\classheader{2013-02-06}


There is a correspondence
\begin{center}
\begin{tikzcd}[row sep=0.1in]
\text{algebra} \ar[leftrightarrow]{r} & \text{geometry}\\
\text{commutative rings} \ar[leftrightarrow]{r} & \text{spaces}
\end{tikzcd}
\end{center}

Let $k$ be an algebraically closed field, and let
\[X=\{x=(x_1,\ldots,x_n)\in k^n\mid f_1(x)=\cdots=f_m(x)=0\}\]
for some $f_i\in k[T_1,\ldots,T_n]$. Let $A$ be the collection of polynomial functions on $X$, i.e. the $k$-valued functions which can be expressed as a polynomial over $k$ in the coordinate functions. Then $A$ is the image of the map
\begin{center}
\begin{tikzcd}[row sep=0.1in]
k[T_1,\ldots,T_n] \ar{r} & \text{maps}(X,k)\\
f(T_1,\ldots,T_n) \ar[mapsto]{r} & (x=(x_1,\ldots,x_n)\mapsto f(x_1,\ldots,x_n))
\end{tikzcd}
\end{center}
If $I$ is the kernel of this map, we therefore have $A\cong k[T_1,\ldots,T_n]/I$.

We claim that $\max(A)=X$. On the homework, you have already shown that $\max(A)=\Hom_k(A,k)$ when $A$ is finitely generated over $k$. Then the bijection from $X$ to $\Hom_k(A,k)$ is given by
\begin{center}
\begin{tikzcd}[row sep=0.1in]
X \ar{r} & \Hom_k(A,k)\\
x\ar[mapsto]{r} & (f\mapsto f(x))
\end{tikzcd}
\end{center}
If $B=k[T_1,\ldots,T_n]/(f_1,\ldots,f_m)$, then we have
\begin{center}
\begin{tikzcd}
B \ar[two heads]{r} \ar[bend right]{rr} & A \ar[hook]{r} & \text{maps}(X,k)
\end{tikzcd}
\end{center}
\begin{center}
\begin{tikzcd}
X=\Hom_k(B,k) & \Hom_k(A,k) \ar{l}[swap]{\text{injective}} & X\ar{l}\ar[bend left]{ll}{\text{identity}}
\end{tikzcd}
\end{center}
hence the injective map is also surjective.


This universe should be the set of prime ideals of some nice commutative ring; we can dream this is possible. This ring should be a very beautiful ring. If this ring is the integers, then we are prime numbers, like 17.

The correspondence between rings and spaces is
\begin{center}
\begin{tikzcd}
A\ar{r}{\varphi} & B
\end{tikzcd}\qquad\qquad\begin{tikzcd}[row sep=0.0in]
\Spec(B) \ar{r} & \Spec(A)\\
\frak{p}\ar[mapsto]{r} & \varphi^{-1}(\frak{p})
\end{tikzcd}
\end{center}


\begin{remark}
If $A$ and $B$ are finitely generated rings over a field $k$, and $\varphi:A\to B$ is a homomorphism of $k$-algebras, or if $A$ and $B$ are finitely generated over $\Z$, then we have an induced map $\max(B)\to\max(A)$, also given by $\frak{p}\mapsto\varphi^{-1}(\frak{p})$.
\end{remark}

\begin{proof}
For the case over $k$, if $\frak{m}\in\max(B)$, then $A/\varphi^{-1}(\frak{m})\subset B/\frak{m}$. Because $B/\frak{m}$ is a finite extension of $k$, we have that $A/\varphi^{-1}(\frak{m})$ is an integral domain that is finite-dimensional over a field, and therefore it is a field.
\end{proof}


\subsection*{Zariski topology on $\Spec(A)$, on $\max(A)$}

Recall that on the homework, you saw that for a compact Hausdorff space $X$, and \[A=\{\text{continuous maps }X\to\C\},\] then there was a natural identification $X=\max(A)$, and for any subset $S\subseteq X$, we have $\overline{S}=V(I(S))$ where
\begin{align*}
I(S) & =\{f\in A\mid f(x)=0 \text{ for all }x\in S\}\\
V(J) & =\{x\in X\mid f(x)=0\text{ for all }f\in J\}=\{\frak{m}\in\max(A)\mid J\subseteq\frak{m}\}
\end{align*}


If $A$ is any commutative ring, then the Zariski topology on $\Spec(A)$ is defined by declaring the closed sets to be those of the form
\[V(J)=\{\frak{p}\in\Spec(A)\mid \frak{p}\supseteq J\}\]
for ideals $J$ of $A$. This gives a topology, because you can check that
\[\bigcap_{i}V(J_i) =V(J)\]
where $J$ is the ideal generated by all the $J_i$, and
\[\bigcup_{i=1}^n V(J_i)=V\left(\bigcap_{i=1}^n J_i\right)\]
Then $\max(A)$ gets the subspace topology, as a subset of $\Spec(A)$.

For example, in $\Spec(\Z)$, the closed sets are $\emptyset$, $\Spec(\Z)$ itself, or a finite set of maximal ideals. This is a stupid topological space, and we can't recover the ring just from this topological space; this is not as nice as the case of the compact Hausdorff space.

%For an algebraically closed field $k$, we can consider $\P^1(k)$ as the union of  $\P^1(k)-\{0\}$ and $\P^1(k)-\{\infty\}$, where we identify $\max(k[T])$ with $\max(k[\frac{1}{T}])$.

We hope to hear the voice of the ring $A$. The first person which appears here is the ring of continuous functions on a compact Hausdorff space. I'm a young guy called ``commutative ring'', but I was originally ``the ring of continuous functions on a compact Hausdorff space''. Now I am an algebraic object, so I must say goodbye to my home village, the space, but I will always keep it in my heart as a set of maximal ideals.



Polynomials are determined by their values at finitely many points; thus
\[\overline{S}=\{x\in\Spec(A)\mid \text{ if }f,g\in A \text{ and }f(y)=g(y)\text{ for all }y\in S, \text{ then }f(x)=g(x)\}\]

At midnight, some people put a candle on their head and \textit{*makes hammering motion with hand*}\ldots\ I'm out of time, so I will explain this next time.