Lemma 59.74.3. Let $X$ be a Noetherian scheme.

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

$\mathcal{F} \longrightarrow \prod \nolimits _{i = 1, \ldots , n} f_{i, *}\underline{E_ i}$

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

2. 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.

3. 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 59.74.2), 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 59.71.2 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 26.12.5) on $\overline{\{ x\} }$. Since $\mathcal{F}$ is constructible, there is a finite separable extension $K/\kappa (x)$ 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 29.53.13 we see that $Y$ is a normal integral scheme. As $K/\kappa (x)$ is a finite extension, 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 59.73.13). 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 finite1.

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 28.22.13 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 59.51.7. By Lemma 59.73.8 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 10.140.9; details omitted). This finishes the proof of (1).

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

[1] If $X$ is a Nagata scheme, for example of finite type over a field, then $Y \to Z$ is finite.

## Comments (3)

Comment #5543 by Harry Gindi on

The corresponding lemma in SGA4 (IX.2.14) asserts (1) for a map F into the product rather than the coproduct. Is this a typo?

Comment #5573 by Harry Gindi on

Also, typo, 'finite separable extension κ(x) c Spec(K)' should be finite separable extension κ(x) c K.

Comment #5730 by on

OMG! Thanks very much. Luckily we never used this elsewhere (of course if we had used it we would've probably noticed how this is nonsense). This is now fixed in this commit.

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