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 $e(X) = d \cdot e(Y)$? 


Let $M$ and $N$ be separated $\mathbf{C}$-schemes of finite type. Let $\pi:M\longrightarrow N$ be a finite etale morphism of degree $d$. Then $e\_c(M) = d \cdot e\_c(N)$. To prove this, we may and do assume that $M$ and $N$ are connected and 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. (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.) 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 $p>d$. But what about the *wild* case? 



Behaviour of euler characteristics in characteristic p for finite etale covers
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 $1\neq 1\times p$.


