\classheader{2013-02-13}

\begin{definition}
A non-empty topological space $X$ is said to be irreducible if, whenever we have $X=Y\cup Z$ where $Y$ and $Z$ are closed subsets of $X$, then either $X=Y$ or $X=Z$.
\end{definition}

For example, the set $V(xy)$ is not irreducible:
\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,ultra 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} \quad\raisebox{0.85in}{ = }\quad
\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,ultra 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} \quad\raisebox{0.85in}{ $\cup$ }\quad
\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,ultra 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{theorem}
Either
\begin{itemize}
\item[(i)] let $A$ be a commutative ring, and $X=\Spec(A)$, or
\item[(ii)] let $A$ be finitely generated commutative ring over a field $k$, and  $X=\max(A)$.
\end{itemize}
Then
\begin{enumerate}
\item In the bijection
\[\text{radical ideals of }A\longleftrightarrow\text{closed subsets of }X,\]
prime ideals correspond to irreducible subsets.
\item In case (i), if a prime ideal $\frak{p}$ corresponds to $Y$, then $Y=\overline{\{\frak{p}\}}$ in $\Spec(A)$.
\item In case (i), assume that $A$ is noetherian; this is already true in case (ii). Then $X$ is a finite union of irreducible closed subsets.
\end{enumerate}
\end{theorem}
\begin{proof}[Proof of 1]
Let $\frak{p}$ be a prime ideal of $A$. We will prove that $V(\frak{p})$ is irreducible.

Suppose for the sake of contradiction that $V(\frak{p})=V(J_1)\cup V(J_2)$. For $i=1,2$, because
\[V(J_i+\frak{p})=V(J_i)\cap V(\frak{p})=V(J_i),\]
WLOG we can assume that $J_i\supseteq \frak{p}$. We want to prove that either $J_1=\frak{p}$ or $J_2=\frak{p}$. If $J_1\neq\frak{p}$, then we can find some $x\in J_1\setminus\frak{p}$. Choose any $y\in J_2$. Then $xy\in J_1\cap J_2$, so
\[V(xy)\supseteq V(J_1\cap J_2)=V(J_1)\cup V(J_2)=V(\frak{p}).\]
In case (i), this shows that $\frak{p}\in V(\frak{p})\subseteq V(xy)$, so that $xy\in\frak{p}$. In case (ii), this shows that for any $\frak{m}\in\max(A)$ with $\frak{m}\supseteq\frak{p}$, we have $xy\in\frak{m}$; but
\[\bigcap_{\frak{n}\in\max(A/\frak{p})}\frak{n}=0\]
because $A/\frak{p}$ has no nilpotents, so
\[\bigcap_{\substack{\frak{m}\in\max(A)\\\frak{m}\supseteq\frak{p}}}\frak{m}=\frak{p},\]
and thus in case (ii) we also have $xy\in \frak{p}$. Thus, in either case, we have $xy\in\frak{p}$, and because $x\notin \frak{p}$ we must have that $y\in\frak{p}$. Thus $J_2\subseteq\frak{p}$, and because $J_2\supseteq \frak{p}$, this shows that $J_2=\frak{p}$.

Now let's prove the converse. Let $Y\subseteq X$ be a closed subset that is irreducible; we want to prove that $I(Y)$ is a prime ideal of $A$. By replacing $X$ by $Y$, it is enough to prove that if $X$ is irreducible, then $I(X)=\nil(A)$ is a prime ideal. By replacing $A$ by $A/\nil(A)$, we may assume that $I(X)=\nil(A)=0$, so that our goal is now to show that $A$ is an integral domain. We can do this because the map $\Spec(A/\nil(A))\to \Spec(A)$ induced by the quotient map $A\to A/\nil(A)$ is a homeomorphism.

Let $f,g\in A$ satisfy $xy\in I(Y)=0$. Thus $V(f)\cup V(g)=X$. Because $X$ is irreducible, either $V(f)=X$ or $V(g)=X$, so that either $f=0$ or $g=0$.
\end{proof}

\begin{proof}[Proof of 2]
Easy.
\end{proof}

We need some preparation for the proof of 3.

\begin{proposition}
If $A$ is a commutative ring, the following two conditions are equivalent:
\begin{itemize}
\item[(i)] $A$ is noetherian.
\item[(ii)] For any non-empty set $\Phi$ of ideals of $A$, there is a maximal element of $\Phi$ (under the ordering given by inclusion).
\end{itemize}
\end{proposition}
\begin{proof}
First, let's show that (i) $\implies$ (ii); thus, let $A$ be noetherian and let $\Phi$ be a non-empty set of ideals of $A$. Let $I_0\in\Phi$. If $I_0$ is not maximal in $\Phi$, then there is some $I_1\in\Phi$ such that $I_0\subsetneq I_1$. If $I_1$ is not maximal in $\Phi$, then there is some $I_2\in\Phi$ such that $I_1\subsetneq I_2$. \textbf{Continuing}, if $\Phi$ doesn't have a maximal ideal, then we have an infinite ascending chain of ideals
\[I_0\subsetneq I_1\subsetneq I_2\subsetneq\cdots\]
which contradicts the assumption that $A$ is noetherian because $J=\bigcup_{n=0}^\infty I_n$ cannot be finitely generated.

Now, let's show that (ii) $\implies$ (i). For any ideal $J$, let $\Phi$ consist of all the finitely generated ideals contained in $J$. If $J$ is not finitely generated, then there is no maximal element in this collection, which would contradict our assumption.
\end{proof}

This proof is a little dangerous. It may take 30 years. We may have to tell our children that, when we were young, we were in a course called ``Algebra 2'' where we tried to find a maximal ideal in $\Phi$, but that we still are not done. This is not such a good thing for the family. We need to use some sort of axiom of choice.

It never happens that, when we go home and open the refrigerator, we see all infinitely many prime numbers there. We will never observe all of an infinite set, but we know it is there.


\begin{proof}[Proof of 3]
We know that there is a bijection
\[\text{radical ideals of }A\longleftrightarrow\text{closed subsets of }\Spec(A).\]
For any non-empty set $\Phi$ of closed subsets of $X$, we know that $\Phi$ has a minimal element.

Now let
\[\Phi=\{\text{closed subsets }Z\subseteq X\mid Z\text{ is not a finite union of irreducible closed subsets of }Z\}.\]
We want to prove that $\Phi=\emptyset$. Suppose for the sake of contradiction that $\Phi\neq\emptyset$. Then $\Phi$ has a minimal element $Z$, and certainly we cannot have that $Z$ is irreducible, so $Z=Y_1\cup Y_2$ where $Y_1,Y_2$ are closed and $Y_1,Y_2\neq Z$. By minimality of $Z$, we must have that
\[Y_1=Y_{11}\cup\cdots\cup Y_{1m},\qquad Y_2=Y_{21}\cup\cdots\cup Y_{2n}\]
for some irreducible closed sets $Y_{ij}$. But then $Z=\bigcup_{i,j} Y_{ij}$, which contradicts the fact that $Z\in\Phi$. Thus, our assumption was incorrect, and we must have $\Phi=\emptyset$.
\end{proof}

Let $k$ be an algebraically closed field. An irreducible algebraic set of $k^n=\max(k[T_1,\ldots,T_n])$ is called an affine algebraic variety.

There are things which we want to call algebraic varieties which are not affine. For example, if we glue two copies of $k$ together, we can make
\begin{center}
\begin{tikzcd}[row sep=-0.1in,column sep=-0.05in]
\P^1(k)=& k & \cup & k\\
 & & {\text{\,\scriptsize$k^\times$}} \ar[start anchor=base,end anchor=base,shift right=0.1in,mapsto]{ru}[swap,pos=0]{x}[swap,pos=1]{\!\!x^{-1}}
 \ar[start anchor=base,end anchor=base,shift  left=0.1in,mapsto]{lu}[pos=0]{x}[near end]{x}
\end{tikzcd}
\end{center}







