Lemma 51.8.3. Let $X$ be a locally Noetherian scheme. Let $j : U \to X$ be the inclusion of an open subscheme with complement $Z$. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ U$-module. Assume

$X$ is Nagata,

$X$ is universally catenary, and

for $x \in \text{Ass}(\mathcal{F})$ and $z \in Z \cap \overline{\{ x\} }$ we have $\dim (\mathcal{O}_{\overline{\{ x\} }, z}) \geq 2$.

Then $j_*\mathcal{F}$ is coherent.

**Proof.**
By Lemma 51.8.1 it suffices to prove $j_*i_{x, *}\mathcal{O}_{W_ x}$ is coherent for $x \in \text{Ass}(\mathcal{F})$. Let $\pi : Y \to X$ be the normalization of $X$ in $\mathop{\mathrm{Spec}}(\kappa (x))$, see Morphisms, Section 29.54. By Morphisms, Lemma 29.53.14 the morphism $\pi $ is finite. Since $\pi $ is finite $\mathcal{G} = \pi _*\mathcal{O}_ Y$ is a coherent $\mathcal{O}_ X$-module by Cohomology of Schemes, Lemma 30.9.9. Observe that $W_ x = U \cap \pi (Y)$. Thus $\pi |_{\pi ^{-1}(U)} : \pi ^{-1}(U) \to U$ factors through $i_ x : W_ x \to U$ and we obtain a canonical map

\[ i_{x, *}\mathcal{O}_{W_ x} \longrightarrow (\pi |_{\pi ^{-1}(U)})_*(\mathcal{O}_{\pi ^{-1}(U)}) = (\pi _*\mathcal{O}_ Y)|_ U = \mathcal{G}|_ U \]

This map is injective (for example by Divisors, Lemma 31.2.10). Hence $j_*i_{x, *}\mathcal{O}_{W_ x} \subset j_*\mathcal{G}|_ U$ and it suffices to show that $j_*\mathcal{G}|_ U$ is coherent.

It remains to prove that $j_*(\mathcal{G}|_ U)$ is coherent. We claim Divisors, Lemma 31.5.11 applies to

\[ \mathcal{G} \longrightarrow j_*(\mathcal{G}|_ U) \]

which finishes the proof. It suffices to show that $\text{depth}(\mathcal{G}_ z) \geq 2$ for $z \in Z$. Let $y_1, \ldots , y_ n \in Y$ be the points mapping to $z$. By Algebra, Lemma 10.72.11 it suffices to show that $\text{depth}(\mathcal{O}_{Y, y_ i}) \geq 2$ for $i = 1, \ldots , n$. If not, then by Properties, Lemma 28.12.5 we see that $\dim (\mathcal{O}_{Y, y_ i}) = 1$ for some $i$. This is impossible by the dimension formula (Morphisms, Lemma 29.52.1) for $\pi : Y \to \overline{\{ x\} }$ and assumption (3).
$\square$

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