The Stacks project

Proof. We may assume $S = \mathop{\mathrm{Spec}}(A)$ where $A$ is an $\mathbf{F}_ p$-algebra and that $\mathcal{E}$ is the quasi-coherent module associated to the free $A$-module $Ae_1 \oplus \ldots \oplus Ae_ n$. We write $F(e_ i) = \sum a_{ij} e_ j$.

Surjectivity of $F - 1$. It suffices to show that any element $\sum a_ i e_ i$, $a_ i \in A$ is in the image of $F - 1$ after replacing $A$ by a faithfully flat étale extension. Observe that

Consider the $A$-algebra

A computation shows that $\text{d}x_ i$ is zero in $\Omega _{A'/A}$ and hence $\Omega _{A'/A} = 0$. Since $A'$ is of finite type over $A$, this implies that $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A)$ is unramified and hence is quasi-finite. Since $A'$ is generated by $n$ elements and cut out by $n$ equations, we conclude that $A'$ is a global relative complete intersection over $A$. Thus $A'$ is flat over $A$ and we conclude that $A \to A'$ is étale (as a flat and unramified ring map). Finally, the reader can show that $A \to A'$ is faithfully flat by verifying directly that all geometric fibres of $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A)$ are nonempty, however this also follows from Étale Cohomology, Lemma 59.63.2. Finally, the element $\sum x_ i e_ i \in A'e_1 \oplus \ldots \oplus A'e_ n$ maps to $\sum a_ i e_ i$ by $F - 1$.

Constructibility of the kernel. The calculations above show that $\mathop{\mathrm{Ker}}(F - 1)$ is represented by the scheme

over $S = \mathop{\mathrm{Spec}}(A)$. Since this is a scheme affine and étale over $S$ we obtain the result from Étale Cohomology, Lemma 59.73.1. $\square$

Proof. We may assume $S$ is affine. Say $S = \mathop{\mathrm{Spec}}(A)$. Then, if we write $A = \bigcup A_ i$ as the union of its finite type $\mathbf{Z}$-subalgebras, we can find an $i$ and a morphism $f_ i : X_ i \to S_ i = \mathop{\mathrm{Spec}}(A_ i)$ of finite type whose base change to $S$ is $f : X \to S$, see Limits, Lemma 32.10.1. After increasing $i$ we may assume $f_ i : X_ i \to S_ i$ is smooth, proper, and of relative dimension $1$, see Limits, Lemmas 32.13.1 32.8.9, and 32.18.4. By More on Morphisms, Lemma 37.53.8 we obtain an open subscheme $U_ i \subset S_ i$ such that the fibres of $f_ i : X_ i \to S_ i$ over $U_ i$ are geometrically connected. Then $S \to S_ i$ maps into $U_ i$. We may replace $X \to S$ by $f_ i : f_ i^{-1}(U_ i) \to U_ i$ to reduce to the case discussed in the next paragraph.

Assume $S$ is Noetherian. We may write $S = U \cup Z$ where $U$ is the open subscheme defined by the nonvanishing of $\ell $ and $Z = V(\ell ) \subset S$. Since the formation of $R^ qf_*\underline{\mathbf{Z}/\ell \mathbf{Z}}$ commutes with arbitrary base change (Étale Cohomology, Theorem 59.91.11), it suffices to prove the result over $U$ and over $Z$. Thus we reduce to the following two cases: (a) $\ell $ is invertible on $S$ and (b) $\ell $ is zero on $S$.

Case (a). We claim that in this case the sheaves $R^ qf_*\underline{\mathbf{Z}/\ell \mathbf{Z}}$ are finite locally constant on $S$. First, by proper base change (in the form of Étale Cohomology, Lemma 59.91.13) and by finiteness (Étale Cohomology, Theorem 59.83.10) we see that the stalks of $R^ qf_*\underline{\mathbf{Z}/\ell \mathbf{Z}}$ are finite. By Étale Cohomology, Lemma 59.94.4 all specialization maps are isomorphisms. We conclude the claim holds by Étale Cohomology, Lemma 59.75.6.

Case (b). Here $\ell = p$ is a prime and $S$ is a scheme over $\mathop{\mathrm{Spec}}(\mathbf{F}_ p)$. By the same references as above we already know that the stalks of $R^ qf_*\underline{\mathbf{Z}/p\mathbf{Z}}$ are finite and zero for $q \geq 2$. It follows from Étale Cohomology, Lemma 59.39.3 that $f_*\underline{\mathbf{Z}/p\mathbf{Z}} = \underline{\mathbf{Z}/p\mathbf{Z}}$. It remains to prove that $R^1f_*\underline{\mathbf{Z}/p\mathbf{Z}}$ is constructible. Consider the Artin-Schreyer sequence

See Étale Cohomology, Section 59.63. Recall that $f_*\mathcal{O}_ X = \mathcal{O}_ S$ and $R^1f_*\mathcal{O}_ X$ is a finite locally free $\mathcal{O}_ S$-module of rank equal to the genera of the fibres of $X \to S$, see Algebraic Curves, Lemma 53.20.13. We conclude that we have a short exact sequence

Applying Lemma 63.13.1 we win. $\square$

Proof. If $n = \ell n'$ for some prime number $\ell $, then we get a short exact sequence $0 \to M[\ell ] \to M \to M' \to 0$ of finite $\Lambda $-modules and $M'$ is annihilated by $n'$. This produces a corresponding short exact sequence of constant sheaves, which in turn gives rise to an exact sequence

Thus, if we can show the result in case $M$ is annihilated by a prime number, then by induction on $n$ we win by Étale Cohomology, Lemma 59.71.6.

Let $\ell $ be a prime number such that $\ell $ annihilates $M$. Then we can replace $\Lambda $ by the $\mathbf{F}_\ell $-algebra $\Lambda /\ell \Lambda $. Namely, the sheaf $R^ qf_*\underline{M}$ where $\underline{M}$ is viewed as a sheaf of $\Lambda $-modules is the same as the sheaf $R^ qf_*\underline{M}$ computed by viewing $\underline{M}$ as a sheaf of $\Lambda /\ell \Lambda $-modules, see Cohomology on Sites, Lemma 21.20.7.

Assume $\ell $ be a prime number such that $\ell $ annihilates $M$ and $\Lambda $. Let us reduce to the case where $M$ is a finite free $\Lambda $-module. Namely, choose a resolution

Recall that $f_*$ has finite cohomological dimension on sheaves of $\Lambda $-modules, see Étale Cohomology, Lemma 59.92.2 and Derived Categories, Lemma 13.32.2. Thus we see that $R^ qf_*\underline{M}$ is the $q$th cohomology sheaf of the object

in $D(S_{\acute{e}tale}, \Lambda )$ for some integer $a$ large enough. Using the first spectral sequence of Derived Categories, Lemma 13.21.3 (or alternatively using an argument with truncations) we conclude that it suffices to prove that $R^ qf_*\underline{\Lambda })$ is constructible.

At this point we can finally use that

by Étale Cohomology, Lemma 59.96.6. Since any module over the field $\mathbf{Z}/\ell \mathbf{Z}$ is flat we obtain

Hence it suffices to prove the result for $R^ qf_*\underline{\mathbf{Z}/\ell \mathbf{Z}}$ by Étale Cohomology, Lemma 59.71.10. This case is Lemma 63.13.2. $\square$