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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?

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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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  • $\begingroup$ This is étale over $\mathbb A^1$, but is ramified at infinity. $\endgroup$
    – Angelo
    Commented Oct 9, 2010 at 10:55
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    $\begingroup$ 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. $\endgroup$
    – Torsten Ekedahl
    Commented Oct 9, 2010 at 10:56
  • $\begingroup$ I think that the equality should hold for projective varieties; I will try to write a proof later. $\endgroup$
    – Angelo
    Commented Oct 9, 2010 at 10:58
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    $\begingroup$ Dear Angelo, yes your argument works for projective varieties as the Lefschetz fixed point formula is true there. $\endgroup$
    – Torsten Ekedahl
    Commented Oct 9, 2010 at 11:57
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    $\begingroup$ 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. $\endgroup$
    – Angelo
    Commented Oct 9, 2010 at 12:11

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