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Let $X$ be an integral scheme, and $g: \eta \to X$ the inclusion of its generic point. If $M$ is an abelian group and $M_\eta$ is a constant sheaf on $\eta$ taking value $M$, then is it true that $g_* M_\eta = M_X$?

My thought is yes, as if $U \subset X$ is open, then $g_* M_\eta (U) = M_\eta( g^{-1}(U)) = M$, as every nonempty open set will contain the generic point. However it seems that one may need that $X$ is normal (or some other hypothesis) for this to be true. For reference, this is Exercise 3.7 (Chap II) in Milne's Etale Cohomology.

DKS
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  • The argument in your third line is perfectly correct and normality (or any other property $X$ may or may not have) plays absolutely no role in this question. – Georges Elencwajg Nov 13 '18 at 22:47
  • @GeorgesElencwajg Glad to know I'm not losing my mind, I also thought this was odd. Milne claims that normality is required for this (he goes on to claim that if $X$ is a curve with a node $z$, then there is an exact sequence $0\to M_X \to g_M_\eta \to i_M_\eta \to 0$, where $i:Z \to X$ is the inclusion of the nodal point). Is it possible he's referring to this scenario in the Etale topology (implicitly, I'm also asking if you know if this is true for the Etale topology)? – DKS Nov 14 '18 at 00:30
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    Dear DKS, since I'm not 100% sure of my answer for the étale topology case, I'd rather say nothing than take the risk of misleading you. – Georges Elencwajg Nov 14 '18 at 20:01

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We need to show $$M_X\rightarrow g_*g^*M_X=g_*M_\eta\tag{$\dagger$}$$ is an isomorphism. Fix a geometric point $x$ of $X$ and let $X_{(x)}$ denote the (spectrum of the) strict henselization of $X$ at $x$; $X_{(x)}$ is a normal domain as $X$ is normal. The map on stalks is $$M\rightarrow\Gamma(\eta\times_X X_{(x)},M_\eta).$$ $\eta_x:=\eta\times_X X_{(x)}$ is the spectrum of the field of fractions of the strict henselization of $X$ at $x$, which is separable algebraic over $k(\eta)$. Letting $I_x:=\operatorname{Gal}(k(\overline\eta)/k(\eta_x))$ ($k(\overline\eta)$ denotes the separable closure of $k(\eta_x)$, which is the same as the separable closure of $k(\eta)$), $$\Gamma(\eta\times_X X_{(x)},M_\eta)=M^{I_x}:=\{m\in M:gm=m\forall g\in I_x\}$$ where here we have identified $M$ with the stalk of $M_\eta$ at a geometric point centered on $\eta$. So asking that the map ($\dagger$) be an isomorphism at the geometric point $x$ is the same as requiring that $I_x$ act trivially on $M$ considered as $\operatorname{Gal}(k(\overline\eta)/k(\eta))$-module. In particular, this condition is satisfied by the constant sheaf, which corresponds to $M$ with trivial action of Galois.

(I am implicitly using here the description of étale morphism with normal integral target in EGA $\text{IV}_4$ 18.10.7.)

Tomo
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