Lemma 54.65.2. Let $S$ be a connected scheme. Let $\ell $ be a prime number. Let $\mathcal{F}$ a finite type, locally constant sheaf of $\mathbf{F}_\ell $-vector spaces on $S_{\acute{e}tale}$. Then there exists a finite étale morphism $f : T \to S$ of degree prime to $\ell $ such that $f^{-1}\mathcal{F}$ has a finite filtration whose successive quotients are $\underline{\mathbf{Z}/\ell \mathbf{Z}}_ T$.

**Proof.**
Choose a geometric point $\overline{s}$ of $S$. Via the equivalence of Lemma 54.64.1 the sheaf $\mathcal{F}$ corresponds to a finite dimensional $\mathbf{F}_\ell $-vector space $V$ with a continuous $\pi _1(S, \overline{s})$-action. Let $G \subset \text{Aut}(V)$ be the image of the homomorphism $\rho : \pi _1(S, \overline{s}) \to \text{Aut}(V)$ giving the action. Observe that $G$ is finite. The surjective continuous homomorphism $\overline{\rho } : \pi _1(S, \overline{s}) \to G$ corresponds to a Galois object $Y \to S$ of $\textit{FÉt}_ S$ with automorphism group $G = \text{Aut}(Y/S)$, see Fundamental Groups, Section 53.7. Let $H \subset G$ be an $\ell $-Sylow subgroup. We claim that $T = Y/H \to S$ works. Namely, let $\overline{t} \in T$ be a geometric point over $\overline{s}$. The image of $\pi _1(T, \overline{t}) \to \pi _1(S, \overline{s})$ is $(\overline{\rho })^{-1}(H)$ as follows from the functorial nature of fundamental groups. Hence the action of $\pi _1(T, \overline{t})$ on $V$ corresponding to $f^{-1}\mathcal{F}$ is through the map $\pi _1(T, \overline{t}) \to H$, see Remark 54.64.2. As $H$ is a finite $\ell $-group, the irreducible constituents of the representation $\rho |_{\pi _1(T, \overline{t})}$ are each trivial of rank $1$ (this is a simple lemma on representation theory of finite groups; insert future reference here). Via the equivalence of Lemma 54.64.1 this means $f^{-1}\mathcal{F}$ is a successive extension of constant sheaves with value $\underline{\mathbf{Z}/\ell \mathbf{Z}}_ T$. Moreover the degree of $T = Y/H \to S$ is prime to $\ell $ as it is equal to the index of $H$ in $G$.
$\square$

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## Comments (1)

Comment #4186 by Nicolas Müller on

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