# Special values of zeta functions and extensions of base fields.

Let $X$ be a scheme of finite type over a finite field $k=\mathbb{F}_{q}$ of $q$ elements.

Then, one can define the zeta function $Z_{X/k}(T)$ of $X$ ovet $k$ as $\prod_{x\in |X|}\frac{1}{1-T^{deg_{k}(x)}}$, where $deg_{k}(x)=[k(x):k]$ is the degree of extension $k(x)/k$ for any closed point $x$ of $X$.

Assume that the structure morphism $X \to Spec(k)$ factors through $Spec(k^{\prime})$, the spectrum of a finite extension field $k^{\prime} \supset k$.

Then, one can define the zeta function $Z_{X/k^{\prime}}(T)$ of $X$ ovet $k^{\prime}$ as $\prod_{x\in |X|}\frac{1}{1-T^{deg_{k^{\prime}}(x)}}$, where $deg_{k^{\prime}}(x)=[k(x):k^{\prime}]$ is the degree of extension $k(x)/k^{\prime}$ for any closed point $x$ of $X$.

My question is: Does the following equation holds?:

(the special value of $Z_{X/k}$ at $T=1$) = $[k^{\prime}:k] \cdot ($the special value of $Z_{X/k^{\prime}}$ at $T=1$)

Under the assumption that the structural morphism factors through $\mathrm{Spec} k'$, all residue fields contain $k'$, so we have the straghtforward relation $Z_{X/k}(T)=Z_{X/k'}(T^{[k':k]})$. –  François Brunault Dec 15 '12 at 11:30
The special value ot $T=1$ is not always well-defined, for example if $X=\mathrm{Spec}(k')$ then $Z_{X/k'}(T)=1/(1-T)$. You probably mean the leading term, in which case there is indeed a relation, but I will let you work it out using the simple relation $Z(T)=Z'(T^d)$. –  François Brunault Dec 15 '12 at 12:18