dimension of a scheme and degree of an L-function

Disclaimer: I first asked this question on Mathoverflow but I was told it was off-topic for that site, so I ask it here.

I try to understand correctly the notion of scheme, as Serre in the second volume of his Oeuvres defines zeta and L-functions in this context. What seems interesting to me is that he states a theorem that says the zeta function $\zeta(X,s)$ of a scheme $X$ converges absolutely in the right half plane $\Re(s)\gt \dim X$, which is interesting for two reasons:

1) the Hasse-Weil zeta function of a rational elliptic curve is, up to normalization, an element of the Selberg class of degree two, and, as far as I know, before the modularity theorem was proved, such an L-function was only known to converge absolutely in the right half plane $\Re(s)>2$,

2) every element of the Selberg class converges absolutely in the right half-plane $\Re(s)\gt 1$, and thus in any right half plane of the form $\Re(s)\gt m$ where $m$ is any positive integer.

Hence my questions:

A) does the notion of fibered product of schemes imply that, if $X$ and $Y$ are two schemes of finite type, the dimension of the fibered product of $X$ and $Y$ is the product of the dimensions of $X$ and $Y$?

B) is any (local factor of an) element of the Selberg class the Zeta function of a scheme of finite type?

C) is the degree conjecture for the Selberg class motivated by the fact that the degree of an element thereof should be the dimension of a scheme?

I only answer to (A) for now. The expected formula is not the product of dimensions, but the sum, and is really not always true. Indeed : take $X$ and $Y$ to be the affine lines over $\mathbf{Z}$, and $Z$ their product. Then the dimension of $Z$ is the sum of dimensions of $X$ and $Y$... minus the dimension of (the spectrum of) $\mathbf{Z}$, which is not $0$. Over a field though, you have the expected formula : if $X,Y$ are (non empty) $k$ schemes locally of finite type over a field $k$, then $X\times_k Y$ (is non empty and) is of dimension $dim(X\times_k Y) = dim(X) + dim(Y)$. (Note that the spectrum of a field is zero dimensional.)
• Thanks a lot. So that, if we want to view an L-function $F$ as "representative" of a scheme $X_F$, and assuming the degree of $F$ is the dimension of $X_F$, the product $F.G$ should be representative of $X_{F}\times_{k} X_{G}$ for some field $k$, right? – Sylvain Julien Mar 6 '15 at 19:57