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Let $X$ be some smooth scheme over $\mathbf C$ equipped with an action of $\mu_n$ (the group of $n$th roots of unity). The étale cohomology groups of X are therefore equipped with an action of $\mu_n$.

Now, let's suppose that the action of $\mu_n$ on $X$ extends to an action of $\mathbf G_m$. Then, the analytic space $X(\mathbf C)$ is equipped with an action of $\mu_n$ which extends to an action og $\mathbf C^\times$. As $\mathbf C^\times$ is connected, one concludes that the action of $\mu_n$ on the singular cohomology of $X(\mathbf C)$ is trivial (the multiplication by each element $\xi\in \mu_n$ is homotopic to the identity).

Going backwards to étale cohomology, using a comparison theorem between étale and singular cohomology, one concludes that the action of $\mu_n$ on the étale cohomology of $X$ is trivial.

Is there a direct proof that the action on the étale cohomology groups is trivial without having to refer to the topological case?

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Smoothness of $X$ is not needed (neither for the comparison isomorphism nor for the result in question). Let $X$ be any quasi-separated scheme over a separably closed field $k$, equipped with an action by a connected $k$-group scheme $G$ of finite type. Let $n > 0$ be an integer not divisible by the characteristic of $k$ and choose an integer $i \ge 0$. Then we want to show that the action of $G(k)$ on ${\rm{H}}^i(X, \mathbf{Z}/(n))$ is trivial (using etale cohomology here).

[The hypothesis on $n$ is necessary because if $n = p = {\rm{char}}(k)>0$ and $X = {\rm{Spec}}(A)$ is affine then the effect of $G(k)$ on ${\rm{H}}^1(X, \mathbf{Z}/(p)) = A/\wp(A)$ (with $\wp(f) = f^p-f$) is the induced action from the $G(k)$-action on $A = \Gamma(X,O_X)$, and this is generally nontrivial (e.g., $X = G = \mathbf{A}^1_k$ with the translation action corresponding to $c.f(t) = f(t+c)$ on global functions (for $c \in G(k)$).]

By a spectral sequence argument using a covering by quasi-compact $G$-stable open subsets we may reduce to the case when $X$ is quasi-compact (and quasi-separated). It is harmless to make the radiciel extension from $k$ to its algebraic closure ("topological invariance" of etale cohomology), so we may assume that the separably closed $k$ is even algebraically closed. We may then replace $G$ with $G_{\rm{red}}$ so that $G$ is smooth.

Let $f:G \times X \rightarrow G$ be the projection map. The hypothesis on $n$ and smoothness of $G$ allow us to apply the smooth base change theorem to conclude that $\mathscr{F} = {\rm{R}}^if_{\ast}(\mathbf{Z}/(n))$ is the constant sheaf on $G$ attached to ${\rm{H}}^i(X,\mathbf{Z}/(n))$. Consider the action automorphism $$\alpha: G \times X \simeq G \times X$$ defined by $(g,x) \mapsto (g, gx)$. This commutes with $f$, and so induces an automorphism $[\alpha]$ of $\mathscr{F}$ on $G$. The effect on the stalk at $g \in G(k)$ is the $g$-action on ${\rm{H}}^i(X, \mathbf{Z}/(n))$ that we want to be trivial. But $\mathscr{F}$ is a constant sheaf on a connected scheme, so the effect on $\mathscr{F}$ of any automorphism is uniquely determined by the effect on a single stalk. Looking on the stalk at $g=1$ thereby shows that $[\alpha]$ is the identity automorphism. Now pass to the effect of $[\alpha]$ on the stalk at any $g \in G(k)$ to conclude.

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Here's an alternative to grghxy's answer. Let me stick to the original assumptions for simplicity, but they can certainly be relaxed. Let $G$ be the image of the $\mu_n$ action on étale cohomology with finite coefficients. In particular, $G$ is finite. Let $g\in G$. Since the action is assumed to extend to a $\mathbb{C}^*$-action, we can solve $g=h^N$ for arbitrary $N>0$. Now choose $N=|G|$.

NB See some comments by grghxy & Yonatan Harpaz below.

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  • $\begingroup$ Since $h$ has even larger order than $g$, and in the end one is only using $N = |G|$ rather than arbitrarily large $N$, why doesn't this "prove" the triviality of the $\mu_p$-action induced by any $\mu_{p^2}$-action (by taking $N=p$)? Please clarify. $\endgroup$
    – grghxy
    Oct 8 '15 at 15:35
  • $\begingroup$ Good point! If there is a fix, it'll have to wait until after I finishing teaching etc. $\endgroup$ Oct 8 '15 at 15:45
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    $\begingroup$ Choosing $N = |G|$ is not enough because the action of $\mu_{nN}$ might involve elements of $Aut(H^n_{et}(X,A))$ which are not in $G$. However, taking $N=|Aut(H^n_{et}(X,A))|$ should be enough (here $A$ is a finite abelian group, and $X$ is sufficiently nice, so this automorphism group is finite). $\endgroup$ Oct 8 '15 at 20:03
  • $\begingroup$ This fix looks very nice. I accepted the previous answer because it appeared first. Thank you anyway! $\endgroup$
    – Oblomov
    Oct 9 '15 at 7:01

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