# Is the integrality of the zeta function easy?

I'm trying to get the gist of the proof of the Weil conjectures. Let $X$ be a variety over $\mathbb{F}_{p^n}$. A priori $Z(X,t)\in \mathbb{Q}[[t]]$. Due to the Grothendieck-Lefschetz fixed point theorem, $Z(X,t)=\prod P_i(t)^{(-1)^{i+1}}$, where $P_i(t)$ is the characteristic polynomial of the Frobenius acting on $H^i(X,\mathbb{Q}_l)$ where $l$ is a fixed prime different from $p$. This implies that $Z(X,t)\in \mathbb{Q}_l(t)\cap \mathbb{Q}[[t]]$ for every prime $l$ different from $p$.

Does this suffice to determine that it is in $\mathbb{Q}(t)$? If not, then how was it proven that it is in $\mathbb{Q}(t)$?

-
You mean rationality. Integrality sounds like $Z(X,t)\in\mathbb Z[[t]]$ or that the Frobenius eigenvalues are algebraic integers. $H^i$ should be $H^i_c.$ – shenghao Oct 26 '11 at 1:49
About $H^i$ being $H^i_c$, you're of course right. I was assuming in my head that $X$ is projective, so they are both isomorphic. I believe, and correct me if I'm wrong, that when people talk about the "rationality" of the zeta function, they mean that it is a rational function. That it is a rational function with coefficients in the rationals, is, I thought, called the integrality of the zeta function. – Makhalan Duff Oct 26 '11 at 4:13

It is not necessarily obvious that this suffices, though it is true (fortunately). You can find a proof in Milne's Lectures on Etale Cohomology. (I happened to just be looking at this while thinking about this question.)

It is Lemma 27.9, phrased as: Let $k\subset K$ be fields and $f(t)\in k[[t]]$. If $f(t)\in K(t)$, then $f(t)\in k(t)$.

-
Wow, that's a pretty strong lemma. – Makhalan Duff Oct 25 '11 at 22:55
From the number of votes, you'd think I'd come up with it myself :) – B R Oct 26 '11 at 20:10

First embedd $\mathbf{Q}_{\ell}\subseteq\mathbf{C}$ and then there is quite an easy argument that I discovered which answers your question, see the following link 1.

In fact this argument shows that for any normal subfield $K\subseteq\mathbf{C}$, if $$\frac{f}{g}\in\mathbf{C}(x_1,x_2,\ldots,x_n)\cap K[[x_1,x_2,\ldots,x_n]]$$ with $\gcd(f,g)=1$ and the constant term of $f$ and $g$ are equal to $1$ then $f,g\in K[x_1,\ldots,x_n]$.

-