Lemma 56.73.3. Let $X$ be a Noetherian scheme.

Let $\mathcal{F}$ be a constructible sheaf of sets on $X_{\acute{e}tale}$. There exist an injective map of sheaves

\[ \mathcal{F} \longrightarrow \coprod \nolimits _{i = 1, \ldots , n} f_{i, *}\underline{E_ i} \]

where $f_ i : Y_ i \to X$ is a finite morphism and $E_ i$ is a finite set.

Let $\mathcal{F}$ be a constructible abelian sheaf on $X_{\acute{e}tale}$. There exist an injective map of abelian sheaves

\[ \mathcal{F} \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} f_{i, *}\underline{M_ i} \]

where $f_ i : Y_ i \to X$ is a finite morphism and $M_ i$ is a finite abelian group.

Let $\Lambda $ be a Noetherian ring. Let $\mathcal{F}$ be a constructible sheaf of $\Lambda $-modules on $X_{\acute{e}tale}$. There exist an injective map of sheaves of modules

\[ \mathcal{F} \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} f_{i, *}\underline{M_ i} \]

where $f_ i : Y_ i \to X$ is a finite morphism and $M_ i$ is a finite $\Lambda $-module.

Moreover, we may assume each $Y_ i$ is irreducible, reduced, maps onto an irreducible and reduced closed subscheme $Z_ i \subset X$ such that $Y_ i \to Z_ i$ is finite étale over a nonempty open of $Z_ i$.

**Proof.**
Proof of (1). Because we have the ascending chain condition for subsheaves of $\mathcal{F}$ (Lemma 56.73.2), it suffices to show that for every point $x \in X$ we can find a map $\varphi : \mathcal{F} \to f_*\underline{E}$ where $f : Y \to X$ is finite and $E$ is a finite set such that $\varphi _{\overline{x}} : \mathcal{F}_{\overline{x}} \to (f_*S)_{\overline{x}}$ is injective. (This argument can be avoided by picking a partition of $X$ as in Lemma 56.70.2 and constructing a $Y_ i \to X$ for each irreducible component of each part.) Let $Z \subset X$ be the induced reduced scheme structure (Schemes, Definition 25.12.5) on $\overline{\{ x\} }$. Since $\mathcal{F}$ is constructible, there is a finite separable extension $\kappa (x) \subset \mathop{\mathrm{Spec}}(K)$ such that $\mathcal{F}|_{\mathop{\mathrm{Spec}}(K)}$ is the constant sheaf with value $E$ for some finite set $E$. Let $Y \to Z$ be the normalization of $Z$ in $\mathop{\mathrm{Spec}}(K)$. By Morphisms, Lemma 28.51.13 we see that $Y$ is a normal integral scheme. As $\kappa (x) \subset K$ is finite, it is clear that $K$ is the function field of $Y$. Denote $g : \mathop{\mathrm{Spec}}(K) \to Y$ the inclusion. The map $\mathcal{F}|_{\mathop{\mathrm{Spec}}(K)} \to \underline{E}$ is adjoint to a map $\mathcal{F}|_ Y \to g_*\underline{E} = \underline{E}$ (Lemma 56.72.12). This in turn is adjoint to a map $\varphi : \mathcal{F} \to f_*\underline{E}$. Observe that the stalk of $\varphi $ at a geometric point $\overline{x}$ is injective: we may take a lift $\overline{y} \in Y$ of $\overline{x}$ and the commutative diagram

\[ \xymatrix{ \mathcal{F}_{\overline{x}} \ar@{=}[r] \ar[d] & (\mathcal{F}|_ Y)_{\overline{y}} \ar@{=}[d] \\ (f_*\underline{E})_{\overline{x}} \ar[r] & \underline{E}_{\overline{y}} } \]

proves the injectivity. We are not yet done, however, as the morphism $f : Y \to Z$ is integral but in general not finite^{1}.

To fix the problem stated in the last sentence of the previous paragraph, we write $Y = \mathop{\mathrm{lim}}\nolimits _{i \in I} Y_ i$ with $Y_ i$ irreducible, integral, and finite over $Z$. Namely, apply Properties, Lemma 27.22.13 to $f_*\mathcal{O}_ Y$ viewed as a sheaf of $\mathcal{O}_ Z$-algebras and apply the functor $\underline{\mathop{\mathrm{Spec}}}_ Z$. Then $f_*\underline{E} = \mathop{\mathrm{colim}}\nolimits f_{i, *}\underline{E}$ by Lemma 56.51.7. By Lemma 56.72.8 the map $\mathcal{F} \to f_*\underline{E}$ factors through $f_{i, *}\underline{E}$ for some $i$. Since $Y_ i \to Z$ is a finite morphism of integral schemes and since the function field extension induced by this morphism is finite separable, we see that the morphism is finite étale over a nonempty open of $Z$ (use Algebra, Lemma 10.138.9; details omitted). This finishes the proof of (1).

The proofs of (2) and (3) are identical to the proof of (1).
$\square$

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