\classheader{2013-02-04}

I have some comments to add about the Hasse zeta function. Recall that the definition is
\[\zeta_A(s)=\prod_{m\in\max(A)}\frac{1}{1-\#(A/m)^{-s}}\]
where $A$ is a finitely generated ring over $\Z$. I mentioned the conjecture of Hasse,
\begin{conjecture-N}
There is a meromorphic continuation of $\zeta_A$ to all of $\C$.
\end{conjecture-N}
But I forgot to mention the generalization of Riemann's hypothesis:
\begin{conjecture-N}
The zeros and poles of $\zeta_A(s)$ satisfy $\Re(s)\in\frac{1}{2}\Z$.
\end{conjecture-N}

\begin{remark}
When $A=\ringofintegersof{K}$ for a number field $K$, the function $\zeta_A(s)$ is called the Dedekind zeta function of $K$. For this case, Conjecture 1 was proved a long time ago; Conjecture 2 is not known.

When $A$ is over a finite field $\F_q$, Conjecture 1 was proved by Dwork, and then later in a deeper way by Grothendieck (in this case, they were proving that $\zeta_A(s)$ is a rational function in $q^{-s}$). Conjecture 2 was proved by Deligne in 1973.
\end{remark}

\begin{conjecture}[Weil conjectures, 1949]
Let $A$ be a ring over $\F_q$, i.e. a ring that is a friend of $\F_q[x]$ like $\F_q[x,y]/(y^2-x^3-x-1)$.
\begin{itemize}
\item $\zeta_A(s)$ is a rational function in $q^{-s}$
\item Conjecture 2
\item $\zeta_A(s)$ tells us the shape of $A=\F_q[T_1,\ldots,T_n]/(f_1,\ldots,f_m)$, or rather the shape of the algebraic variety
\[\{x=(x_1,\ldots,x_n)\mid f_1(x)=\cdots=f_m(x)=0\}.\]
Over $\F_q$, this does not have a shape - we can't draw it. But we can think it is over $\C$, and then we can see the shape.
\end{itemize}
\end{conjecture}


\subsection*{Great encounters in history}

\begin{itemize}
\item Young algebraic geometry met Riemann's hypothesis.
\item Young Dante met Beatrice.
\end{itemize}


If $A$ is an integral domain over $\Z$, i.e. $A\supset \Z$, the Langlands conjectures tell us that
\[\zeta_A(s)=\frac{\prod_{i=1}^m(s,f_i)}{\prod_{j=1}^nL(s,g_j)}\]
where $f_i,g_j$ are modular forms for $\GL_n(\Q)$. Each of these $L$-functions has a meromorphic continuation to $\C$, so if we knew the Langlands conjectures, Conjecture 1 would be done.


If $K=\Q(i)$, then $\ringofintegersof{K}=\Z[i]$, and
\[\zeta_K(s)=\zeta(s)L(s,\chi)\]
where $\chi:(\Z/4\Z)^\times\to\{\pm 1\}$.

If $K=\Q(\sqrt[3]{2})$, then $\ringofintegersof{K}=\Z[\sqrt[3]{2}]$, and
\[\zeta_K(s)=\zeta(s)L(s,f)\]
where
\[f(z)=\eta(6z)\eta(18z)=\sum_{n=1}^\infty a_nq^n,\qquad q=e^{2\pi i z}.\]
Using the nice analytic properties of the function of
\[L(s,f)=\sum_{n=1}^\infty \frac{a_n}{n^s},\]
we can show that $\zeta_K$ also has a meromorphic continuation.

Now, we will get back to algebra.

\begin{proposition}
Let $A$ be a finitely generated ring over $k$, an algebraically closed field. Then there are bijections
\[\max(A)\longleftrightarrow \Hom_k(A,k)\longleftrightarrow \{x=(x_1,\ldots,x_n)\in k^n\mid f_1(x)=\cdots=f_m(x)=0\}\]
where $A$ is isomorphic as a $k$-algebra to $k[T_1,\ldots,T_n]/(f_1,\ldots,f_m)$.
\end{proposition}

We know that there is always such a presentation of $A$ because if $A$ is generated by $h_1,\ldots,h_n$ over $k$, then we have a surjective $k$-algebra map $k[T_1,\ldots,T_n]\to A$ defined by sending $T_i$ to $h_i$. The kernel $I$ of this map is an ideal of $k[T_1,\ldots,T_n]$, and therefore it is finitely generated because this ring is Noetherian. If $I=(f_1,\ldots,f_m)$, then we get
\[A\cong k[T_1,\ldots,T_n]/(f_1,\ldots,f_m).\]

\begin{proof}
First, we do the case when $A=k[T_1,\ldots,T_n]$. Clearly,
\begin{center}
\begin{tikzcd}[row sep=0.1in]
\Hom_k(k[T_1,\ldots,T_n],k) \ar[leftrightarrow]{r} & k^n \ar[leftrightarrow]{r} & \max(k[T_1,\ldots,T_n])\\
\varphi & (\varphi(T_i))_i \ar[mapsto]{l} & \\
 & a=(a_i)_i \ar[mapsto]{r} & \{f\mid f(a)=0\}\\
\varphi \ar[mapsto]{rr} & & \{f\mid \phi(f)=0\}
\end{tikzcd}
\end{center}
Now note that there are bijections
\[\Hom_k(k[T_1,\ldots,T_n]/(f_1,\ldots,f_m),k) \longleftrightarrow \{x=(x_1,\ldots,x_n)\in k^n\mid f_i(x)=0\}\]
and
\[\Hom_k(k[T_1,\ldots,T_n]/(f_1,\ldots,f_m),k) \longleftrightarrow \{\varphi\in \Hom_k(k[T_1,\ldots,T_n]/(f_1,\ldots,f_m),k) \mid \varphi(f_1)=\cdots=\varphi(f_m)=0\}\]
and
\[\max(k[T_1,\ldots,T_n]/(f_1,\ldots,f_m))\longleftrightarrow \{\frak{m}\in\max(k[T_1,\ldots,T_n])\mid \frak{m}\ni f_i\text{ for all }i\}\]
\end{proof}


For any commutative ring $A$, any $f\in A$ and $\frak{p}\in\Spec(A)$, we define $f(\frak{p})$, the value of $f$ at $\frak{p}$, to be the image of $f$ in the fraction field $\kappa(\frak{p})$ of $A/\frak{p}$ (this field is called the residue field at $\frak{p}$). If $\frak{p}\in\max(A)$, then $\kappa(\frak{p})=A/\frak{p}$.

The idea here is to consider any ring $A$ as a ring of functions on the space $\Spec(A)$. However, the values at different points can take values in different fields. For example, given $f\in\Z$ and $\frak{p}=(p)$ a prime ideal of $\Z$, then $f(\frak{p})=f\bmod p\in\F_p$.

\begin{theorem}
Let $A$ be a commutative ring. Then
\[f(\frak{p})=0\text{ for all }\frak{p}\in\Spec(A)\iff f\text{ is nilpotent},\]
i.e. $f^n=0$ for some $n\geq 1$. In the case that $A$ is finitely generated over a field, then in fact
\[f(\frak{m})=0\text{ for all }\frak{m}\in\max(A)\iff f\text{ is nilpotent}.\]
\end{theorem}

If $A=k[[T]]$ is the ring of formal power series, then $\Spec(A)=\{(T),(0)\}$, and $\max(A)=\{(T)\}$.

For any ring homomorphism $f:A\to B$, there is a corresponding map $\Spec(B)\to\Spec(A)$. You should think of $f$ as taking a function on $\Spec(A)$ and pulling it back via this map to a function on $\Spec(B)$.

There is not a corresponding map on sets of maximal ideals; for example, if $f:\Z\hookrightarrow\Q$, then $(0)\in\max(\Q)$, but $f^{-1}(0)=(0)$ is not a maximal ideal of $\Z$.

The world of rings and the world of spaces correspond very nicely.