In his lecture "Zeta functions and $L$-functions", Serre presents a very elegant proof of the convergence of the zeta function 
$ \zeta (X,s) = \prod\_{x \in |X|} (1- N(x)^{-s})^{-1}$ in the half plane $R(s) > dim(X)$, where $X$ is a scheme of finite type over $\mathbb{Z}$, $|X|$ the set of closed points of $X$ and $N(x)$ the number of elements in the residue field $k(x)$.


He reduces the claim to the case where $X = Spec \, A[x\_1, \ldots x\_n]$ and $A$ is either $\mathbb{Z}$ or $\mathbb{F}\_p$.


The decisive input is the following lemma:


a) If $X$ is the finite union of the schemes $X\_i$, and the claim holds for all $X\_i$, then it holds for $X$. 


b) If $f: X \to Y$ is finite and the claim holds for $Y$, then it holds for $X$ as well.


I've been trying to prove b) but I seem to be missing something. Here's what I've tried so far:
I was considering $\zeta(X,s) = \prod\_{y \in |Y|} \zeta(X\_y \, ,s)$, where $X\_y$ is the fiber of $f$ at $y$. I now the fibers are finite but I don't know how to connect this with the fact that $\zeta(Y,s)$ converges. Is it true that the residue field $k(y)$ is a finite extension of $k(x)$ for all $x \in X\_y$ (of degree $\deg f$)? I know this is the case for the function fields.


Any help is very appreciated!



Convergence of zeta functions for schemes of finite type over the integers
I think part of your confusion lies in the definition of a finite morphism (see e.g. wikipedia). By definition, there is a finite affine open cover $U\_i = \mathrm{Spec}(B\_i)$ of $Y$ such that $f^{-1}(U\_i) = \mathrm{Spec}(A\_i)$ is affine and $A\_i$ is a finite $B\_i$-module.


Therefore there exists some $d$ such that for all closed points $y \in |Y|$, the fibre $f^{-1}(y)$ is a finite scheme of degree at most $d$ over the residue field of $y$. In particular, there are at most $d$ points above $y$, and each such point $x$ satisfies $N(x) \geq N(y)$.


One therefore obtains
\begin{align\*}
 |\zeta (X,s)| & = \prod\_{x \in |X|} \left|1- N(x)^{-s}\right|^{-1} \\
 & = \prod\_{y \in |Y|} \prod\_{\substack{x \in |X| \\ f(x) = y}} 
 \left|1- N(x)^{-s}\right|^{-1} \\
 & \leq \prod\_{y \in |Y|} \left|1- N(x)^{-s}\right|^{-d} \\
 & = |\zeta (Y,s)|^d.
\end{align\*}
Hence the absolute convergence of $\zeta (X,s)$ for $\mathrm{re}(s) > \mathrm{dim}(X)$ follows from the absolute convergence of $\zeta (Y,s)$ for $\mathrm{re}(s) > \mathrm{dim}(X)=\mathrm{dim}(Y)$.


