\classheader{2013-02-11}


The rest of the course will cover
\begin{itemize}
\item the correspondence between ideals and geometry
\item local rings, Dedekind domains, and regular local rings
\item completion, $p$-adic numbers
\item algebraic curves
\item projective curves
\item Weil conjectures and recent big theorems in number theory
\end{itemize}

Last time, we proved that
\[\nil(A)=\{f\in A\mid f(\frak{p})=0\text{ for all }\frak{p}\in\Spec(A)\}=\bigcap_{\frak{p}\in\Spec(A)}\frak{p}.\]
Now, let's prove that if $A$ is finitely generated over a field, then in fact
\[\nil(A)=\{f\in A\mid f(\frak{p})=0\text{ for all }\frak{p}\in\max(A)\}=\bigcap_{\frak{m}\in\max(A)}\frak{m}.\]
The $\subseteq$ inclusion follows from what we proved last time, so let's prove the $\supseteq$ inclusion.

Assume that $f(\frak{p})=0$ for all $\frak{p}\in\max(A)$. We will prove that $A[\frac{1}{f}]=0$, which suffices to show that $f$ is nilpotent.

If $A[\frac{1}{f}]\neq 0$, then there is some $\frak{m}\in\max(A[\frac{1}{f}])$. Because $A$ is finitely generated over a field, the inverse image of $\frak{m}$ in $A$ is also a maximal ideal, say $\frak{p}\in\max(A)$. Then $\frac{f}{1}\notin \frak{m}$ because $f\in A[\frac{1}{f}]^\times$, but then $f\notin\frak{p}$, which contradicts our assumption.

\subsection*{Ideal $\longleftrightarrow$ Geometry}

Let $A$ be a commutative ring. For an ideal $I$ of $A$, we use $\sqrt{I}$ or $\rad(I)$ to denote
\[\{f\in A\mid \text{there is some }n\geq 1\text{ such that }f^n\in I\},\]
which is called the radical of $I$. Note that $\sqrt{I}$ is the inverse image of $\nil(A/I)$ under the quotient map $A\to A/I$. If $I=\sqrt{I}$, then we say that $I$ is a radical ideal.

Note that $\sqrt{\!\!\sqrt{I}}=\sqrt{I}$; this is easy to see from the definition.

\begin{theorem}
Let $A$ be a commutative ring.
\begin{enumerate}
\item The radical ideals $I$ of $A$ are in bijection with the closed subsets $V(I)$ of $\Spec(A)$.
\item If $A$ is finitely generated over a field, then $I=\sqrt{I}$ if and only if $V(I)$ is a closed subset of $\Spec(A)$.
\end{enumerate}
\end{theorem}
\pagebreak
\begin{proposition}
Let $A$ be a commutative ring.
\begin{enumerate}
\item For any subset $S\subseteq\Spec(A)$, we have $\overline{S}=V(I(S))$, where recall that
\begin{align*}
I(S)&=\{f\in A\mid f(\frak{p})=0\text{ for all }\frak{p}\in S\},\\
V(I)&=\{\frak{p}\in\Spec(A)\mid f(\frak{p})=0\text{ for all }f\in I\}.
\end{align*}
\item For any ideal $J$ of $A$, we have $\sqrt{J}=I(V(J))$.
\item[1' {\small\&} 2'.] If $A$ is finitely generated over a field, then $\Spec(A)$ can be replaced by $\max(A)$.
\end{enumerate}
\end{proposition}
\begin{proof}[Proof of part 2 of proposition]
By replacing $A$ by $A/J$, it is enough to prove that
\[\nil(A)=I(V(0))\]
because $\nil(A)=\sqrt{0}$ and there is a bijection
\[\Spec(A/J)\longleftrightarrow\{\frak{p}\in\Spec(A)\mid \frak{p}\supseteq J\}.\]But we know this is true, because $V(0)=\Spec(A)$, so
\[I(V(0))=\{f\in A\mid f(\frak{p})=0\text{ for all }\frak{p}\in\Spec(A)\}=\nil(A).\qedhere\]
\end{proof}

Note that the theorem follows from the proposition; the bijection is given by
\begin{center}
\begin{tikzcd}
\text{} \ar[bend left]{r}{V(\;)} & \text{} \ar[bend left]{l}{I(\;)}
\end{tikzcd}
\end{center}

\begin{definition}

Let $k$ be an algebraically closed field. A subset of $k^n$ of the form
\[V(f_1,\ldots,f_m)=\{x=(x_1,\ldots,x_n)\in k^n\mid f_1(x)=\cdots=f_m(x)=0\}\]
where $f_1,\ldots,f_m\in k[T_1,\ldots,T_n]$ is called an algebraic subset of $k^n$. We know there is an identification $k^n=\max(k[T_1,\ldots,T_n])$. Then algebraic subsets of $k^n$ are just the closed subsets in the Zariski topology; the above set is just $V(J)$ for $J=(f_1,\ldots,f_m)$.
\end{definition}

Let $S$ be an algebraic subset of $k^n$, and let $A$ be the ring of polynomial functions on $S$, i.e. the functions on $S$ that can be written as polynomials in the coordinate functions. Then we can identify $S=\max(A)$, and there is a bijection
\begin{center}
\begin{tikzcd}
\{\text{radical ideals }J\text{ of }A\} \ar[bend left]{r}{V(\;)} & \{\text{algebraic subsets of }k^n\text{ contained in }S\} \ar[bend left]{l}{I(\;)}
\end{tikzcd}
\end{center}

\subsection*{Prime Ideals}

In the ring $k[T_1,T_2]$, the ideals $(T_1)$, $(T_2)$, and $(T_1^2-T_2^3-1)$ are prime ideals, but $(T_1T_2)$ is not a prime ideal.

\begin{center}
\begin{tikzpicture}[scale=0.8]
\begin{axis}[title={$y^2-x^3-1=0$}, no markers, axis lines=middle,axis y line=center,axis equal,axis y line=middle, axis x line=middle, xmin=-2.5, xmax=2.5, ymin=-2.5, ymax=2.5,xtick={-2,-1,...,2},ytick={-2,-1,...,2},axis equal,width=7cm, height=7cm,tick label style={color=gray}]
\addplot +[no markers,raw gnuplot,very thick] gnuplot {
set contour base;
set cntrparam levels discrete 0.00;
unset surface;
set view map;
set isosamples 1000;
set xrange[-4:3];
set yrange[-3:3];
splot y^2-x^3-1;
};
\end{axis}
\end{tikzpicture}
\begin{tikzpicture}[scale=0.8]
\begin{axis}[title={$y=0$}, no markers, axis lines=middle,axis y line=center,axis equal,axis y line=middle, axis x line=middle, xmin=-2.5, xmax=2.5, ymin=-2.5, ymax=2.5,xtick={-2,-1,...,2},ytick={-2,-1,...,2},axis equal,width=7cm, height=7cm,tick label style={color=gray}]
\addplot +[no markers,raw gnuplot,very thick] gnuplot {
set contour base;
set cntrparam levels discrete 0.00;
unset surface;
set view map;
set isosamples 1000;
set xrange[-4:3];
set yrange[-3:3];
splot y;
};
\end{axis}
\end{tikzpicture}
\begin{tikzpicture}[scale=0.8]
\begin{axis}[title={$x=0$}, no markers, axis lines=middle,axis y line=center,axis equal,axis y line=middle, axis x line=middle, xmin=-2.5, xmax=2.5, ymin=-2.5, ymax=2.5,xtick={-2,-1,...,2},ytick={-2,-1,...,2},axis equal,width=7cm, height=7cm,tick label style={color=gray}]
\addplot +[no markers,raw gnuplot,very thick] gnuplot {
set contour base;
set cntrparam levels discrete 0.00;
unset surface;
set view map;
set isosamples 1000;
set xrange[-4:3];
set yrange[-3:3];
splot x;
};
\end{axis}
\end{tikzpicture}
\end{center}
\begin{center}
\begin{tikzpicture}[scale=0.8]
\begin{axis}[title={$xy=0$}, no markers, axis lines=middle,axis y line=center,axis equal,axis y line=middle, axis x line=middle, xmin=-2.5, xmax=2.5, ymin=-2.5, ymax=2.5,xtick={-2,-1,...,2},ytick={-2,-1,...,2},axis equal,width=7cm, height=7cm,tick label style={color=gray}]
\addplot +[no markers,raw gnuplot,very thick] gnuplot {
set contour base;
set cntrparam levels discrete 0.00;
unset surface;
set view map;
set isosamples 1000;
set xrange[-4:3];
set yrange[-3:3];
splot x*y;
};
\end{axis}
\end{tikzpicture}
\end{center}
Prime ideals in $k[T_1,\ldots,T_n]$ correspond to irreducible algebraic sets, also known as algebraic varieties. These cannot be written as a non-trivial union of smaller algebraic sets. In the same way that a prime number is a number that cannot be divided into two other numbers, an algebraic variety is an algebraic subset that cannot be divided into two other algebraic subsets.

The prime ideal is a princess of the world of ideals. Her father is the prince ``Point'' in the world of geometry. Her mother is the princess ``Prime Numbers'' in the world of numbers. She inherits the purity from her parents.