Lemma 59.66.4. Let $S$ be an irreducible, geometrically unibranch scheme. Let $\ell $ be a prime number and $n \geq 1$. Let $\Lambda $ be a Noetherian ring. Let $\mathcal{F}$ be a finite type, locally constant sheaf of $\Lambda $-modules on $S_{\acute{e}tale}$ which is annihilated by $\ell ^ n$. 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 of the form $\underline{M}_ T$ for some finite $\Lambda $-modules $M$.

**Proof.**
Choose a geometric point $\overline{s}$ of $S$. Via the equivalence of Lemma 59.65.2 the sheaf $\mathcal{F}$ corresponds to a finite $\Lambda $-module $M$ with a continuous $\pi _1(S, \overline{s})$-action. Let $G \subset \text{Aut}(V)$ be the image of the homomorphism $\rho : \pi _1(S, \overline{s}) \to \text{Aut}(M)$ giving the action. Observe that $G$ is finite as $M$ is a finite $\Lambda $-module (see proof of Lemma 59.65.2). 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 58.7. 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 $M$ corresponding to $f^{-1}\mathcal{F}$ is through the map $\pi _1(T, \overline{t}) \to H$, see Remark 59.65.3. Let $0 = M_0 \subset M_1 \subset \ldots \subset M_ t = M$ be as in Lemma 59.66.3. This induces a filtration $0 = \mathcal{F}_0 \subset \mathcal{F}_1 \subset \ldots \subset \mathcal{F}_ t = f^{-1}\mathcal{F}$ such that the successive quotients are constant with value $M_{i + 1}/M_ i$. Finally, 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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