Convergence of zeta functions for schemes of finite type over the integers 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!
 A: Let's see.  The key point is that the formula for the zeta function is that, if we consider the case of finite type over a finite field $X/k={\mathbb F}_q)$, there is an equivalence between the 'absolute' zeta function as you define it and the relative zeta function:
$$Z(X,k;t)=exp\{\sum_{m\leq 1} \frac{N_m}{m} t^m\},$$
 where $N_m=card(X({\mathbb F}_{q^m}))$ is the number of rational points of $X$ over the unique extension of degree $m$ over the base field $k$.
The fact is that, if we have a rational point over an extension $k_m$, then the image is also defined over such extension!  That's the rationale behind the bound provided (simple as hell!).  That's why, if $x\in X(k_m)$, then so does $f(x)$, and there are clearly no more $k_m$-defined points on $X$ above $f(x)$ than $deg(f)$.  And that's that!
In fact, the "absolute" zeta function is derived from the former by the substitution $t=q^{-s}.$
This is key, for in our case, to every finite morphism $f:X\to Y$ (in the case where $X, Y$ are defined over a finite field $k$) corresponds an easy bound
$$N_m(X)\leq deg(f)\cdotp N_m(Y).$$  We do not think in terms of the field generated by the coordinates of our points, but we merely ask that these belong to a fixed field $k_m.$  This facilitates our count enormously, and enables us to use the degree of $f$ efficiently.
With this bound, you can obtain the desired convergence result by taking an Euler product over all (finite) characteristics.
I am pretty sure that Serre's paper contained this kind of background (don't have it here with me), but in any case Mircea Mustata has a lovely set of notes on the matter:
http://www.math.lsa.umich.edu/~mmustata/zeta_book.pdf
Needless to say, but I'll just remind that the dimension of an algebraic scheme is its Kronecker dimension, i.e. an elliptic curve over $\mathbb{Z}$ is of dimension $1+1=2$ (that's why it's called an arithmetic surface!).  This does indeed count when you write bounds on the product, Euler-style.
Let us deal with the case where $X \to Spec(\mathbb{Z})$ misses a finite number of points of its target.
Taking logarithms, one sees that $\log \zeta(X_p,s)$ is equivalent to $C_p p^{-(s-d)}$, where $d$ is the fibre dimension of the structure map ($C_p$ is controlled essentially by $deg(f)$ and by $Y$, and is $\leq deg(f)$ if our $f$ has the affine space over $\mathbb{Z}$ as its target).  It suffices to argue as in the case of the zeta function so as to establish that the infinite product converges for $Re(s-d)>1$, and since $\dim X=d+1$, we are done.  I can imagine, though, that using the existence of a finite $f:X\to Y$ does imply, through the above bounds, that the absolute zeta function of $X$ converges whenever $Re(s)>\dim X$.
In the case where we have a finite morphism $f:X\to \mathbb{A}^n_{\mathbb{Z}},$ (or finite over an open subset of $Spec(\mathbb{Z}$) the zeta function of $Y$ corresponds to $\zeta(s-n)$, and the lower bound for $Re(s)$ is $n+1$, i.e. the Kronecker dimension of the schemes involved.
That's how I did it, way back when.  Should you need further clarification, just ask.
