
## 54.65 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 54.78.8 below.

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

$f_!, f^{-1}, f_*$

of adjoint functors between $\textit{Ab}(X_{\acute{e}tale})$ and $\textit{Ab}(Y_{\acute{e}tale})$. The functor $f_!$ is discussed in Section 54.69. 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 54.69.5) and we can view this as a map $f_*f^{-1} \to \text{id}$.

Definition 54.65.1. Let $f : Y \to X$ be a finite étale morphism of schemes. The map $f_* f^{-1} \to \text{id}$ described above and below is called the trace.

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

1. it commutes with étale localization on $X$ and

2. 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 40.18.3 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

$\mathcal{F} \xrightarrow {res} f_* f^{-1} \mathcal{F} \xrightarrow {trace} \mathcal{F}$

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

$H^ n_{\acute{e}tale}(X, \mathcal{F}) \longrightarrow H^ n_{\acute{e}tale}(Y, f^{-1}\mathcal{F})$

is injective. Namely, we have

$H^ n_{\acute{e}tale}(Y, f^{-1}\mathcal{F}) = H^ n_{\acute{e}tale}(X, f_*f^{-1}\mathcal{F})$

by the vanishing of the higher direct images (Proposition 54.54.2) and the Leray spectral sequence (Proposition 54.53.2). Thus we can consider the maps

$H^ n_{\acute{e}tale}(X, \mathcal{F}) \to H^ n_{\acute{e}tale}(Y, f^{-1}\mathcal{F})= H^ n_{\acute{e}tale}(X, f_*f^{-1}\mathcal{F}) \xrightarrow {trace} H^ n_{\acute{e}tale}(X, \mathcal{F})$

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 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 $G$. $\square$

Comment #3396 by Dongryul Kim on

At the end of the second paragraph, it is stated that $f_{!} = f_\ast$ if $f$ is finite etale. Can we refer to Lemma 03S7 (54.69.5) here? I had a hard time finding out that this is proved later.

Comment #3462 by on

OK, yes, good point. I also added some remarks explaining how to define the trace map without using the existence of the functor $f_!$ whose construction comes later in the Stacks project. This seems adequate for now, but if there are any complaints then we can always move the sections around. See changes to latex here.

Comment #3859 by Emmanuel Kowalski on

Is the French title intentional?

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