dots,c_{n})=0$ so $(c_{1},\dots,c_{n})\in X$
\end{lyxlist}
\end{proof}
\begin{namedthm}[Weak Nullstellensatz]
 Let $k=\overline{k}$. Suppose $I\subseteq k[x_{1},\dots,x_{n}]$
satisfies $V(I)=\emptyset$, then $I=\left\langle 1\right\rangle =k[x_{1},\dots,x_{n}]$.\end{namedthm}
\begin{proof}
We use an induction proof on $n$.
\begin{lyxlist}{00.00.0000}
\item [{$n=1$:}] $I=\left\langle f\right\rangle \subseteq k[x_{1}]$.
If $f\notin k$ there exists $\alpha\in k$ with $f(\alpha)=0$ so
$V(I)\neq\emptyset$. Thus if $V(I)=\emptyset$, $I=\left\langle f\right\rangle =\left\langle 1\right\rangle $.
\item [{$n>1$:}] Let $I=\left\langle f_{1},\dots,f_{s}\right\rangle $
and suppose $V(I)=\emptyset$. We may assume $\deg(f_{i})>0$ for
all $i$. Let the degree of $f_{1}$ be $N$. Consider the morphism
$\phi:\aa^{n}\to\aa^{n}$ given by $\phi^{*}:k[x_{1},\dots,x_{m}]\to k[z_{1},\dots,z_{n}]$
with $\phi^{*}(x_{i})=z_{i}+a_{i}z_{1}$ with $a_{1}=0$ and $a_{i}\in K$
for $i>1$.\\
Note: $\phi^{*}$ is an isomorphism, since the matrix is
\[
\begin{pmatrix}1 & 0 & 0 & \dots & 0\\
a_{2} & 1 & 0 & \dots & 0\\
a_{3} & 0 & 1 & \dots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
a_{n} & 0 & 0 & \dots & 1
\end{pmatrix}
\]
is invertible. $(\phi^{*})^{-1}(z_{i})=x_{i}-a_{i}x_{1}$. This means
that $1\in I$ if and only if $1\in\phi^{*}(I)$, and $\phi^{-1}(X)=V(\phi^{*}(I))=\emptyset$.
(This is because $V(\phi^{*}(I))=\{y:\phi^{*}f(y)=0\,\forall f\in I\}=\{y:f\circ\phi(y)=0\,\forall f\in I\}=\{y:\phi(y)\in V(I)\}$.
)\\
Let $f_{1}=\sum c_{u}x^{u}$. Note that $\phi^{*}(f_{1})=c(a_{2},\dots,a_{n})z_{1}^{N}+\mathrm{l.o.t\, in\:}z_{1}$
where $c(a_{2},\dots,a_{n})$ is the non-zero polynomial in $a_{2},\dots,a_{n}$,
i.e., $c(a_{2},\dots,a_{n})=\sum_{|u|=N}c_{u}\prod a_{i}^{u_{i}}$.
Thus we can choose $(a_{2},\dots,a_{n})\in k^{n-1}$ with $c(a_{2},\dots,a_{n})\neq0$.
(Exercise: this holds because the field is infinite)\\
Then $g_{1}\in k$, so $V(g_{1},\dots,g_{s})=\emptyset$ for $\phi_{1}^{*}(f_{i})=g_{i}z_{1}^{N_{i}}+\mathrm{l.o.t}.$
Let $J=\phi^{*}(I)\cap k[z_{1},\dots,z_{n}]$, then by the extension
theorem, if $(c_{2},\dots,c_{n})\in V(J)$ then there exists $c_{1}\in k$
with $(c_{1},\dots,c_{n})\in V(\phi^{*}(I))$. Thus $V(J)=\emptyset$
and by induction $J=\left\langle 1\right\rangle $, so $1\in\phi^{*}(I)$
and so $1\in I$.
\end{lyxlist}
\end{proof}
Note that is $1\in I$, we can write $1=\sum A_{i}f_{i}$ with $A_{i}\in k[x_{1},\dots,x_{n}]$.
\begin{namedthm}[Nullstellensatz]
 Let $k=\overline{k}$. Then $I(V(I))=\sqrt{I}$.\end{namedthm}
\begin{proof}
Let $f^{m}\in I$, then $f^{m}(x)=0\,\forall x\in V(I)$ so $f(x)=0$
for all $x\in V(I)$. Hence $f\in I(V(I))$, thus $\sqrt{I}\subseteq I(V(I))$

For the reverse inclusion, suppose $f\in I(V(I))$ and let $I=\left\langle f_{1},\dots,f_{s}\right\rangle $
and $\widetilde{I}=\left\langle f_{1},\dots,f_{s},1-yf\right\rangle \subseteq k[x_{1},\dots,x_{n},y]$.
Now that $V(\widetilde{I})=\emptyset$ since if $f_{1}(x_{1},\dots,x_{n})=\dots=f_{s}(x_{1},\dots,x_{n})=0$
then $f(x_{1},\dots,x_{n})=0$ so $1-yf(x_{1},\dots,x_{n})\neq0\,\forall y$.
So by the Weak Nullstellensatz we have that $1\in\widetilde{I}$.
So there exists $p_{1},\dots,p_{s},q\in k[x_{1},\dots,x_{n},y]$ with
$1=\sum p_{i}f_{i}+q(1-yf)$. Regard this as an expression in $k(x_{1},\dots,x_{n},y)$
and substitute $y=\frac{1}{f}$, then $1=\sum p_{i}(x_{1},\dots,x_{n}\frac{1}{f})f_{i}$.
Choose $m>0$ for which $p_{i}(x_{1},\dots,x_{n}\frac{1}{f})f^{m}\in k[x_{1},\dots,x_{n}]$
then $f^{m}=\sum(p_{i}(x_{1},\dots,x_{n}\frac{1}{f})f^{m})f_{i}$,
hence $f^{m}\in I$ and thus $f\in\sqrt{I}$.

\newpage{}
\end{proof}

\section{Irreducible Components}

(There is some cross-over with commutative algebra here, revise both!)
\begin{defn}
A variety $X\subseteq\aa^{n}$ is \emph{reducible }if $X=X_{1}\cup X_{2}$
with $X_{1},X_{2}$ non-empty varieties in $\aa^{n}$ and $X_{1},X_{2}\subsetneqq X$

$X$ is \emph{irreducible }if it is not reducible.\end{defn}
\begin{example*}
{}
\begin{itemize}
\item $X=V(x,y)\subseteq\aa^{2}$ then $X=V(x)\cup V(y)$.
\item $V(x^{2}+y^{2}-1)\subseteq\aa^{2}$ is irreducible but it is not trivial
to prove. We will prove this later.
\item $X\subseteq\aa^{1}$ is reducible if and only if $1<|X|<\infty$
\item $X=V(f)$ is a hypersurface in $\aa^{n}$. Let $f=cf_{1}^{\alpha_{1}}\dots f_{r}^{\alpha_{r}}$where
$c\in k$ and $f_{i}$ are distinct irreducible polynomials. Then
$V(f)=V(f_{1})\cup\dots\cup V(f_{r})$. Claim: If $r>1$ then $V(f)$
is reducible. We just need too show that $V(f_{i})\neq\emptyset,X$
for all $i$. Now $V(f_{i})\neq\emptyset$ since $1\notin\left\langle f_{i}\right\rangle $.
If $V(f_{i})=X$ then $V(f_{j})\subseteq V(f_{i})$ for some $j\neq i$.
Hence $f_{i}\in I(V(f_{j}))=\sqrt{f_{j}}=\left\langle f_{j}\right\rangle $
(exercise). So $f_{j}|f_{i}$ which contradicts $f_{i},f_{j}$ being
distinct irreducible. Actually $V(f_{i})$ are all irreducible, so
$X=V(f_{1})\cup\dots\cup V(f_{r})$ is a decomposition into irreducible.
\end{itemize}
\end{example*}
\begin{thm}
Let $X\subseteq\aa^{n}$ be a variety. Then $X=X_{1}\cup\dots\cup X_{r}$,
where each $X_{r}$ is irreducible. This representation is unique
up to permutation provided it is irredundant (i.e., $X_{i}\nsubseteq X_{j}$
for any $i\neq j$)\end{thm}
\begin{proof}
For this theorem, we use the fact that $k[x_{1},\dots,x_{n}]$ is
Noetherian, in particular, that there is no infinite ascending chain
$I_{1}\subsetneq I_{2}\subsetneq I_{3}\subsetneq\dots$ of ideals
in $k[x_{1},\dots,x_{n}]$. 

Existence: If $X$ is irreducible then we are done. Otherwise write
$X=X_{1}\cup X_{2}$ where $X_{1},X_{2}$ are proper subvarieties.
Again if both are irreducible then we are done. Otherwise we can write
$X_{1}=X_{11}\cup X_{12}$ and $X_{2}=X_{21}\cup X_{22}$ where $X_{ij}$
are proper subvarieties of $X_{i}$. Iterate this process. We claim
that this process terminates with $X=\cup X_{j}$ (Finite union).
If not we have an infinite descending chain $X\supsetneq X_{1}\supsetneq X_{11}\supsetneq X_{111}\supsetneq\dots$.
This gives a reverse containment $I(X)\subsetneq I(X_{1})\subsetneq I(X_{11})\subsetneq\dots$.
This chain must stabilize, so $I(X_{111\dots11})=I(X_{111\dots11111})$
but $V(I(X_{11\dots111})=X_{11\dots11}$which contradicts the proper
inclusion of varieties. Since $V(I(V(I)))=V(I)$. Hence the decomposition
process must terminates.

Uniqueness: Suppose $X=X_{1}\cup X_{2}\cup\dots\cup X_{r}=X_{1}'\cup\dots\cup X_{s}'$
are two irredundant irreducible decompositions. Consider 
\begin{eqnarray*}
X\cap(X_{i}') & = & X_{i}'\\
 & = & (X_{1}\cup\dots\cup X_{r})\cap X_{i}'\\
 & = & (X_{1}\cap(X_{i}'))\cup\dots\cup(X_{r}\cap(X_{i}'))
\end{eqnarray*}
Since $X_{i}'$ is irreducible, there must be $j$ with $X_{j}\cap(X_{i}')=X_{i}'$,
so $X_{i}'\subseteq X_{j}$. The same argument shows that there is
$k$ with $X_{j}\subseteq X_{k}'$, so we have $X_{i}'\subseteq X_{j}\subseteq X_{k}'$.
Since the decomposition is irredundant, $X_{i}'=X_{k}'=X_{k}$. This
construct a bijection between $\{X_{j}\}$ and $\{X_{i}'\}$, hence
$r=s$ and the decomposition is unique up to permutation.
\end{proof}
Note: This was a topological proof. A topological space with no infinite
descending chain of closed set is called Noetherian (note how this
is the ``opposite'' condition to Noetherian ring). Noetherian topological
spaces have irreducible decompositions.
\begin{thm}
Let $X\subseteq\aa^{n}$ be a variety. The following are equivalent:
\begin{enumerate}
\item $X$ is irreducible
\item The coordinate ring $k[X]$ is a domain
\item $I(X)\subseteq k[x_{1},\dots,x_{n}]$ is prime.
\end{enumerate}
\end{thm}
\begin{proof}
{}
\begin{lyxlist}{00.00.0000}
\item [{$2\iff3$:}] Recall $k[X]=k[x_{1},\dots,x_{n}]/I(X)$. So if $f,g\in k[X]$
satisfy $fg=0$ then there are lifts $\widetilde{f},\widetilde{g}\in k[x_{1},\dots,x_{n}]$
such that $\widetilde{f},\widetilde{g}\notin I(X)$ and $\widetilde{f}\widetilde{g}\in I(X)$.
Same argument works the other way round.
\item [{$1\Rightarrow3$:}] Suppose $I(X)$ is not prime, that is, there
exists $f,g\notin I(X)$ with $fg\in I(X)$. Let $X_{1}=V(f)\cap X$
and $X_{2}=V(g)\cap X$. Since $f,g\notin I(X)$ then $X_{1},X_{2}\subsetneq X$.
However $X_{1}\cup X_{2}=(V(f)\cap X)\cup(V(g)\cap X)=((V(f)\cup V(g))\cap X=V(fg)\cap X=X$
since $fg\in I(X)$. So $X$ is reducible.
\item [{$3\Rightarrow1$:}] Suppose $X=X_{1}\cup X_{2}$ with $X_{1}$
and $X_{2}$ proper subvarieties. Then $I(X)\subsetneq I(X_{1}),I(X)\subsetneq I(X_{2})$
(To see this take $V(\_)$ of both side then $V(I(X_{i}))=X_{i}$).
So we may choose $f\in I(X_{1})\backslash I(X)$ and $g\in I(X_{2})\backslash I(X)$.
Now $fg\in I(X_{1})\cap I(X_{2})$, so $V(I(X_{1})\cap I(X_{2}))\subseteq V(fg)$.
But $V(I(X_{1})\cap I(X_{2}))=V(I(X_{1}))\cup V(I(X_{2}))=X_{1}\cup X_{2}=X$.
So $fg\in I(X)$ so $I(X)$ is not prime.
\end{lyxlist}
\end{proof}
\begin{rem*}
Some text reserve the word ``variety'' for irreducible varieties
and call what we call varieties ``algebraic sets''.
\end{rem*}
Warning: If $X=V(I)$ is irreducible, this does not imply that $I$
is prime, just that $I(X)$ is. This about $I=\left\langle x^{2},xy^{2}\right\rangle \subseteq k[x,y]$.
\begin{thm}
Let $k=\overline{k}$ (this condition is unnecessary as there exists
a commutative algebra proof which show this theorem holds for $k\neq\overline{k}$.
See Commutative Algebra notes, this is the whole theory of Primary
Decomposition). Let $I=\sqrt{I}$ (a radical ideal) in $k[x_{1},\dots,x_{n}]$,
then $I=P_{1}\cap\dots\cap P_{r}$ where each $P_{i}$ is prime. This
decomposition is unique up to order if irredundant. \end{thm}
\begin{proof}
Let $X=V(I)$ and let $X=X_{1}\cup\dots\cup X_{r}$ be an irredundant
irreducible decomposition. Let $P_{i}=I(X_{i})$ which is prime by
the previous theorem and let $P=\cap P_{i}$. Then $V(P)=\cup V(P_{i})=\cup X_{i}=X$.
So $\sqrt{P}=I(X)=I$. If $f^{m}\in P$ for some $m>0$ then $f^{m}\in P_{i}$
for all $i$, so $f\in P_{i}$ for all $i$. Hence $f\in\cap P_{i}=P$
and thus $\sqrt{P}=P$. So $I=\cap P_{i}$

Uniqueness follows from the uniqueness of primary decomposition.
\end{proof}
The next question to come up is how can we determine the $P_{i}$,
that is the prime decomposition of radical ideals.
\begin{defn}
Let $I,J$ be ideals. Then the \emph{colon }(or \emph{quotient}) ideal
is $(I:J)=\{f\in k[x_{1},\dots,x_{n}]:fg=I\,\forall g\in J\}\subseteq I$.\end{defn}
\begin{example*}
Let $I=\left\langle x^{2},xy^{2}\right\rangle $ and $J=\left\langle x\right\rangle $.
Then $(I:J)=\{f:fg\in\left\langle x^{2},xy^{2}\right\rangle \,\forall g\in\left\langle x\right\rangle \}=\{f:fx\in\left\langle x^{2},xy^{2}\right\rangle \}=\left\langle x,y^{2}\right\rangle $\end{example*}
\begin{thm}
Let $I=\sqrt{I}$ and let $I=\cap P_{i}$ be an irredundant primary
decomposition. Then the $P_{i}$ are precisely the prime ideals of
the form $(I:f)$ for $f\in k[x_{1},\dots,x_{n}]$.\end{thm}
\begin{proof}
Notice: $(I:f)=(\cap P_{i}:f)=\cap(P_{i}:f)$. Now for any prime $P$
we have $(P:f)=\begin{cases}
\left\langle 1\right\rangle  & f\in P\\
P & f\notin P
\end{cases}.$ So $(I:f)=\cap_{f\notin P_{i}}P_{i}$. Fix $P_{i}$, since $P_{j}\nsubseteq P_{i}$
for any $j\neq i$, we can find $f_{j}\in P_{j}\backslash P_{i}$.
Let $f=\prod_{i\neq j}f_{j}$ then $f\in\cap_{j\neq i}P_{j}\backslash P_{i}$.
So $(I:f)=P_{i}.$

Conversely if $(I:f)=P$ is prime for some $f$, then $P=\cap_{f\notin P_{i}}P_{i}$
(as $\cap P_{i}=\prod P_{i}$ so $P=P_{i}$ for some $i$. In more
details $P\subseteq P_{i}$ for all $i$. If $P\subsetneq P_{i}$for
all $i$ then we can find $f_{i}\in P_{i}\backslash P$, so $f=\prod f_{i}\in\cap P_{i}\backslash P$
which is a contradiction. So $P=P_{i}$for some $i$)\end{proof}
\begin{example*}
Let $I=\left\langle xy,xz,yx\right\rangle $, then $V(I)=$ union
of $x,y,z$ axes $=V(x,y)\cup V(x,z)\cup V(y,z)$. So $I=\left\langle x,y\right\rangle \cap\left\langle x,z\right\rangle \cap\left\langle y,x\right\rangle $.
We want to see the theorem in action, so notice that $(I:z)=\left\langle x,y\right\rangle ,$$(I:y)=\left\langle x,z\right\rangle $
and $(I:x)=\left\langle y,z\right\rangle $. Warning: $(I:x+y)=\left\langle z,xy\right\rangle $
(not as obvious.)

Let $I=\left\langle x^{3}-xy^{2}-x\right\rangle $. Then $(I:x^{2}+y^{2}-1)=\left\langle x\right\rangle $
and $(I:x)=$$\left\langle x^{2}+y^{2}-1\right\rangle $.\end{example*}
\begin{note*}
If $X,Y$ are varieties in $\aa^{n}$ then $(I(X):I(Y))=I(X\backslash Y)$.
To see this: fix $x\in X\backslash Y$, since $x\notin Y$ there is
$g\in I(Y)$ with $g(x)\neq0$. So if $f\in(I(X):I(Y))$ then $f(x)g(x)=0$,
so $f(x)=0$ and thus $f\in I(X\backslash$Y) . Conversely if $f\in I(X\backslash Y)$
and $g\in I(Y)$ then $fg\in I(X)$ so $f\in(I(X):I(Y))$. Hence $\overline{(X\backslash Y)}=V(I(X):I(Y))$.
\end{note*}

\subsection{Rational maps}

How can we decide if $X$ is irreducible? This is hard in general!
We use the following trick. If $\phi:Y\to X$ is surjective and $X=X_{1}\cup X_{2}$
then $Y=\phi^{-1}(X_{1})\cup\phi^{-1}(X_{2})$. Now both sides are
closed and proper if both $X_{1}$ and $X_{2}$ are. So if $X$ is
reducible then so is $Y$. Or if $Y$ is irreducible then so is $X$.
\begin{defn}
A morphism $\phi:X\to Y$ of affine varieties is \emph{dominant} if
$\overline{\phi(X)}=Y$\end{defn}
\begin{example*}
Take $\phi:V(xy-1)\to\aa^{1}$ defined by $(x,y)\mapsto x$. This
is not surjective but it is dominant.\end{example*}
\begin{prop}
A morphism $\phi:X\to Y$ is dominant if and only if $\phi^{*}:k[Y]\to k[X]$
is injective.\end{prop}
\begin{example*}
$k[x]\to k[x,y]/\left\langle xy-1\right\rangle ,$ $x\mapsto x$ (linked
to the previous exampled) is injective.\end{example*}
\begin{proof}
A morphism $\phi:X\to Y$ induces a homomorphism $\phi^{*}:k[Y]\to k[X]$.
Now $\phi(X)\subseteq Z\subsetneq Y$ ($Z$ a variety) if and only
if there exists $g\in I(X)\backslash I(Y)$ with $g(\phi(x))=0\,\forall x\in X$,
so $\phi^{*}(g(x))=0\,\forall x\in X$. Hence $\phi^{*}g\in I(X)$
and thus the image of $g$ in $k[Y]$ is non-zero but is mapped to
zero by $\phi^{*}$ so $\phi^{*}$ is not injective.\end{proof}
\begin{prop}
If $\phi:X\to Y$ is dominant and $X$ is irreducible then so is $Y$.\end{prop}
\begin{proof}
Since $\phi$ is dominant, the map $\phi^{*}:k[Y]\to k[X]$ is injective.
Since $X$ is irreducible, we have $k[X]$ is a domain, and thus so
is $k[Y]$. Hence $Y$ is also irreducible.\end{proof}
\begin{defn}
A \emph{rational map} $\phi:\aa^{n}\dashrightarrow\aa^{m}$ is defined
by $\phi(x_{1},\dots,x_{n})=(\phi_{1}(x_{1},\dots,x_{n}),\dots,\phi_{m}(x_{1},\dots x_{n}))$
with $\phi_{i}\in k(x_{1},\dots,x_{n})$ (the field of rational functions)\end{defn}
\begin{example*}
$\phi:\aa^{1}\dashrightarrow\aa^{1}$ defined by $\phi(x)=\frac{1}{x}$.
\end{example*}
Warning: $\phi$ is not necessarily a function defined on all of $\aa^{n}$.
Write $\phi_{i}=\frac{f_{i}}{g_{i}}$ for $f_{i},g_{i}\in k[x_{1},\dots,x_{n}]$
and let $U=\{x\in\aa^{n}:g_{i}(x)\neq\}$. Then $\phi:U\to\aa^{m}$
is well defined. Notice that $U$ is an open set $(U=\aa^{n}\backslash V(\prod g_{i}))$.
In the above example $\phi$ is defined on $U=\{x\in\aa^{1}:x\neq0\}$.
\begin{note*}
A rational map induces a $k$-algebra homomorphism $\phi^{*}:k[z_{1},\dots,z_{m}]\to k(x_{1},\dots,x_{n})$
defined by $z_{i}\mapsto\phi_{i}$. Conversely any such $k$-algebra
homomorphism determines a rational map.\end{note*}
\begin{example*}
Let $\phi:\aa^{1}\to\aa^{2}$ be the inverse stereographic projection,
that is, defined by $\phi(t)=(\frac{t^{2}-1}{t^{2}+1},\frac{2t}{t^{2}+1})$.
It is a rational map from $\aa^{1}$ to $V(x^{2}+y^{2}-1)$. It is
defined on $\aa^{1}\setminus\pm i$ and the image $V(x^{2}+y^{2}-1)\backslash\{(1,0)\}$. \end{example*}
\begin{defn}
Let $Y\subseteq\aa^{m}$, a \emph{rational map }$\phi:\aa\dashrightarrow Y$
is a rational map $\phi:\aa^{n}\to\aa^{m}$ with $\phi^{*}(I(Y))=0$,
so $\phi^{*}:k[Y]\to k(x_{1},\dots,x_{n})$.\end{defn}
\begin{example*}
Let $\phi:\aa^{1}\to V(x^{2}+y^{2}-1)$ be the inverse stereographic
projection. Then $\phi^{*}(x)=\frac{t^{2}-1}{t^{2}+1},\phi^{*}(y)=\frac{2t}{t^{2}+1}$.
Hence $\phi^{*}(x^{2}+y^{2}-1)=(\frac{t^{4}-2t^{2}+1+4t^{2}}{t^{4}+2t^{2}+1}-1)=0$,
so $\phi$ is indeed a rational map.
\end{example*}
What about rational maps $\phi:X\dashrightarrow\aa^{m}$? We recall
that a morphism $X\to\aa^{m}$ was an equivalence class of morphism
$\aa^{n}\to\aa^{m}$. But we have some problems: consider $X=V(x)\subseteq\aa^{2}$
and $\phi:\aa^{2}\dasharrow\aa^{3}$ defined by $\phi(x,y)=(x^{2},\frac{1}{xy},y^{3})$.
Then $\phi$ is define on $U=\{(x,y):x,y\neq0\}$, $\phi$ is not
defined at $(x,y)$ for any $(x,y)\in X$. The solution to this is
to allow rational maps that are defined on enough of $X$.
\begin{defn}
Let $R$ be a commutative ring with identity. An element $f\in R$
is a \emph{zero-divisor} if there exists $g\in R$ with $g\neq0$
such that $fg=0$.
\end{defn}

\begin{defn}
Let $\phi:\aa^{n}\dashrightarrow\aa^{m}$ be a rational map with $\phi_{i}=\frac{f_{i}}{g_{i}}$
where $f_{i}$ and $g_{i}$ have no common factors. Then $\phi$ is
\emph{admissible} on $X$ if the image of each $g_{i}$ in $k[X]$
is a non-zero divisor.\end{defn}
\begin{example*}
$\phi:\aa^{2}\dashrightarrow\aa^{3}$, $\phi(x,y)=(x^{2},\frac{1}{xy},y^{3})$
is not admissible on $V(x)$. 

Let $X=V(x^{2}-y^{2})\subseteq\aa^{2}$ and $\phi:\aa^{2}\dashrightarrow\aa^{1}$
defined by $(x,y)\mapsto\frac{1}{x+y}$ . Then $\phi$ is not admissible
on $X$ as $x+y$ is a zero divisor in $k[x,y]/(x^{2}-y^{2})$. On
the other hand $\psi:\aa^{2}\dasharrow\aa^{2}$ defined by $\psi(x,y)=(\frac{1}{x},\frac{1}{y})$
is.\end{example*}
\begin{defn}
Let $U$ be the set of non-zero divisor on a ring $R$. Note $U\neq0$
since $1\in U$. The \emph{total quotient ring} (ring under ``obvious''
multiplication and addition) is as follow
\[
Q(R)=R[U^{-1}]=\frac{\{\frac{r}{s}:r\in R,s\in U\}}{\frac{r_{1}}{s_{1}}=\frac{r_{2}}{s_{2}}\mathrm{\, if\,}r_{1}s_{2}=r_{2}s_{1}}
\]


(This like the localization in Commutative Algebra)\end{defn}
\begin{example*}
$R=\zz,U=\zz\setminus\{0\}$ then $Q(R)=\qq$.

$R=k[x_{1},\dots,x_{n}]$ then $Q(R)=k(x_{1},\dots,x_{n})$.\end{example*}
\begin{defn}
If $X$ is a variety, the total quotient ring of $k[X]$ is written
$k(X)$ and is called the \emph{ring of rational functions of }$X$.\end{defn}
\begin{note*}
If $R$ is a domain, $U=R\backslash\{0\}$, so $Q(R)$ is the field
of fractions of $R$. So if $X$ is irreducible, $k(X)$ is the field
of fractions of $k[X]$.\end{note*}
\begin{prop}
Let $\phi:\aa^{n}\dasharrow\aa^{m}$ be a rational map admissible
on a affine variety $X\subseteq\aa^{n}$. Then $\phi^{*}$ induces
a $k$-algebra homomorphism $\phi^{*}:k[z_{1},\dots,z_{n}]\to k(X)$.
Conversely each such homomorphism arise from a rational map.\end{prop}
\begin{proof}
Write $\phi_{i}=f_{i}=g_{i}$ with $f_{i}$ and $g_{i}$ not sharing
any irreducible factors. By hypothesis each $g_{i}$ is a non-zero
divisor on $k[X]$. So $\frac{f_{i}}{g_{i}}$ is a well defined element
of $k(X)$. Thus $\phi^{*}:k[z_{1},\dots,z_{m}]\to k(X)$ given by
$\phi^{*}(z_{i})=\frac{f_{i}}{g_{i}}$ is well defined.

Conversely given $\phi^{*}:k[z_{1},\dots,z_{m}]\to k(X)$ write $\phi^{*}(z_{i})=\frac{f_{i}}{g_{i}}$
for some $f_{i},g_{i}\in k[x_{1},\dots,x_{n}]$ with $g_{i}$ a non-zero
divisor on $k[X]$. Then $\phi:\aa^{n}\dasharrow\aa^{m}$ defined
by $\phi_{i}(x)=f_{i}(x)/g_{i}(x)$ is admissible on $X$.\end{proof}
\begin{defn}
Let $X\subseteq\aa^{n}$ be an affine variety. Two rational maps $\phi,\psi:\aa^{n}\dashrightarrow\aa^{m}$
admissible on $X$ are said to be \emph{equivalent on }$X$ if the
induced homomorphism $\phi^{*},\psi^{*}:k[z_{1},\dots,z_{m}]\to k(X)$
are equal.\end{defn}
\begin{example*}
Let $X=V(x+y)\subseteq\aa^{2}$. Let $\phi:\aa^{2}\dasharrow\aa^{2}$
be defined by $\phi(x,y)=(\frac{3x}{2y^{2}},\frac{2x}{3x+5y})$. This
is defined on $U_{\phi}=\{(x,y):y^{2}\neq0,3x+5y\neq0\}$. Let $\psi:\aa^{2}\dasharrow\aa^{2}$
be defined by $\psi(x,y)=(\frac{3}{2x},-1)$. This is defined on $U_{\psi}=\{(x,y):x\neq0\}$.
These are clearly not the same rational maps but we will show that
they are equivalent on $X$.

$\phi^{*}:k[z_{1},z_{2}]\to k(x,y)$ is defined by $\phi^{*}(z_{1})=\frac{3x}{2y^{2}}$
and $\phi^{*}(z_{2})=\frac{2x}{3x+5y}$. And $\psi^{*}:k[z_{1},z_{2}]\to k(x,y)$
is defined by $\psi^{*}(z_{1})=\frac{3}{2x}$ and $\psi^{*}(z_{2})=-1$.
Now in $k(X)=k[x,y]/(x+y)$ we have
\begin{eqnarray*}
\frac{3x}{2y^{2}} & = & \frac{3x}{2x^{2}}=\frac{3}{2x}\\
\frac{2x}{3x+5y} & = & \frac{2x}{2y}=-1
\end{eqnarray*}
So $\phi^{*},\psi^{*}:k[z_{1},z_{2}]\to k(X)$ are equal so $\phi,\psi$
are equivalent on $X$.

(Check $\phi=\psi$ on $U_{\psi}\cap U_{\phi}\cap X$)\end{example*}
\begin{defn*}
Let $X\subseteq\aa^{n}$ and $Y\subseteq\aa^{m}$ be affine varieties.
\emph{A rational map} $\phi:X\dasharrow Y$ is an equivalence class
of rational maps $\phi:\aa^{n}\dasharrow Y$ admissible on $X$.\end{defn*}
\begin{cor}
Let $X\subseteq\aa^{n}$ and $Y\subseteq\aa^{m}$ be affine varieties.
Then there is a one to one correspondence between rational maps $X\dasharrow Y$
and $k$-algebra homomorphism $k[Y]\to k(X)$.\end{cor}
\begin{defn}
A rational map $\phi:X\dasharrow Y$ is \emph{dominant} if $\phi^{*}:k[Y]\to k(X)$
is injective.\end{defn}
\begin{example*}
Let $\phi:\aa^{1}\dasharrow V(x^{2}+y^{2}-1)$ defined by $\phi(t)=(\frac{t^{2}-1}{t^{2}+1},\frac{2t}{t^{2}+1})$.
Then $\phi$ is dominant.\end{example*}
\begin{lem}
If $\phi:X\dasharrow Y$ is dominant and $X$ is irreducible, then
so is $Y$\end{lem}
\begin{proof}
Since $\phi$ is dominant we have by definition $\phi^{*}:k[Y]\to k(X)$
is injective. Since $X$ is irreducible, $k[X]$ is a domain, so $k(X)$
is a field. Hence $k[Y]$ is also a domain and thus $Y$ is irreducible.\end{proof}
\begin{cor}
$V(x^{2}+y^{2}-1)$ is irreducible.\end{cor}
\begin{defn}
Let $Y\subseteq\aa^{n}$ be an affine variety. A \emph{rational parametrisation}
of $Y$ is a rational map $\phi:\aa^{n}\dasharrow Y$ such that $Y=\overline{\im(\phi)}$,
i.e., a dominant rational map $\phi:\aa^{n}\dasharrow Y$. Such $Y$
are called \emph{unirational.}\end{defn}
\begin{note*}
Unirational varieties are irreducible, by the lemma, and we have $k(Y)\hookrightarrow k(x_{1},\dots,x_{n})$.\end{note*}
\begin{defn}
A variety $X$ is \emph{rational} if it admits a rational parametrisation
$\phi:\aa^{n}\dasharrow X$ such that the induced field extension
$\phi^{*}:k(X)\hookrightarrow k(x_{1},\dots,x_{n})$ is an isomorphism.\end{defn}
\begin{cor}
$X$ is rational if and only if $k(X)\cong k(x_{1},\dots,x_{n})$\end{cor}
\begin{proof}
If $X$ is rational then $k(X)\cong k(x_{1},\dots,x_{n})$ by definition,
so we just need to show the converse. Suppose we have $\phi^{*}:k(X)\to k(x_{1},\dots,x_{n})$.
Then $\phi^{*}|_{k[X]}$ is injective, so defines a dominant rational
map $\phi:\aa^{n}\dasharrow X$. Hence $X$ is rational.\end{proof}
\begin{defn*}
Let $X,Y$ be irreducible varieties. We say $X,Y$ are \emph{birational
}if $k(X)\cong k(Y)$ as $k$-algbera.\end{defn*}
\begin{prop}
If $X,Y$ are irreducible varieties and $k(X)\cong k(Y)$ then there
exists dominant rational maps $X\dashrightarrow Y$ and $Y\dashrightarrow X$
that are inverses.\end{prop}
\begin{proof}
If $\phi^{*}:k(X)\overset{\cong}{\to}k(Y)$, then $\phi^{*}|_{k[X]}$is
injective, so the corresponding rational map $\phi:Y\dasharrow X$
is dominant. Similarly $\phi^{*-1}$ induces a dominant rational map
$\phi^{-1}:X\dasharrow Y$. By construction $\phi^{*}\circ\phi^{*-1}=\id|_{Y}$.

\newpage{}
\end{proof}

\section{Projective Varieties.}
\begin{defn}
\emph{Projective Space} $\pp^{n}$ over a field $k$ is $(k^{n+1}\backslash\{0\})/\sim$
where $\underline{v}\sim\lambda\underline{v}$ for $\lambda\in k^{*}=k\backslash\{0\}$.
A point in $\pp^{n}$ correspond to a line through the origin in $k^{n+1}$.\end{defn}
\begin{notation*}
$[x_{0}:x_{1}:\dots:x_{n}]$ is the equivalence class of $(x_{0},x_{1},\dots,x_{n})\in k^{n+1}$.
\end{notation*}
Recall: A polynomial $f=\sum c_{u}x^{u}$ is \emph{homogeneous }if
$|u|=d$ for all $u$ with $c_{u}\neq0$ for some $d$.
\begin{defn}
An ideal $I\subseteq k[x_{0},x_{1},\dots,x_{n}]$ is homogeneous if
$I=\left\langle f_{1},\dots,f_{s}\right\rangle $, where each $f_{i}$
is homogeneous.\end{defn}
\begin{example*}
$\left\langle 7x_{0}^{2}+8x_{1}x_{2}+9x_{1}^{2},3x_{1}^{3}+x_{2}^{3}\right\rangle $
is, $\left\langle x+y^{2},y^{2}\right\rangle =\left\langle x,y^{2}\right\rangle $
is.\end{example*}
\begin{defn}
Let $f\in k[x_{0},\dots,x_{n}]$. Then $f=\sum f_{i}$ where each
$f_{i}$ is a homogeneous polynomial of degree $i$. The $f_{i}$
are called the \emph{homogeneous components }of $f$.\end{defn}
\begin{example*}
Let $I$ be a homogeneous ideal and let $f\in I$. Then each homogeneous
component of $f$ is in $I$. Idea: we choose $g_{1},\dots,g_{s}$
homogeneous with $I=\left\langle g_{1},\dots,g_{s}\right\rangle $.
Then we can write $f=\sum c_{u_{i}}x^{u_{i}}f_{i}$, where the $f_{i}$could
be repeated. Then $f_{i}=\sum_{j:\deg(x^{u_{j}})+\deg(g_{i})=i}c_{u_{i}}x^{u_{i}}g_{i}\in I$.\end{example*}
\begin{defn}
Let $I$ be a homogeneous ideal in $k[x_{0},x_{1},\dots,x_{n}]$.
The \emph{projective variety }defined by $I$ is $\vv(I)=\{[x]\in\pp^{n}:f(x)=0\mathrm{\, for\, all\, homogenous\,}f\in I\}$\end{defn}
\begin{example*}
{}
\begin{itemize}
\item Let $I=\left\langle 2x_{0}-x_{1},3x_{0}-x_{2}\right\rangle $. Then
$\vv(I)=\{[1:2:3]\}\subseteq\pp^{3}$.
\item $I=\left\langle x_{0}x_{2}-x_{1}^{2}\right\rangle $. Then $\vv(I)=\{[1:t:t^{2}]:t\in k\}\cup\{[0:0:1]\}$
\item $I=\left\langle x_{0},x_{1},x_{2}\right\rangle \subseteq k[x_{0},x_{1},x_{2}]$.
$\vv(I)=\emptyset$. Note that the weak Nullstellenzatz does not apply
here.
\item $I=\left\langle x_{0}x_{3}-x_{1}x_{2},x_{0}x_{2}-x_{1}^{2},x_{1}x_{3}-x_{2}^{2}\right\rangle \subseteq k[x_{0},x_{1},x_{2},x_{3}]$.
Then $\vv(I)=([1,t,t^{2},t^{3}]:t\in k\cup\{[0:0:0:1]\}$. ``The
twisted cubic''
\end{itemize}
\end{example*}
\begin{note*}
Points in $\vv(I)$ correspond to lines through $\underline{0}$ in
$V(I)\subseteq\aa^{n+1}$. $V(I)$ is called the affine cover over
$I$.\end{note*}
\begin{defn}
\emph{The Zariski topology} on $\pp^{n}$ has closed sets $\vv(I)$
for $I\subseteq k[x_{0},\dots,x_{n}]$.
\end{defn}

\subsection{Affine Charts}
\begin{defn}
Let $U_{i}=\{[u]\in\pp^{n}:x_{i}\neq0\}$. We can write $x\in U_{i}$
uniquely as $[x_{0}:\dots:1:\dots:x_{n}]$ ($1$ in $i^{\mathrm{th}}$
position). $U_{i}$ bijection with $\aa^{n}$. $\pp^{n}=\cup_{i=1}^{n}U_{i}$
``affine cover of $\pp^{n}$''. We can think of $\pp^{n}=U_{0}\cup\{[x]:x_{0}=0\}$.
See that $U_{0}$ is a kind of like $\aa^{n}$ while the set is $\pp^{n-1}$
sometime called ``hyperplane at infinity''. Fix $I$ homogeneous
in $k[x_{0},\dots,x_{n}]$ and let $X=\vv(I)\subseteq\pp^{n}$. Let
$X\cap U_{i}=\{[x]\in\pp^{n}:f(x)=0\,\forall f\in I\}=\{[x_{0}:\dots:1:\dots,x_{n}]\in\pp^{n}:f(x_{0},\dots,1,\dots,x_{n})=0\,\forall f\in I\}=V(I_{i})\subseteq\aa^{n}$
where $I_{i}=\left\langle f(x_{0},\dots,1,\dots,x_{n}):f\in I\right\rangle =1|_{x_{1}=1}$.
\[
X=\bigcup_{i=0}^{n}X\cap U_{i}.
\]
Is a union of affine varieties. This is called an \emph{affine cover},
let $X\cup U_{i}$ are called \emph{affine charts}. \end{defn}
\begin{example*}
$X=\vv(x_{0}x_{2}-x_{1}^{2})\subseteq\pp^{n}$. Then:
\begin{itemize}
\item $X\cap U_{0}=V(x_{2}-x_{1}^{2})=\{(t,t^{2}):t\in k\}$. 
\item $X\cap U_{1}=V(x_{0}x_{2}-1)=\{(t,\frac{1}{t}):t\in k\}$. 
\item $X\cap U_{2}=V(x_{0}-x_{1}^{2})=\{(t^{2},t):t\in k\}$
\end{itemize}
Actually $X=(X\cap U_{0})\cup(X\cap U_{2})$ in this case. We can
think of $X$ as created by ``gluing together'' three affine varieties
$ $$X\cap U_{0},X\cap U_{1},X\cap U_{2}$. This is how abstract varieties
are defined (not covered in this module).
\end{example*}
Given an affine variety $X\subseteq\aa^{n}$, we can embed it into
$\pp^{n}$ by identifying $\aa^{n}$ with $U_{i}$ for some $i$ (normally
$i=0$).
\begin{defn}
The \emph{projective closure} of $X\subseteq\aa^{n}$ in $\pp^{n}$
is the Zariski closure of $X\subseteq U_{i}\subseteq\pp^{n}$ in $\pp^{n}$
(Assume by default $U_{0}$)\end{defn}
\begin{example*}
$X=V(x_{2}-x_{1}^{2})=\{(t,t^{2}):t\in k\}\subseteq\aa^{2}$ . The
projective closure is the Zariski closure of $\{[1:t:t^{2}]:t\in k\}$.
This adds $[0:0:1]$
\end{example*}
Question: Given $X$ how can we compute the projective closure in
$\pp^{n}$?
\begin{defn}
Let $f=\sum c_{u}x^{u}\in k[x_{1},\dots,x_{n}]$. The\emph{ homogenization}
of $f$ is $\widetilde{f}=\sum_{u}x^{u}x_{0}^{d-|u|}$ where $d=\max|u|$

Let $I\subseteq k[x_{1},\dots,x_{n}]$ be an ideal. Its \emph{homogenization}
is $\widetilde{I}=\left\langle \widetilde{f}:f\in I\right\rangle $
\end{defn}
Warning: If $I=\left\langle f_{1},\dots,f_{s}\right\rangle $ then
we do not always have $\widetilde{I}=\left\langle \widetilde{f_{1}},\dots,\widetilde{f_{s}}\right\rangle .$
For example, consider $I=\left\langle x_{1}-1,x_{1}\right\rangle \subseteq k[x_{1}]$.
We have $I=\left\langle 1\right\rangle $, $\widetilde{I}=\left\langle 1\right\rangle \neq\left\langle x_{1}-x_{0},x_{1}\right\rangle =\left\langle x_{1},x_{0}\right\rangle $
\begin{prop}
Let $k=\overline{k}$, $I=\sqrt{I}\subseteq k[x_{1},\dots,x_{n}]$.
The projective closure of $V(I)\subseteq\aa^{n}$ via the identification
$\aa^{n}=U_{0}$ is $\vv(\widetilde{I})\subseteq\pp$\end{prop}
\begin{proof}
If $x\in V(I)$, $f(x)=0\,\forall f\in I$ then $\widetilde{f}(I,x)=0\,\forall\widetilde{f}\in\widetilde{I}$.
So $[1:x]\in\vv(\widetilde{I})$. So the projective closure of $\vv(I)$
is contained in $\vv(\widetilde{I})$

Conversely, suppose that $f\in k[x_{0},\dots,x_{n}]$ is homogeneous
with $f([1:x])=0\,\forall x\in V(I)$. Then $g=f(1,x)\in I(V(I))=\sqrt{I}=I$.
Then $f=x_{0}^{k}\widetilde{g}$ for some $k\geq0$ so since $g\in I,$$f\in\widetilde{I}$
and thus $\vv(\widetilde{I})$ is contained in the projective closure
of $V(I)$.
\end{proof}
Question: How can we compute $\widetilde{I}$? Answer: Let $<$ be
any term order with $\deg(X^{u})>\deg(X^{v})\Rightarrow X^{u}>X^{v}$
(for example we can revlex). Let $G=\{g_{1},\dots,g_{s}\}$ be a Grobner
basis for $I$ with respect to $<$. We claim that $\widetilde{I}=\left\langle \widetilde{g_{1}},\dots,\widetilde{g_{s}}\right\rangle $. 
\begin{proof}[Proof of above claim]
 Extend $<$ to a term order $\widetilde{<}$ on $k[x_{0},\dots,x_{n}]$
by setting $x_{0}^{a}x^{u}\widetilde{<}x_{0}^{b}x^{v}$ if $\begin{cases}
x^{u}<x^{v} & x^{u}\neq x^{v}\\
a<b & x^{u}=x^{v}
\end{cases}$. Note that $\mathrm{in}_{\widetilde{<}}(\widetilde{f})=\ld(f)$.
Let $F\in\widetilde{I}$ be a homogeneous polynomial in $k[x_{0},\dots,x_{n}]$.
Then $f=\sum A_{i}\widetilde{f}_{i}$ for some $f_{i}\in I$, $A_{i}\in l[x_{0},\dots,x_{n}]$.
Write $f(x_{1},\dots,x_{n})=F(1,x_{1},\dots,x_{n})\in k[x_{1},\dots,x_{n}]$.
Then $f=\sum A_{i}(1,x_{1},\dots,x_{n})f_{i}$ so $f\in I$. We know
that $F=x_{0}^{k}\widetilde{f}$ for some $k\geq0$ so $\mathrm{in}_{\widetilde{<}}(\widetilde{f})=x_{0}^{k}\mathrm{in}_{\widetilde{<}}(\widetilde{f})=x_{0}^{k}\ld(f)$.
Since $G$ is a Grobner basis for $I$ with respect to $<$, we have
that $\ld(f_{j})|x_{0}^{k}\ld(f)$ for some $g$. Hence $\mathrm{in}_{\widetilde{<}}(f)\in\left\langle \mathrm{in}_{\widetilde{<}}(\widetilde{g_{1}}),\dots,\mathrm{in}_{\widetilde{<}}(\widetilde{g_{s}})\right\rangle $.
So $\{\widetilde{g_{1}},\dots,\widetilde{g_{s}}\}$ is a Grobner basis
for $\widetilde{I}$, hence it generates $\widetilde{I}$.\end{proof}
\begin{prop}
Let $k=\overline{k}$. $\vv(I)=\emptyset$ if and only if $\left\langle x_{0},\dots,x_{n}\right\rangle \subseteq\sqrt{I}$. \end{prop}
\begin{proof}
Let $X=V(I)\subseteq\aa^{n+1}$. Then $V(I)=\emptyset$, implies either
$X=\emptyset$ so $1\in I$ or $X=\{0\}$ so $\left\langle x_{0},\dots,x_{n}\right\rangle \subseteq\sqrt{I}$. 

Conversely if $\left\langle x_{0},\dots,x_{n}\right\rangle \subseteq\sqrt{I}$
then $V(I)\subseteq\{0\}$, so $\vv(I)=\emptyset$.\end{proof}
\begin{defn}
The ideal $\left\langle x_{0},\dots,x_{n}\right\rangle $ is called
the \emph{irrelevant ideal.}

Let $X\subseteq\pp^{n}$. \emph{The ideal $I(X)$ }is $I(X)=\left\langle \mathrm{homogeneous\,}f\in k[x_{0},\dots,x_{n}]:f([x])=0\,\forall x\in X\right\rangle $

\emph{Homogeneous coordinate ring} of $X\subseteq\pp^{n}$ is $k[x_{0},\dots,x_{n}]/I(X)$.\end{defn}
\begin{thm}[Projective Nullstellensatz]
 Let $k=\overline{k}$. Let $I$ be a homogeneous ideal in $[x_{0},\dots,x_{n}]$
with $\left\langle x_{0},\dots,x_{n}\right\rangle \nsubseteq\sqrt{I}$.
Then $I(\vv(I))=\sqrt{I}$.\end{thm}
\begin{proof}
Let $X=\vv(I)$ and let $Y=V(I)\subseteq\aa^{n+1}$ be affine cover
of $X$. Then $I(\vv(I))=\{f\,\mathrm{homogeneous:}f([x])=0\,\forall[x]\in X\}=\{f:f(x)=0\,\forall x\in Y\}=I(Y)=\sqrt{I}$\end{proof}
\begin{defn}
A projective variety $X$ is \emph{reducible} if $X=X_{1}\cup X_{2}$
with $X_{1},X_{2}\subsetneq X$ and $X_{1},X_{2}$ are subvarietes
of $X$.
\end{defn}
Exercise: If $X$ is a non-empty irreducible variety then $I(X)$
is prime.


\subsection{Morphisms of projective varieties.}

A rational map of degree $d$, $\phi:\pp^{n}\dasharrow\pp^{m}$ is
given by $\phi([x_{0}:\dots:x_{n}])=[\phi_{0}(x_{0},\dots,x_{n}):\dots:\phi_{m}(x_{0},\dots,x_{n})]$
where $\phi_{i}$ are homogeneous polynomials of degree $d$ in $k[x_{0},\dots,x_{n}]$.
\begin{example*}
$\phi([s:t])=[s^{2}:st:t^{2}]$ a rational map $\pp^{1}\dasharrow\pp^{2}$
of degree 2. (actually a morphism)

$\phi([s:t])=[s^{3}:s^{2}:st^{2}]$ a rational map $\pp^{1}\dasharrow\pp^{2}$
of degree $3$. This is not defined at $[s:t]=[0:1]$ (not a morphism)
\end{example*}
A rational map $\phi:\pp^{n}\dasharrow\pp^{m}$ is a \emph{morphism
}if $\phi$ is defined for all $[x]\in\pp^{n}$.
\begin{note*}
We don't need to use rational functions as we can clear denominators.
The polynomials need to be homogeneous of the same degree to make
the map well defined, i.e., independent of representative of $[x]\in\pp^{n}$.
\end{note*}
$\phi$ is a morphism if and only if $V(\phi_{0},\dots,\phi_{m})\subseteq\pp^{n}$
is empty if and only if $\left\langle x_{0},\dots,x_{n}\right\rangle \subseteq\sqrt{\left\langle \phi_{0},\dots,\phi_{m}\right\rangle }$
\begin{defn}
A rational map $\phi_{i}:\pp^{n}\dasharrow\pp^{m}$ is \emph{linear
}if degree $\phi_{i}=1\,\forall i$.
\end{defn}
In that case $\phi$ is determined by a $m+1\times n+1$ matrix $A=(a_{ij})$.
Then $\phi$ is a morphism when $\mathrm{rank}A=n+1$ (or $\ker(A)=0$).
In the case $n=m$, then $\phi$ is a morphism if and only if $A$
is invertible.

The set $\{\phi:\phi([x])=[Ax]\mathrm{\, for\, an\, invertible\,}(n+1)\times(n+1)\,\mathrm{matrix\,}A\}=\mbox{Aut}(\pp^{n})$
forms a group under composition. Note that $A_{1}=\begin{bmatrix}1 & 0\\
0 & 1
\end{bmatrix}$ and $A_{2}=\begin{bmatrix}2 & 0\\
0 & 2
\end{bmatrix}$ define the same morphism. If fact $\mbox{Aut}(\pp^{n})=\GL_{n+1}/k^{*}=:\mbox{PGL}_{n+1}$
where $k^{*}=\{\lambda I\}$


\subsubsection{Veronese Embedding}
\begin{defn}
The morphism $\phi:\pp^{1}\dasharrow\pp^{d}$ given by $\phi([x_{0}:x_{1}])=[x_{0}^{d}:x_{0}^{d-1}x_{1}:\dots:x_{1}^{d}]$
is called the $d^{\mathrm{th}}$ \emph{Veronese embedding.}

$Y=\im(\phi)=\vv(y_{1}y_{j+1}-y_{i+1}y_{j}:0\leq i,j\leq n-1)$. $Y$
is called the \emph{rational normal curve of degree $d$.}\end{defn}
\begin{example*}
There are ${n+d \choose d}$ monomials of degree $d$. To see this,
notice that any string of $d$ $*$ and $n$ $|$ correspond uniquely
to a monomial of degree $d$, for example, $***|*|***$ correspond
to $x_{0}^{3}x_{1}x_{2}^{3}$ while $||**|*$ correspond to $x_{2}^{3}x_{3}$.
So the number of the monomials is the number of such strings.
\end{example*}
The $d^{\mathrm{th}}$ \emph{Veronese embedding }of $\pp^{n}$ is
$\pp^{n}\dasharrow\pp^{{n+d \choose d}-1}$ defined by $[x_{0}:\dots:x_{n}]\mapsto[x_{0}^{d}:x_{0}^{d-1}x_{1}:\dots:x_{n}^{d}]$
(all monomial of degree $d$). The image of $\phi$ is $\vv(z_{\alpha}z_{\beta}-z_{\gamma}z_{\delta}:\alpha+\beta=\gamma+\delta)$
where $z$ are coordinates on $k[z_{\alpha}:\alpha\in\nn^{n+1},\sum\alpha_{i}=d]$.
This is a generalization of $y_{i}\leftrightarrow z_{d-i,i}$. We
prove all of this in the following proposition
\begin{prop}
$\im(\phi)$ is closed and equals $\vv(z_{\alpha}z_{\beta}-z_{\gamma}z_{\delta}:\alpha+\beta=\delta+\gamma)$,
where $z$ are coordinates $k[z_{\alpha}:\alpha\in\nn^{n}:\sum\alpha_{i}=d]$. \end{prop}
\begin{proof}
Let $Z=\vv(z_{\alpha}z_{\beta}-z_{\gamma}z_{\delta}:\alpha+\beta=\delta+\gamma)$.
If $z=\phi(x)$ then $z_{\alpha}z_{\beta}-z_{\gamma}z_{\delta}=x^{\alpha}x^{\beta}-x^{\delta}x^{\gamma}=0$,
so if $\alpha+\beta=\gamma+\delta$ we have $\im(\phi)\subseteq\zz$.

Conversely we want to consider $z\in Z$ and want to find $[x]\in\pp^{n}$
with $\phi([x])=[z]$. We first show there is $i$ with $z_{de_{i}}\neq0$.
To see this, suppose $z_{\alpha}\neq0$ for some $\alpha$ (there
must be some such $\alpha$). Without loss of generality $\alpha_{0}>0$.
If $\alpha_{0}<\frac{d}{2}$ we write $2\alpha=(2\alpha e_{0}+\widetilde{\alpha})+\alpha''$
where $\widetilde{\alpha}_{0},\alpha_{0}''=0$ and $\sum\alpha_{i}''=d$
(For example if $d=5,\alpha=(2,2,1)$ then $(4,4,2)=(4,1,0)+(0,3,2))$

Then $z_{\alpha}^{2}=z_{2\alpha_{0}e_{0}+\widetilde{\alpha}}z_{\alpha''}$
(here we have $\alpha=\beta=\alpha,\gamma=2\alpha_{0}e_{0}+\widetilde{\alpha},\delta=\alpha'')$.
So $z_{\alpha}\neq0$ implies that $z_{2\alpha_{0}e_{0}+\widetilde{\alpha}}\neq0$.
So after repeated applications, we may assume $\alpha_{0}>\frac{d}{2}$.
Then we write $2\alpha=de_{0}+(2\alpha-de_{0})$, then $z_{2}^{\alpha}=z_{de_{0}}z_{2\alpha-de_{0}}$,
so $z_{de_{0}}\neq0$. Now set $x_{i}=\frac{z_{(d-1)e_{0}+e_{i}}}{z_{de_{0}}}$
for $1\leq i\leq n$ and $x_{0}=1$. Set $[z']=\phi([x])$. 

We now show that $[z']=[z]$ to show this we show $z_{\alpha}'z_{de_{0}}=z_{\alpha}$.
We do this by a proof on induction on $\sum_{i=1}^{n}\alpha_{i}$.
The base case is by definition/construction. The general case follows
from $z_{\alpha}z_{de_{0}}=z_{\alpha-e_{i}+e_{0}}z_{(d-1)e_{0}+e_{i}}$
for $\alpha_{i}>0$. Note that this is also true for $z'$. Hence
$z_{de_{0}}z'_{\alpha}=z_{de_{0}}z'_{de_{0}}z'_{\alpha}=z_{de_{0}}z'_{\alpha-e_{1}+e_{0}}z'_{(d-1)e_{0}+e_{i}}=z_{\alpha-e_{i}+e_{0}}z'_{(d-1)e_{0}+e_{i}}=\frac{z_{\alpha-e_{1}+e_{0}}z_{(d-1)e_{0}+e_{p}}}{z_{de_{0}}}=z_{\alpha}$.
\end{proof}

\subsubsection{Segre Embedding}

The Segre embedding realizer $\pp^{n}\times\pp^{m}$ as a subvarieties
of $\pp^{(n+1)(m+1)-1}$. We map $([x],[y])\in\pp^{n}\times\pp^{m}$
to $\phi([x],[y])=([x_{i}y_{j}]:0\leq i\leq n,0\leq j\leq m).$
\begin{prop}
$\im(\phi)=\vv(z_{ij}z_{kl}-z_{il}z_{kj}:0\leq i,k\leq n,0\leq j,l\leq m)$
where the $z$ are coordinates on $k[z_{ij}:0\leq i\leq n,0\leq j\leq m]$.
(Notice that they are the 2 by 2 minors of a generic $(n+1)\times(m+1)$
matrix $(z_{ij})$\end{prop}
\begin{proof}
Let $Y=\vv(z_{ij}z_{kl}-z_{il}z_{kj}:0\leq i,k\leq n,0\leq j,l\leq m)$.
If $z=\phi([x],[y])$ then $z_{ij}z_{kl}-z_{il}z_{kj}=x_{i}y_{j}x_{k}y_{l}-x_{i}y_{l}x_{k}y_{j}=0$
so $\im(\phi)\subseteq Y$.

Conversely, given $[z]\in Y$, without loss of generality, we may
assume $z_{00}\neq0$. Set $x_{i}=z_{i0}/z_{00}$ and $y_{j}=z_{0j}/z_{00}$,
$x_{0}=y_{0}=1$, and let $z'=\phi([x],[y])$. Then the equation on
$z_{ij}z_{00}-z_{i0}z_{0j}$ implies that $[z']=[z]$.
\end{proof}

\subsubsection{Grassmannian}

The Grassmannian $G(d,n)$ parametrizes all $d$-dimensial subspace
of $k^{n}$
\begin{example*}
$G(1,n)=\pp^{n-1}$ (a one dimensional subspace is a line through
$0$)

$G(n-1,n)=\pp^{n-1}$
\end{example*}
We'll describe $G(d,n)$ as a projective variety by the Pucker $G(d,n)\hookrightarrow\pp^{{n \choose d}-1}$.
Let $V\subseteq k^{n}$ be a $d$-dimensial subspace. Choose a basis
$v_{1},\dots,v_{d}$ for $V$ and write $A_{v}=\begin{pmatrix}v_{1}\\
\vdots\\
v_{d}
\end{pmatrix}$ for the $d\times n$ matrix with rows the $v_{i}$. Map $V$ to the
vector of $d\times d$ of $A_{V}$ in $\pp^{{n \choose d}-1}$, for
example $\phi:V\mapsto(1:3:4:-1:-2:-2)$. We name the coordinates
on $\pp^{{n \choose d}-1}$, $x_{I}$ where $I\subseteq\{1,\dots,n\}$
and $|I|=d$. $I$ indexes the columns of the $d\times d$ submatrix
of $A_{v}$ whose determinant is $\phi(V)_{I}$
\begin{note*}
{}
\begin{enumerate}
\item $\phi(V)$ is not the zero vectors, since rank $A_{V}=d$, so $A_{v}$
has a non-vanishing minor of size $d$
\item If we choose a different basis $v_{1}',\dots,v_{d}'$ for $V$, then
$A'_{v}=UA_{V}$ where $U$ is a $d\times d$ invertible matrix (in
fact the change of basis matrix). So the $I$th minor of $A_{v}'$
is $\det(U)$. So $A_{V},A_{V}'$ gives the same point in $\pp^{{n \choose d}-1}$.
This means the map $\phi:V\mapsto\phi(V)\in\pp^{{n \choose d}-1}$is
well defined
\item We can recover $V$ from $\phi(V)$.

\begin{example*}
If $\phi(V)=[1:0:0:0:0:0]$ then $V=\mathrm{span}\left(\begin{pmatrix}1\\
0\\
0\\
0
\end{pmatrix},\begin{pmatrix}0\\
1\\
0\\
0
\end{pmatrix}\right)$. Since $\phi(V)_{12}=1$ we can assume $A_{V}=\begin{pmatrix}1 & 0 & * & *\\
0 & 1 & * & *
\end{pmatrix}$. 
\end{example*}
\end{enumerate}
\end{note*}
Let $I$ be an index with $\phi(V)_{I}\neq0$ (this exists by $1$).
Let $B$ be the $d\times d$ submatrix of $A_{V}$ indexed by $I$.
Then $\det(B)\neq0$. So $A'_{V}=B^{-1}A_{V}$ has an identity matrix
in the column indexed by $I$. But then for $j\notin I$, $(A_{V}')_{ji}=\pm\phi(V)_{I\setminus\{1\}\cup\{j\}}$. 

Question: What does $\im(\phi)$ look like?
\begin{example*}
$G(2,4)$ assume that $\phi(V)_{12}\neq0$ so that we can take $A_{V}=\begin{pmatrix}1 & 0 & a & b\\
0 & 1 & c & d
\end{pmatrix}$, $\phi(V)=[1:c:d:-a:-b:ad-bc]$. Note $\phi(V)\subseteq\vv(x_{12}x_{34}-x_{13}x_{24}+x_{14}x_{23})$.
The equation $x_{12}x_{34}-x_{13}x_{24}+x_{14}x_{23}$ is invariant
(up to sign) under the $S_{4}$ action on the labels, where we set
$x_{21}=-x_{12}$. This says that $\phi(V)\subseteq\vv(x_{12}x_{34}-x_{13}x_{24}+x_{14}x_{23})$
any $V$. Alternatively, we could check the other row reduced forms.

Conversely, if $[z]\in\vv(x_{12}x_{34}-x_{13}x_{24}+x_{14}x_{23})$
with $z_{12}\neq0$, then $[z]=\phi(V)$ for $V=\mbox{row}\begin{pmatrix}1 & 0 & -\frac{z_{23}}{z_{12}} & -\frac{z_{23}}{z_{12}}\\
0 & 1 & \frac{z_{13}}{z12} & \frac{z_{14}}{z_{12}}
\end{pmatrix}$

The formula \foreignlanguage{english}{$x_{12}x_{34}-x_{13}x_{24}+x_{14}x_{23}$
is called a \emph{Plucker relation}.}
\end{example*}
\selectlanguage{english}%
For the embedding $G(d,n)\hookrightarrow\pp^{{n \choose d}-1}$ we
get a Plucker relation $PJ_{1}J_{2}$ for all $J_{1},J_{2}\subseteq\{1,\dots,n\}$
with $|J_{1}|=d-1$, $|J_{2}|=d+1$. 
\[
PJ_{1}J_{2}=\sum_{j\in J_{2}}(-1)^{\sgn(j,J_{1})}X_{J_{1}\cup j}X_{J_{2}\backslash j}
\]
where $X_{J_{1}\cup j}=0$ if $j\in J_{1}$ and $\sgn(j,J_{1})=\#(i\in J_{i}:i>j)+\#(i\in J_{2}:i<j)$
\selectlanguage{british}%
\begin{example*}
$n=4$, $d=2$, $J_{1}=\{1\}$ and $J_{2}=\{2,3,4\}$. Then $PJ_{1}J_{2}=x_{12}x_{34}-x_{13}x_{24}+x_{14}x_{23}$\end{example*}
\begin{defn}
Let $I_{d,n}=\left\langle PJ_{1}J_{2}:J_{1},J_{2}\subseteq\{1,\dots,n\},|J_{1}|=d-1,|J_{2}|=d+1\right\rangle \subseteq k[X_{1}:|I|=d]$\end{defn}
\begin{thm}
$G(d,n)=\im(\phi)=\vv(I_{d,n})$\end{thm}
\begin{proof}
Assignment sheet.
\end{proof}
Question; What are the affine charts for $G(d,n)$? 

Answer: $G(d,n)\cap U_{I}$ is $V$ which look like $A_{V}=(I_{d}|\widetilde{A})$
(where $I_{d}$ are the columns of $I$ and $\widetilde{A}$ the other
columns, not that $\widetilde{A}$ is an arbitary $d\times(n-d)$
matrix). So $G(d,n)\cap U_{I}\cong\aa^{d(n-d)}$.

Check: $G(2,4)\cap U_{1,2}=V(x_{12}x_{34}-x_{13}x_{24}+x_{14}x_{23})\subseteq\aa^{5}$.
This is isomorphic to $\aa^{4}$ since $k[x_{13},x_{14},x_{23},x_{24},x_{34}]/(x_{12}x_{34}-x_{13}x_{24}+x_{14}x_{23})\cong k[x_{13},x_{14},x_{23},x_{24}]\cong k[\aa^{4}]$

We can think of $G(d,n)$ as ${n \choose d}$ copies of $\aa^{d(n-d)}$
``glue together''. This worked for any field, e.g., the real Grassmannian
is the manifold of dimension $d(n-d)$. Similarly for $\cc$.

\newpage{}


\section{Dimension and Hilbert Polynomial}
\begin{defn}
A ring $R$ is \emph{$\zz$-graded }if there is a decomposition (as
groups) $R\cong\oplus_{i\in\zz}R_{i}$ with $R_{i}R_{j}\subseteq R_{i+j}$.
The $R_{i}$ are called the \emph{graded pieces} and $f\in R_{i}$
is homogeneous of degree $i$.

A \emph{graded} $k$-algebra is a $k$-algebra $R$ with $cf\in R_{i}\,\forall f\in R_{i}$
(so each $R_{i}$ is a $k$-vector space). Then $k\subseteq R_{0}$
(normally for our examples $R_{0}=k$)\end{defn}
\begin{example*}
$R=k[x_{0},\dots,x_{n}]$, $R_{i}$ are polynomials of degree $i$.

$S=k[x_{0},\dots,x_{n}]$, $I$ homegenous ideal. $R=S/I$ then $R_{i}=S_{i}/I_{i}$.

$X\subseteq\pp^{n}$ a projective variety, $R(X):=k[x_{0},\dots,x_{n}]/I(X)$
is the projective coordinate ring of $X$.\end{example*}
\begin{defn}
Let $R$ be a graded $k$-algbera with $\dim_{k}(R_{i})<\infty$ for
all $i$. The \emph{Hilbert Function} of $R$ is $H_{R}(d)=\dim_{k}R_{d}$
(note $H_{R}:\zz\to\nn$).\end{defn}
\begin{example*}
Let $R=k[x_{0},\dots,x_{n}]$. Then $H_{R}(d)={n+d \choose d}={n+d \choose n}$
since a basis for $R_{d}$ is the set of monomials of degree $d$.

$S=k[x_{0},x_{1},x_{2}]$ and $f$ homogeneous of degree $3$. Let
$R=S/\left\langle f\right\rangle $.

\begin{tabular}{|c|c|}
\hline 
$i$ & $\dim_{k}(S/\left\langle f\right\rangle )_{i}=\dim_{k}(S)-\dim_{k}\left\langle f\right\rangle _{i}$\tabularnewline
\hline 
\hline 
$<0$ & $0$\tabularnewline
\hline 
$0$ & $1$\tabularnewline
\hline 
$1$ & $3$\tabularnewline
\hline 
$2$ & $6$\tabularnewline
\hline 
$3$ & ${2+3 \choose 2}-1=9$\tabularnewline
\hline 
$4$ & ${2+4 \choose 2}-3=12$\tabularnewline
\hline 
\end{tabular}

In general we have the following graded short exact sequence:
\[
\xymatrix{0\ar[r] & S\ar[r] & S\ar[r] & S/\left\langle f\right\rangle \ar[r] & 0\\
 & g\ar@{|->}[r] & gf\, g\ar@{|->}[r] & \overline{g}
}
\]
So $\dim_{k}(S/\left\langle f\right\rangle )_{d}=\dim_{k}S_{d}-\dim_{k}S_{d-3}$
(assuming $d\geq1$)
\end{example*}
How can we compute $H_{R}$?
\begin{prop}
Let $I\subseteq S=k[x_{1},\dots,x_{n}]$ and let $<$ be a term order.
The (image of the) monomials in $S$ not in the initial ideal of $I$
form a $k$-basis for $S/I$. Thus if $I$ is homogeneous $H_{S/I}(d)=H_{S/\ld(I)}(d)$\end{prop}
\begin{proof}
Let $f$ be polynomial in $S$. Then the remainder of dividing $f$
be a Grobner basis for $I$ with respect to $<$ is a polynomial $g$
with $f-g\in I$ and $g=\sum c_{u}x^{u}$ where $c_{u}\neq0$ implies
$c^{u}\notin\ld(I)$. So $f=g$ in $S/I$ and $g\in\mathrm{span}\{x^{U}:x^{U}\notin\ld(I)\}$,
so this set spans $S/I$. If $I$ is homogeneous and $f$ has degree
$d$, then so does $g$, so the set of monomials not in $\ld(I)$
of degree $d$ spans $(S/I)_{d}$. To see that these sets are linearly
independent, note that if $f=\sum c_{u}x^{u}$ is a linear dependence,
then $f\neq0$ but $f\equiv0$ in $S/I$, hence $f\in I$ and $c_{u}\neq0\Rightarrow x^{u}\notin\ld(I)$.
Then $\ld(f)\notin\ld(I)$ which contradicts $f\in I$. So we conclude
that $\{x^{u}:x^{u}\notin\ld(I)\}$ is linearly independent so is
a basis for $S/I$. If $I$ is homogeneous then $\{x^{u}:\deg(x^{u})=d,x^{u}\notin\ld(I)\}$
is a basis for $(S/I)_{d},$ as well as a basis for $(S/\ld(I))_{d}$.
So the Hilbert functions are equal.
\end{proof}
This reduces the question to ``how can we compute $H_{S/m}$for $m$
a monomial ideal?'' The key point is the following short exact sequence.
Let $I$ be a homogeneous ideal and $f$ homogeneous of degree $d$.
Then we have the following s.e.s
\[
0\to S/(I:f)\overset{\phi}{\to}S/I\overset{\psi}{\to}S/(I,f)\to0
\]
 where $(I:f)=\{g\in S:fg\in I\}$ and $(I,f)=I+\left\langle f\right\rangle $.
The map $\phi$ is defined by $g\mapsto fg$ while the map $\psi$
is defined by $g\mapsto\overline{g}$. We check that this sequence
is exact.
\begin{enumerate}
\item $\phi$ is injective: If $fg\in I$ then $g\in(I:f)$
\item $\im(\phi)=\ker(\psi)$: 


``$\subseteq$'': $g\in S,fg\in I+\left\langle f\right\rangle $
so $\im(\phi)\subseteq\ker(\psi)$.\\
``$\supseteq$'': Suppose $g\in\ker(\psi)$ then $g=i+hf$ for $i\in I$.
Hence $g-hf=i\in I$, so $g\equiv hf$ in $S/I$, so $g=\phi(h)\in\im(\phi)$.

\item $\psi$ is surjective since $I+\left\langle f\right\rangle \supseteq I$.
\end{enumerate}
This short exact sequence is graded, i.e., $0\to(S/(I:f))_{m}\to(S/I)_{m+d}\to(S/(I,f))_{m+d}\to0$.
Recall: given an exact sequence $0\to U\to V\to W\to0$ of vector
space we have $V\cong U\oplus W$. So $\dim V=\dim U+\dim W$. In
our case $\dim_{k}(S/I)_{m+d}=\dim_{k}(S/(I:f))_{m}+\dim_{k}(S/(I,f))_{m+d}$.
Apply this when $I$ is a monomial ideal and $f$ a variable. Then
$(I:f),(I,f)\supseteq I$. If $f$ is chosen carefully we have strict
containment, so eventually $(I:f),(I,f)$ are monomial prime ideals,
which we know the Hilbert function of.
\begin{lem}
Let $I=\left\langle x_{i_{1}},\dots,x_{i_{s}}\right\rangle \subseteq S=k[x_{0},\dots,x_{n}]$
be prime. Then $H_{S/I}(m)={m+n-s \choose m-s}={m+n-s \choose n}$\end{lem}
\begin{proof}
$S/I\cong k[x_{j}:j\neq i_{k}\,\mathrm{for\, any\,}k]$. This has
Hilbert function ${n-s+m \choose m}={n-s+m \choose n-s}$. This is
a polynomial in $m\geq-(n-s)$\end{proof}
\begin{example*}
Let $I=\left\langle x_{0}x_{3},x_{0}x_{2},x_{1}x_{3}\right\rangle \subseteq S=k[x_{0},x_{1},x_{2},x_{3}]$.
Let us choose $f=x_{0}$. Then $(I:f)=\left\langle x_{2},x_{3}\right\rangle $
and $(I,f)=\left\langle x_{0},x_{1}x_{3}\right\rangle $. So $H_{S/I}(d)=H_{S/\left\langle x_{2},x_{3}\right\rangle }(d-1)+H_{S/\left\langle x_{0},x_{1}x_{3}\right\rangle }(d)$.
Take $f=x_{1}$. Then $(\left\langle x_{0},x_{1}x_{3}\right\rangle :x_{1})=\left\langle x_{0},x_{3}\right\rangle $and
$(\left\langle x_{0},x_{1}x_{3}\right\rangle ,x_{1})=\left\langle x_{0},x_{1}\right\rangle $.
So $H_{S/I}(d)=H_{S/\left\langle x_{1},x_{3}\right\rangle }(d-1)+H_{S/\left\langle x_{0},x_{3}\right\rangle }(d-1)+H_{S/\left\langle x_{0},x_{1}\right\rangle }(d)=d+d+d+1=3d+1$.
This is valid for $d\geq0$.\end{example*}
\begin{thm}
Let $I$ be a homogeneous ideal in $S=k[x_{0},\dots,x_{n}]$. Then
there exists a polynomial $P\in\qq[t]$ such that $H_{S/I}(d)=P(d)$
for $d\gg0$.\end{thm}
\begin{proof}
Since $H_{S/I}=H_{S/\ld(I)}$, we may assume that $I$ is a monomial
ideal. The case $I$ is a monomial prime was the lemma. The proof
is by Noetherian induction.

Given a monomial ideal $I$ that is not prime, we may assume that
the theorem is true for all monomial ideals containing $I$. Choose
a variable $x_{i}$ properly dividing a generator of $I$. This must
exists since $I$ is not prime. Then $(I:x_{i}),(I,x_{i})\supsetneq I$.
By induction there exists $P_{1},P_{2}\in\qq[t]$ with $H_{S/(I:x_{i})}(d)=P_{1}(d)$
for $d\gg0$ and $H_{S/(I,x_{i})}(d)=P_{2}(d)$ for $d\gg0$. Then
$H_{S/I}(d)=P_{1}(d-1)+P_{2}(d)$ for $d\gg0$ and this is a polynomial
in $d$.\end{proof}
\begin{defn}
Let $X\subseteq\pp^{n}$ be a projective variety. The polynomial $P:=P_{x}$
of the theorem for $I=I(X)$ is called the \emph{Hilbert Polynomial}.

Let $X\subseteq\pp^{n}$ be a projective variety. Then the \emph{dimension}
of $X$ is the degree of the Hilbert Polynomial.\end{defn}
\begin{example*}
{}
\begin{enumerate}
\item If $V\subseteq k^{n+1}$ is a subspace of $\dim(d+1)$ then $\pp(V)\subseteq\pp^{n}$
is a subsvariety of dimension $d$
\item $X=\mathrm{twisted\, cubic}=$image of $\phi:\pp^{1}\to\pp^{3}$ defined
by $[t_{0},t_{1}]\mapsto[t_{0}^{3},t_{0}^{2}t_{1},t_{0}t_{1}^{2},t_{1}^{3}]=\vv(x_{0}x_{3}-x_{1}x_{2},x_{0}x_{2}-x_{1}^{2},x_{1}x_{3}-x_{2}^{2})=\vv(I)$.
$\ld(I)=\left\langle x_{0}x_{3},x_{0}x_{2},x_{1}x_{3}\right\rangle $
(where $x_{0}>x_{1}>x_{2}>x_{3})$. Then from the previous work we
have $H_{k}[x_{0},\dots,x_{3}]/\ld(I)=3d+1$ for $d\geq1$, so $\dim(X)=1$
\end{enumerate}
\end{example*}
There are many different (equivalent) definition of dimension. Proving
they are equivalent is non-trivial. For example, for $X\subseteq\aa^{n}$
we can define $\dim(X)$ to be the dimension of the projective closure
of $X$ (See Eisenbud Commutative Algebra, Chapters 8-13)


\subsection{Singularities}

How close is a variety to a manifold?

Let $X=V(f)\subseteq\aa^{n}$. Fix $\underline{a}\in X$. What is
the tangent plane to $X$ at $\underline{a}$? 
\begin{example*}
{}
\begin{itemize}
\item $n=2$

\begin{itemize}
\item $X=V(y-f(x))$, for example $X=V(y-x^{2})$. The tangent line at $a$
is $(y-a_{2})=\frac{df}{dx}|_{a}(x-a_{1})$
\item $X=V(f(x,y))$, for example $X=V(y^{2}+x^{2}-1)$. The slope is $\frac{dy}{dx}=-\frac{\frac{df}{dx}}{\frac{df}{dy}}$.
Tangent line is $y-a_{2}=\frac{dy}{dx}|_{a}(x-a_{1})$. So $\frac{df}{dy}(y-a_{2})=-\frac{df}{dx}(x-a_{1})\Rightarrow\frac{df}{dy}(y-a_{2})+\frac{df}{dx}(x-a_{1})=0$.
\end{itemize}
\item $n=3$. $X=V(x^{2}+y^{2}+z^{2}-1)$, $\underline{a}=(1,0,0)$, the
tangent plane to $X$ at $\underline{a}$ is spanned by $(0,1,0),(0,0,1)$,
i.e., $\{x=1\}$. $\triangledown f(\underline{a})=\begin{pmatrix}2x\\
2y\\
2z
\end{pmatrix}|_{1,0,0}=\begin{pmatrix}2\\
0\\
0
\end{pmatrix}\Rightarrow2x-x=0$.
\end{itemize}
\end{example*}
The tangent space to the variety of $f$ at $\underline{a}$ is $T_{\underline{a}}(X)=\{(y_{1},\dots,y_{n}):\sum\frac{df}{dx_{i}}|_{a}(y_{i}-a_{i})=0\}$.
This is a hyperplane with normal vector $\triangledown f(a)$ unless
$\triangledown f(a)=0$
\begin{example*}
$X=V(x^{3}-y^{2})$, $\underline{a}=(0,0)$. Then $\triangledown f(\underline{a})=\begin{pmatrix}3x^{2}\\
-2y
\end{pmatrix}|_{a}=\begin{pmatrix}0\\
0
\end{pmatrix}$. So $T_{0,0}(X)=\{(y_{1},y_{2}):0y_{1}+0y_{2}=0\}=\aa^{2}$. If $a=(1,1)$
then $\triangledown f(\underline{a})=\begin{pmatrix}3\\
-2
\end{pmatrix}$. so $T_{(1,1)}(X)=\{(y_{1},y_{2}):3y_{1}-2y_{2}=0\}=\{y_{2}=\frac{2}{3}y_{1}\}$\end{example*}
\begin{defn}
$X=V(f)$ is \emph{singular} at a point $\underline{a}\in X$ if the
tangent space to $X$ at $ $$\underline{a}$ is not a hyperplane.

Let $X\subseteq\aa^{n}$, fix $\underline{a}\in X$. The \emph{tangent
space} to $X$ at $\underline{a}$ is $ $$T_{\underline{a}}(X)=\underline{a}+\{(y_{1},\dots,y_{n}):\sum\frac{df}{dx_{i}}|_{a}(y_{i}-a_{i})=0\,\forall f\in I(X)\}$,
i.e., $T_{\underline{a}}(X)=\cap_{f:X\subseteq V(\left\langle f\right\rangle )}T_{\underline{a}}(V(f))$.\end{defn}
\begin{example*}
{}
\begin{itemize}
\item $X=V(x^{2}-y,x^{3}-2)$ and $\underline{a}=(1,1,1)$. Then $T_{\underline{a}}(X)=\{(y_{1},y_{2},y_{3}):2y_{1}-y_{2},3y_{1}-y_{3}=0\}+(1,1,1)=(1,1,1)+$
span of $(1,2,3)$
\item $X=V(x^{2}-y^{2},xz,yz)=V(x-y,z)\cup V(x+y,z)\cup V(x,y)$.

\begin{itemize}
\item $\underline{a}=(1,1,0)$ then $T_{\underline{a}}=\{(y_{1},y_{2},y_{3}):2y_{1}-2y_{2}=0,y_{3}=0\}+(1,1,0)=\mathrm{span}(1,1,0)$
\item $\underline{a}=(1,-1,0)$ then $T_{\underline{a}}(X)=\{(y_{1},y_{2},y_{3}):2y_{1}+2y_{2}=0,y_{3}=0\}+\{1,-1,0\}=\mathrm{span}(1,-1,0)$
\item $\underline{a}=(0,0,1)$ then $T_{\underline{a}}(X)=\mbox{span}(0,0,1)$
\item $\underline{a}=(0,0,0)$ then $T_{\underline{a}}(X)=\aa^{3}$.\end{itemize}
\end{itemize}
\end{example*}

\end{document}
