Behaviour of euler characteristics in characteristic p for finite etale covers

Let $k$ be an algebraic closure of a finite field of characteristic $p$. Fix an integer $l\neq p$. For a separated $k$-scheme $X$ of finite type, we define the (compactly supported) Euler characteristic of $X$ to be $$e(X) =\sum_i (-1)^i \dim_\mathbf{Q_l} H^i_c(X,\mathbf{Q}_l).$$ Here $H^i_c(-,\mathbf{Q}_l)$ denotes the $l$-adic cohomology with compact support.

For example, if $X$ is smooth and projective over $k$, we have that $e(X)$ equals the degree of the top Chern class of $X$.

Let $X$ and $Y$ be separated $k$-schemes of finite type. Let $\pi:X\longrightarrow Y$ be a finite etale morphism of degree $d$.

Question. Is it true that $e(X) = d \cdot e(Y)$?

Here is what I "know":

For any separated $\mathbf{C}$-scheme $M$ of finite type, we define the (compactly supported) Euler characteristic of $M$ to be $$e(M) =\sum_i (-1)^i \dim_\mathbf{Q} H^i_c(M,\mathbf{Q}).$$ Here $H^i_c(-,\mathbf{Q})$ denotes the cohomology with compact support and coefficients in $\mathbf{Q}$ on the category of para-compact Hausdorff spaces. (Just to be clear, we use the analytification of $M$ here.)

Let $M$ and $N$ be separated $\mathbf{C}$-schemes of finite type. Let $\pi:M\longrightarrow N$ be a finite etale morphism of degree $d$. One can show that $e_c(M) = d \cdot e_c(N)$ quite easily as follows:

We may assume that $M$ and $N$ are connected and we may assume that $\pi$ is Galois. Let $G$ be the Galois group. Let $K_0(\mathbf{Q}[G])$ be the Grothendieck group of finitely generated $\mathbf{Q}[G]$-modules. Since the action of $G$ is free, a nontrivial element $g\in G$ has no fixed points. By the Lefschetz trace formula (see the paper by Deligne-Lusztig), we have that $$\sum (-1)^i \textrm{Tr}(H^i_c(g)) = 0.$$ Therefore, by character theory or some result in loc. cit, we have that the class of $H^\cdot_c(M,\mathbf{Q})$ in $K_0(\mathbf{Q}[G])$, defined to be the alternating sum of the classes of $H^i_c(M,\mathbf{Q})$, is an integer multiple of the regular representation. The result then follows from an easy computation.

Question. The same proof works to answer my above question positively when the cover $\pi:X\longrightarrow Y$ above is tame. In particular, if $p$ does not divide $d$. But what about the wild case? Are there some comparison theorems which allow us to simply reduce to the complex case?

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It's not true in general. Over a field of characteristic $p>0$, the map $f:\mathbf{A}^1\to\mathbf{A}^1$ defined by $f(z)=z^p+z$ is etale because its derivative is $1$. The degree of $f$ is $p$, and the Euler characteristic of $\mathbf{A}^1$ is $1$, but $1\neq 1\times p$.

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This is étale over $\mathbb A^1$, but is ramified at infinity. – Angelo Oct 9 2010 at 10:55
You must replace $\mathbf P^1$ by the affine line, your map is (very heavily) ramified at infinity. You do have $1\neq p$ so you still win. – Torsten Ekedahl Oct 9 2010 at 10:56
I think that the equality should hold for projective varieties; I will try to write a proof later. – Angelo Oct 9 2010 at 10:58
Dear Angelo, yes your argument works for projective varieties as the Lefschetz fixed point formula is true there. – Torsten Ekedahl Oct 9 2010 at 11:57
Dear Torsten, the argument is not mine, it ia Aryan's. But yes, after thinking about it, the order of $G$ should not be relevant in the Lefschetz fixed point formula, the only condition should be that the action is tame, that is, that the order of the stabilizers be prime to the characteristic. – Angelo Oct 9 2010 at 12:11