Definition 59.66.1. Let $f : Y \to X$ be a finite étale morphism of schemes. The map $f_* f^{-1} \to \text{id}$ described above and explicitly below is called the *trace*.

## 59.66 Méthode de la trace

A reference for this section is [Exposé IX, §5, SGA4]. The material here will be used in the proof of Lemma 59.83.9 below.

Let $f : Y \to X$ be an étale morphism of schemes. There is a sequence

of adjoint functors between $\textit{Ab}(X_{\acute{e}tale})$ and $\textit{Ab}(Y_{\acute{e}tale})$. The functor $f_!$ is discussed in Section 59.70. The adjunction map $\text{id} \to f_* f^{-1}$ is called *restriction*. The adjunction map $f_! f^{-1} \to \text{id}$ is often called the *trace map*. If $f$ is finite étale, then $f_* = f_!$ (Lemma 59.70.7) and we can view this as a map $f_*f^{-1} \to \text{id}$.

Let $f : Y \to X$ be a finite étale morphism of schemes. The trace map is characterized by the following two properties:

it commutes with étale localization on $X$ and

if $Y = \coprod _{i = 1}^ d X$ then the trace map is the sum map $f_*f^{-1} \mathcal{F} = \mathcal{F}^{\oplus d} \to \mathcal{F}$.

By Étale Morphisms, Lemma 41.18.3 every finite étale morphism $f : Y \to X$ is étale locally on $X$ of the form given in (2) for some integer $d \geq 0$. Hence we can define the trace map using the characterization given; in particular we do not need to know about the existence of $f_!$ and the agreement of $f_!$ with $f_*$ in order to construct the trace map. This description shows that if $f$ has constant degree $d$, then the composition

is multiplication by $d$. The “méthode de la trace” is the following observation: if $\mathcal{F}$ is an abelian sheaf on $X_{\acute{e}tale}$ such that multiplication by $d$ on $\mathcal{F}$ is an isomorphism, then the map

is injective. Namely, we have

by the vanishing of the higher direct images (Proposition 59.55.2) and the Leray spectral sequence (Proposition 59.54.2). Thus we can consider the maps

and the composition is an isomorphism (under our assumption on $\mathcal{F}$ and $f$). In particular, if $H_{\acute{e}tale}^ q(Y, f^{-1}\mathcal{F}) = 0$ then $H_{\acute{e}tale}^ q(X, \mathcal{F}) = 0$ as well. Indeed, multiplication by $d$ induces an isomorphism on $H_{\acute{e}tale}^ q(X, \mathcal{F})$ which factors through $H_{\acute{e}tale}^ q(Y, f^{-1}\mathcal{F})= 0$.

This is often combined with the following.

Lemma 59.66.2. Let $S$ be a connected scheme. Let $\ell $ be a prime number. Let $\mathcal{F}$ be 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 59.65.1 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 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 $V$ corresponding to $f^{-1}\mathcal{F}$ is through the map $\pi _1(T, \overline{t}) \to H$, see Remark 59.65.3. 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 59.65.1 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 $G$.
$\square$

Lemma 59.66.3. Let $\Lambda $ be a Noetherian ring. Let $\ell $ be a prime number and $n \geq 1$. Let $H$ be a finite $\ell $-group. Let $M$ be a finite $\Lambda [H]$-module annihilated by $\ell ^ n$. Then there is a finite filtration $0 = M_0 \subset M_1 \subset \ldots \subset M_ t = M$ by $\Lambda [H]$-submodules such that $H$ acts trivially on $M_{i + 1}/M_ i$ for all $i = 0, \ldots , t - 1$.

**Proof.**
Omitted. Hint: Show that the augmentation ideal $\mathfrak m$ of the noncommutative ring $\mathbf{Z}/\ell ^ n\mathbf{Z}[H]$ is nilpotent.
$\square$

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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