The Stacks project

Lemma 62.14.4. Let $Y$ be an affine scheme. Let $\Lambda $ be a Noetherian ring. Let $\mathcal{F}$ be a constructible sheaf of $\Lambda $-modules on $\mathbf{A}^1_ Y$ which is torsion. Then $Rf_!\mathcal{F}$ has constructible cohomology sheaves where $f : \mathbf{A}^1_ Y \to Y$ is the structure morphism.

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 √Čtale 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 62.12.2. This is a finite $\Lambda $-module by Lemma 62.12.4.

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 62.9.4). 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 62.10.5 and the result for $Z \to Y$ (Lemma 62.14.1) we reduce to showing that $R(U \to Y)_!\mathcal{F}$ has constructible cohomology sheaves. This is Lemma 62.14.3. $\square$


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