The Stacks project

Proof. Since the functor $f_!$ is exact, it suffices to show that $f_!\mathcal{F}$ is constructible for any constructible sheaf $\mathcal{F}$ of $\Lambda $-modules on $X_{\acute{e}tale}$. The question is local on $Y$ and hence we may and do assume $Y$ is affine. Then $X$ is quasi-compact and quasi-separated, see Morphisms, Definition 29.21.1. Say $X = \bigcup _{i = 1, \ldots , n} X_ i$ is a finite affine open covering. By Lemma 63.4.7 we see that it suffices to show that $f_{i, !}\mathcal{F}|_{X_ i}$ and $f_{ii', !}\mathcal{F}|_{X_ i \cap X_{i'}}$ are constructible where $f_ i : X_ i \to Y$ and $f_{ii'} : X_ i \cap X_{i'} \to Y$ are the restrictions of $f$. Since $X_ i$ and $X_ i \cap X_{i'}$ are quasi-compact and separated this means we may assume $f$ is separated. By Zariski's main theorem (in the form of More on Morphisms, Lemma 37.43.4) we can choose a factorization $f = g \circ j$ where $j : X \to X'$ is an open immersion and $g : X' \to Y$ is finite and of finite presentation. Then $f_! = g_! \circ j_!$ by Lemma 63.3.13. By Étale Cohomology, Lemma 59.73.1 we see that $j_!\mathcal{F}$ is constructible on $X'$. The morphism $g$ is finite hence $g_! = g_*$ by Lemma 63.3.4. Thus $f_!\mathcal{F} = g_!j_!\mathcal{F} = g_*j_!\mathcal{F}$ is constructible by Étale Cohomology, Lemma 59.73.9. $\square$

Proof. We will prove the result by induction on $d = \dim (S)$.

Base case. If $d = 0$, then the only thing to show is that the stalks of $R^ qf_!\underline{M}$ are finite $\Lambda $-modules. If $\overline{s}$ is a geometric point of $S$, then we have $(R^ qf_!\underline{M})_{\overline{s}} = H^ q_ c(X_{\overline{s}}, \underline{M})$ by Lemma 63.12.2. This is a finite $\Lambda $-module by Lemma 63.12.4.

Induction step. It suffices to find a dense open $U \subset S$ such that $Rf_!\underline{M}|_ U$ has constructible cohomology sheaves. Namely, the restriction of $Rf_!\underline{M}$ to the complement $S \setminus U$ will have constructible cohomology sheaves by induction and the fact that formation of $Rf_!\underline{M}$ commutes with all base change (Lemma 63.9.4). In fact, let $\eta \in S$ be a generic point of an irreducible component of $S$. Then it suffices to find an open neighbourhood $U$ of $\eta $ such that the restriction of $Rf_!\underline{M}$ to $U$ is constructible. This is what we will do in the next paragraph.

Given a generic point $\eta \in S$ we choose a diagram

as in More on Morphisms, Lemma 37.56.1. We will show that $Rf_!\underline{M}|_ U$ is constructible. First, since $V \to U$ is finite and surjective, it suffices to show that the pullback to $V$ is constructible, see Étale Cohomology, Lemma 59.73.3. Since formation of $Rf_!$ commutes with base change, we see that it suffices to show that $R(X_ V \to V)_!\underline{M}$ is constructible. Let $W \subset X_ V$ be the open subscheme given to us by More on Morphisms, Lemma 37.56.1 part (4). Let $Z \subset X_ V$ be the reduced induced scheme structure on the complement of $W$ in $X_ V$. Then the fibres of $Z \to V$ have dimension $0$ (as $W$ is dense in the fibres) and hence $Z \to V$ is quasi-finite. From the distinguished triangle

of Lemma 63.10.5 and from Lemma 63.14.1 we conclude that it suffices to show that $R(W \to V)_!\underline{M}$ has constructible cohomology sheaves. Next, we have

because the morphism $\nu : \nu ^{-1}(W) \to W$ is a thickening and we may apply Lemma 63.10.6. Next, we let $Z' \subset \coprod \overline{Y}_ i$ denote the complement of the open $j(\nu ^{-1}(W))$. Again $Z' \to V$ is quasi-finite. Again use the distinguished triangle

to conclude that it suffices to prove

has constructible cohomology sheaves (second equality by Lemma 63.9.2). The result for $R(\overline{Y}_ i \to T_ i)_!\underline{M}$ is Lemma 63.13.3 and we win because $T_ i \to V$ is finite étale and we can apply Lemma 63.14.1. $\square$

Proof. We may decompose $\Lambda $ as a product $\Lambda = \Lambda _1 \times \ldots \times \Lambda _ r$ where $\Lambda _ i$ is $\ell _ i$-primary for some prime $\ell _ i$. Thus we may assume there exists a prime $\ell $ and an integer $n > 0$ such that $\ell ^ n$ annihilates $\Lambda $ (and hence $\mathcal{F}$).

Since $U$ is Noetherian, we see that $U$ has finitely many connected components. Thus we may assume $U$ is connected. Let $g : U' \to U$ be the finite étale covering constructed in Étale Cohomology, Lemma 59.66.4. The discussion in Étale Cohomology, Section 59.66 gives maps

whose composition is an isomorphism. Hence it suffices to prove the result for $g_*g^{-1}\mathcal{F}$. On the other hand, we have $Rf_!g_*g^{-1}\mathcal{F} = R(f \circ g)_!g^{-1}\mathcal{F}$ by Lemma 63.9.2. Since $g^{-1}\mathcal{F}$ has a finite filtration by constant sheaves of $\Lambda $-modules of the form $\underline{M}$ for some finite $\Lambda $-module $M$ (by our choice of $g$) this reduces us to the case proved in Lemma 63.14.2. $\square$

Proof. Say $\mathcal{F}$ is annihilated by $n > 0$. Then we can replace $\Lambda $ by $\Lambda /n\Lambda $ without changing $Rf_!\mathcal{F}$. Thus we may and do assume $\Lambda $ is a torsion ring.

Say $Y = \mathop{\mathrm{Spec}}(R)$. Then, if we write $R = \bigcup R_ i$ as the union of its finite type $\mathbf{Z}$-subalgebras, we can find an $i$ such that $\mathcal{F}$ is the pullback of a constructible sheaf of $\Lambda $-modules on $\mathbf{A}^1_{R_ i}$, see Étale Cohomology, Lemma 59.73.10. Hence we may assume $Y$ is a Noetherian scheme of finite dimension.

Assume $Y$ is a Noetherian scheme of finite dimension $d = \dim (Y)$ and $\Lambda $ is torsion. We will prove the result by induction on $d$.

Base case. If $d = 0$, then the only thing to show is that the stalks of $R^ qf_!\mathcal{F}$ are finite $\Lambda $-modules. If $\overline{y}$ is a geometric point of $Y$, then we have $(R^ qf_!\mathcal{F})_{\overline{y}} = H^ q_ c(X_{\overline{y}}, \mathcal{F})$ by Lemma 63.12.2. This is a finite $\Lambda $-module by Lemma 63.12.4.

Induction step. It suffices to find a dense open $V \subset Y$ such that $Rf_!\mathcal{F}|_ V$ has constructible cohomology sheaves. Namely, the restriction of $Rf_!\mathcal{F}$ to the complement $Y \setminus V$ will have constructible cohomology sheaves by induction and the fact that formation of $Rf_!\mathcal{F}$ commutes with all base change (Lemma 63.9.4). By definition of constructible sheaves of $\Lambda $-modules, there is a dense open subscheme $U \subset \mathbf{A}^1_ Y$ such that $\mathcal{F}|_ U$ is a finite type, locally constant sheaf of $\Lambda $-modules. Denote $Z \subset \mathbf{A}^1_ Y$ the complement (viewed as a reduced closed subscheme). Note that $U$ contains all the generic points of the fibres of $\mathbf{A}^1_ Y \to Y$ over the generic points $\xi _1, \ldots , \xi _ n$ of the irreducible components of $Y$. Hence $Z \to Y$ has finite fibres over $\xi _1, \ldots , \xi _ n$. After replacing $Y$ by a dense open (which is allowed), we may assume $Z \to Y$ is finite, see Morphisms, Lemma 29.51.1. By the distinguished triangle of Lemma 63.10.5 and the result for $Z \to Y$ (Lemma 63.14.1) we reduce to showing that $R(U \to Y)_!\mathcal{F}$ has constructible cohomology sheaves. This is Lemma 63.14.3. $\square$

Proof. The question is local on $Y$ hence we may and do assume $Y$ is affine. By the induction principle and Lemma 63.10.4 we reduce to the case where $X$ is also affine.

Assume $X$ and $Y$ are affine. Since $X$ is of finite presentation, we can choose a closed immersion $i : X \to \mathbf{A}^ n_ Y$ which is of finite presentation. If $p : \mathbf{A}^ n_ Y \to Y$ denotes the structure morphism, then we see that $Rf_! = Rp_! \circ Ri_!$ by Lemma 63.9.2. By Lemma 63.14.1 we have the result for $Ri_! = i_!$. Hence we may assume $f$ is the projection morphism $\mathbf{A}^ n_ Y \to Y$. Since we can view $f$ as the composition

we may assume $n = 1$.

Assume $Y$ is affine and $X = \mathbf{A}^1_ Y$. Since $Rf_!$ has finite cohomological dimension (Lemma 63.10.2) we may assume $K$ is bounded below. Using the first spectral sequence of Derived Categories, Lemma 13.21.3 (or alternatively using an argument with truncations), we reduce to showing the result of Lemma 63.14.4. $\square$