Lemma 51.8.9. 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 universally catenary,

for every $z \in Z$ the formal fibres of $\mathcal{O}_{X, z}$ are $(S_1)$.

In this situation the following are equivalent

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

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

**Proof.**
Let $x \in \text{Ass}(\mathcal{F})$. By Proposition 51.8.7 it suffices to check that $A = \mathcal{O}_{\overline{\{ x\} }, z}$ satisfies the condition of the proposition on associated primes of its completion if and only if $\dim (A) \geq 2$. Observe that $A$ is universally catenary (this is clear) and that its formal fibres are $(S_1)$ as follows from More on Algebra, Lemma 15.50.10 and Proposition 15.50.5. Let $\mathfrak p' \subset A^\wedge $ be an associated prime. As $A \to A^\wedge $ is flat, by Algebra, Lemma 10.64.3, we find that $\mathfrak p'$ lies over $(0) \subset A$. The formal fibre $A^\wedge \otimes _ A F$ is $(S_1)$ where $F$ is the fraction field of $A$. We conclude that $\mathfrak p'$ is a minimal prime, see Algebra, Lemma 10.155.2. Since $A$ is universally catenary it is formally catenary by More on Algebra, Proposition 15.100.5. Hence $\dim (A^\wedge /\mathfrak p') = \dim (A)$ which proves the equivalence.
$\square$

## Comments (0)