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\begin{document}
\newpage
\centerline{\bf{Lecture 32}}
\vskip .4cm
\centerline{\bf{IV. Tate's thesis}}

\section{Overview of zeta functions}

\begin{Def} \label{D:IV.1.1}
Let $K$ be a number field. The Dedekind zeta function of $K$ is the function of complex variable defined by:
\[
\displaystyle \zeta_K(s)=\sum_{\mathcal{A}}\frac{1}{N\mathcal{A}^s}
\]
\\
where $\mathcal{A}$ ranges over all non-zero ideals in $O_K$.
\end{Def}
Remember that $N\mathcal{A}=|(O_K/\mathcal{A})|$. From now on assume $Re(s)>1$.\\
If we denote by $F(n)=|\{\mathcal{A}\triangleleft O_K:\ N\mathcal{A}=n\}|$ then we can rewrite\\
\[
\displaystyle \zeta_K(s)=\sum_{n=1}^{\infty}\frac{F(n)}{n^s}.
\]
\par
\begin{Def} \label{D:IV.1.2}
A Dirichlet series is an expression of the form:
\[
\displaystyle \sum_{n=1}^{\infty}\frac{a_n}{n^s},\ a_n\in\C,\ s\in\C.
\]
\end{Def}
\par
\begin{exa} For $K=\Q$ the ideals have the form $n\Z$, with $n$ a non-negative integer. Clearly $N(n\Z)=n$, so the corresponding Dedekind zeta function (called in this particular case the Rieman zeta function) is:
\[
\displaystyle \zeta_K(s)=\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}.
\]
\end{exa}
\par
\begin{pro} \label{P:IV.1.1} The series for $\zeta_K(s)$ converges absolutely and uniformly on compact sets of the region $\{s\in \C : \ Re(s)>1\}$. Therefore, $\zeta_K(s)$ is a holomorphic function of $s$ in this region.
\end{pro}
\par In the proof of this propositon we shall use the following lemma:
\begin{lem} \label{L:IV.1.1} Let $a_n \in \C$ and suppose that $\displaystyle \sum_{n\leq t}a_n=O(t^r),\ r>0$. Then $\displaystyle \sum_{n=1}^{\infty}\frac{a_n}{n^s}$ converges absolutely and uniformly on the region where $Re(s)>r$ and it is a holomorphic function there.
\end{lem} 
\noindent{\em Proof}: Each $\displaystyle \frac{a_n}{n^s}$ is holomorphic in $s$. So $\displaystyle \Phi_M(s)=\sum_{n=1}^{M}\frac{a_n}{n^s}$ is a holomorphic function for any $M\geq 1$. A sequence of holomorphic functions that converges absolutely and uniformly on compact sets of a region $D\subset \C$ converges to a function $\Phi(s)$ holomorphic in $D$. Fix now two integers $M\geq m>1$. Let $A_k=\sum_{n=1}^{k}a_n$. Then:
\[
\Phi_M(s)-\Phi_{m-1}(s)=\sum_{n=m}^{M}\frac{a_n}{n^s}=\sum_{n=m}^{M}\frac{A_n-A_{n-1}}{n^s}=\sum_{n=m}^{M}\frac{A_n}{n^s}-\sum_{n=m}^{M}\frac{A_{n-1}}{n^s}=
\]
\[
=\frac{A_M}{M^s}-\frac{A_{m-1}}{m^s}+\sum_{n=m}^{M-1}A_n(\frac{1}{n^s}-\frac{1}{(n+1)^s}).
\]
Since $A_n=O(n^r)$, we see that $|A_n|\leq B\cdot n^r$, for some constant $B>0$, and for large values on $n$.
Then, when $m$ is large enough:
\[
|\Phi_M(s)-\Phi_{m-1}(s)|\leq \frac{|A_M|}{|M^s|}+\frac{|A_{m-1}|}{|m^s|}+\sum_{n=m}^{M-1}|A_n|\cdot |\frac{1}{n^s}-\frac{1}{(n+1)^s}|\leq
\]
\[
B(\frac{M^r}{|M^s|}+\frac{(m-1)^r}{|m^s|}+\sum_{n=m}^{M-1}n^r\cdot |\frac{1}{n^s}-\frac{1}{(n+1)^s}|).
\]
Since $|T^s|=T^{\sigma}$, for any $T>0$, where $\sigma=Re(s)$, we obtain:
\[
|\frac{1}{n^s}-\frac{1}{(n+1)^s}|\leq |s|\int_n^{n+1}\frac{dt}{|t^{s+1}|} = |s|\int_n^{n+1}\frac{dt}{t^{\sigma+1}}\leq \frac{|s|}{n^{\sigma+1}}.
\] 
(the last inequality follows from the fact that, when $t>1$, the map $\sigma \rightarrow \frac{1}{t^{\sigma+1}}$ is decreasing). Then:
\[
|\Phi_M(s)-\Phi_{m-1}(s)|\leq B(M^{r-\sigma}+m^{r-\sigma}+|s|\sum_{n=m}^{M-1}n^{r-\sigma-1}).
\]
Assume now that $\sigma>r$. Then:
 \[
|\Phi_M(s)-\Phi_{m-1}(s)|\leq B(M^{r-\sigma}+m^{r-\sigma}+|s|\int_{m-1}^{\infty}t^{r-\sigma-1}dt)=\frac{(m-1)^{r-\sigma}}{\sigma-r}.
\]
So:
\[
|\sum_{n=m}^{M}\frac{a_n}{n^s}|\leq B(M^{r-\sigma}+m^{r-\sigma}+|s|\frac{(m-1)^{r-\sigma}}{\sigma-r}).
\]
Clearly, right hand side of the above inequality goes to 0 as $M,m$ go to infinity.
Now if $K$ is a compact subset of the half-plane $Re(s)>r$, then it is a bounded set of complex numbers, so we can find $B'>0$ such that $|s|<B'$ on $K$. Moreover, since $K$ is a compact set, we can choose an $\epsilon>0$ such that $Re(s)-r\geq \epsilon$, for any $s\in K$. Since $M\geq m$ and $Re(s)-r\geq \epsilon$, we get: $M^{r-\sigma}\leq m^{r-\sigma}\leq m^{-\epsilon}$, hence:
\[
|\sum_{n=m}^{M}\frac{a_n}{n^s}|\leq B(2m^{-\epsilon}+B'\frac{(m-1)^{-\epsilon}}{\epsilon}).
\] 
The constants $B'$ and $\epsilon$ only depend on the compact set $K$ and not on $s$, so the convergence is uniform on $K$.
\vskip .3cm Now we can prove Proposition \ref{prop:IV.1.1}{}: Write $\displaystyle \zeta_K(s)=\sum_{n=1}^{\infty}\frac{F(n)}{n^s}$. Then $\sum_{n\leq t}F(n)=O(t)$. A proof of this fact can be found in Daniel Marcus' book "Number fields", on page 158, Theorem 39. The claim follows directly from Lemma \ref{lem:IV.1.1}{}, with $r=1$.
\par
\begin{pro} \label{P:IV.1.2} The following equality holds for $Re(s)>1$:
\[
\zeta_K(s)=\prod_{\mathcal{P}}\frac{1}{1-N\mathcal{P}^{-s}}
\]
where the product is taken over all non-zero prime ideals $\mathcal{P}$ of $K$. 
\end{pro}
(the above equality is called the "Euler product formula")
\\
\noindent{\em Proof}: We will use the following expansion:
\[
\frac{1}{1-x}=1+x+x^2+x^3+... , \ \ \ |x|<1.
\]
Since the norm of any non-zero proper prime ideal of $K$ is larger than 1, set $x=N\mathcal{P}^{-s}$ and use the above expansion (note that since $\sigma:=Re(s)>1, |x|<1$). We get:
\[
\frac{1}{1-N\mathcal{P}^{-s}} = 1+\frac{1}{N\mathcal{P}^s}+\frac{1}{N\mathcal{P}^{2s}}+ ... + \frac{1}{N\mathcal{P}^{\nu s}}+...
\]
hence
\[
\prod_{\mathcal{P}}\frac{1}{1-N\mathcal{P}^{-s}}= \prod_{\mathcal{P}}(1+\frac{1}{N\mathcal{P}^s}+...)=
\sum \frac{1}{N\mathcal{P}_1^{\nu_1 s}}\cdot ...\cdot \frac{1}{N\mathcal{P}_j^{\nu_j s}}=
\]
\[
= \sum \frac{1}{N(\mathcal{P}_1...\mathcal{P}_j)^s}=\sum_{\mathcal{A}\triangleleft O_K}\frac{1}{N \mathcal{A}^s}
\] 
where the last equality is based on the fact that every ideal $\mathcal{A}\triangleleft O_K$ has a unique factorization into prime ideals.
\begin{cor} \label{C:IV.1.1} $\zeta_K(s)\not= 0$ when $Re(s)>1$.
\end{cor}
\noindent{\em Proof}: By definition of convergence of infinite products, namely that the sum of the logarithms of the factors converges, a convergent infinite product has a non-zero value (limit).
\end{document}