Stalks of etale sheaves

Question

Let $\pi:X \rightarrow Y$ be a finite morphism of schemes and $\mathfrak{F}$ be an etale sheaf on $X$. Then for a $y \in Y$ we have the stalk $(\pi_{*}\mathfrak{F})_{y}=\prod_{\pi(x)=y}\mathfrak{F}_{x}^{d(x)}$ where $d(x)$ is the separable degree of the field extension $k(x)/k(y)$ (Corollary 3.5.(c) in Chapter II. in Milne: Etale cohomology). The proof in the book would be completely clear for me, if there were no separable degrees (and Milne does not mention either why are they there)....So my question is where from these separable degrees come into the picture?

Answer 1

It is enough to prove this in the case where $Y$ is $Spec$ of a strictly Henselian ring. I think, one sees the main point in the argument already in the following special case:

Let $Y=Spec(K)$ and $X=Spec(E)$ where $E/K$ is a finite separable extension of fields. Denote $\pi: X\to Y$ the canonical map. Let $\overline{E}$ be an algebraic closure of $E$ and $\overline{x}$ (resp. $\overline{y}$) be the corresponding geometric point of $X$ (resp $Y$). Let $L/K$ be a finite separable extension contained in $\overline{E}$ and containing the Galois closure of $E$. If $F$ is a sheaf on $X$, then $\pi_*F(Spec(L))=F(Spec(L\otimes_K E))$ and $Spec(L\otimes_K E)= \coprod_{i=1}^d Spec(L)$, where $d=[E:K]$. Hence $\pi_* F(Spec(L))=F(Spec(L))^d$. Now take an inductive limit over such $L$, in order to obtain $\pi_*F_{\overline{y}}=F_{\overline{x}}^d$.

Hope this is of some use...