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\title{Math 631, Lecture 3}
\author{Notes by Bryden Cais}
\date{Sept. 11, 2002}
\begin{document}
\maketitle

As usual, let $K$ be an algebraically closed field and $k\subset K$ a subfield.  Let $X,Y$ be affine $k$ varieties and
let $X(K)$ denote the $K$ points of $X$.
We say that $X\subset Y$ if $X(K)\subset Y(K)$ for all $k$ algebras K.  Denote by $I(X)$ the ideal
of all polynomials vanishing on $X$.  Clearly,
$$I(X)=\rad(I(X))$$
since if $f^N$ vanishes on $V$ so does $f$ (since $K$ is a field).

\begin{defn}
    An ideal $I\subset k[T]$ is called \textbf{radical} if $\rad(I)=I$.
\end{defn}
Observe that for any radical ideal $I$, the ring $k[T]/I$ is \textbf{reduced}, that is, has no nilpotent elements.
For any ideal $I$, define $V(I):=\sol(I,K).$

\begin{thm}
    There is a one to one correspondence between radical ideals $I$ and algebraic $k$ sets $V$
    given by
    \begin{align*}
        I&\mapsto V(I)\\
        V&\mapsto I(V).
    \end{align*}
\end{thm}

\begin{proof}
    It is not difficult to see that $V\mapsto I(V)$ is an injective mapping of $k$ varieties into the set of radical
    ideals: if $V_1\neq V_2$, there exists $x\in V_1$ not in $V_2$.  If $V_i$ is the zero set of
    $\{f_{1i},\ldots,f_{m(i)i}\}$ for $i=1,2$, then there exists some $n$ such that $f_{n2}(x)\neq 0$.  Consequently,
    $f_{n2}\not\in I(V_2)$.  Indeed, this argument shows that $V(I(V))=V$ for any variety $V$.  Now Hilbert's Nullstellensatz
    tells us that $V(I)=V(J)$ if and only if $\rad(I)=\rad(J)$, which shows that the reverse mapping $I\mapsto V(I)$
    is injective since $I$ is radical.  Consequently $I(V(I))=I$ and we are done.
\end{proof}

\begin{thm}\label{idealcor}
    Let $K$ be algebraically closed.  Then
    the maximal ideals of $K[T_1,\ldots,T_n]$ are in bijective correspondence with points of $K^n$.
\end{thm}

\begin{proof}
    Let $M$ be a maximal ideal of $K[T]$ and suppose that $M\neq (1)$.  Then $V(M)$ is nonempty (since $K$ is algebraically closed),
    so pick $x=(c_1,c_2,\ldots c_n)\in V(M)$.
    Now let $I=(T_1-c_1,\ldots,T_n-c_n)$.  Clearly $I$ is a maximal ideal since $K[T]/I_x$ is a field (each $T_i$ is sent to $c_i$ so the image
    is just $K$).  Moreover, we see that $x\in V(I)$ and that if $f\not\in M$ then $f$ does not vanish at $x$ (if it did than every polynomial
    would vanish at $x$ since $M$ is maximal).  Therefore, we have $I\subset M$ and since $I$ is maximal, $I=M$ and we are done.
\end{proof}

Theorem \ref{idealcor} is a very important result: it enables us to forget about the ambient space $K^n$ and
work only with the ring $k[T_1,\ldots,T_n]$ and its maximal ideals.  We shall see later how to exploit this to the fullest extent
possible.

We now define a topology on $K^n$ called the $k$-\textbf{Zariski topology} by defining the closed sets to be
the algebraic $k$ sets.  Clearly $K^n$ and $\emptyset$ are closed.  We must verify that arbitrary intersections
and finite unions are closed.
\begin{enumerate}
\item Let $S$ be a finite index set and let $V_s$ for $s\in S$ be algebraic $k$ sets and let $I_s=I(V_s)$.  Define $I=\prod_{s\in S} I_s$.  Then
    we claim that $$V(I)=\bigcup_{s\in S} V_s.$$  To see this, observe that if $x\in V_s$ for some $s$ then
    every $f\in I_s$ vanishes at $x$, so any finite sum of products of elements of all the $I_r$ vanishes at $x$ (since one of the factors
     \textit{must} be in $I_s$); i.e. $x\in V(I)$.
    Conversely, let $x\in V(I)$ and suppose that $x\not\in V_s$ for all $s$.  Then for each $s$ there exists $f_s$ such that $f_s(x)\neq 0$,
    whence $f=\prod_{s\in S}f_s(x)\neq 0$, but $f\in I,$ a contradiction.

\item Now let $S$ be an arbitrary index set and let $V_s,I_s$ be as above.  Set $I=\sum_{s\in S} I_s$.  Now we claim
that $$\bigcap_{s\in S}V_s=V(I).$$  Let $x\in V_s$ for all $s$.  Then $I_S$ vanishes at $x$ for each $s$, so that $I$
also vanishes at $x$; that is, $x\in V(I).$  Conversely, since $I_s\subset \sum_{s\in S} I_s$, we see that
$V(I)\subset V(I_s)$ (since the $V$ operator is inclusion reversing) and hence that $V(I)\subset \bigcap V_s$.
\end{enumerate}

It is not difficult to see that in general, the Zariski topology is not Hausdorff: when $n=1$ for example, the open sets are just the
complements of finite sets, so that it is impossible to separate points.




\end{document}
