Lemma 51.15.1. Let $X$ be a Noetherian scheme. Let $T \subset X$ be a subset stable under specialization. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. Then there is a unique map $\mathcal{F} \to \mathcal{F}'$ of coherent $\mathcal{O}_ X$-modules such that

$\mathcal{F} \to \mathcal{F}'$ is surjective,

$\mathcal{F}_ x \to \mathcal{F}'_ x$ is an isomorphism for $x \not\in T$,

$\text{depth}_{\mathcal{O}_{X, x}}(\mathcal{F}'_ x) \geq 1$ for $x \in T$.

If $f : Y \to X$ is a flat morphism with $Y$ Noetherian, then $f^*\mathcal{F} \to f^*\mathcal{F}'$ is the corresponding quotient for $f^{-1}(T) \subset Y$ and $f^*\mathcal{F}$.

**Proof.**
Condition (3) just means that $\text{Ass}(\mathcal{F}') \cap T = \emptyset $. Thus $\mathcal{F} \to \mathcal{F}'$ is the quotient of $\mathcal{F}$ by the subsheaf of sections whose support is contained in $T$. This proves uniqueness. The statement on pullbacks follows from Divisors, Lemma 31.3.1 and the uniqueness.

Existence of $\mathcal{F} \to \mathcal{F}'$. By the uniqueness it suffices to prove the existence and uniqueness locally on $X$; small detail omitted. Thus we may assume $X = \mathop{\mathrm{Spec}}(A)$ is affine and $\mathcal{F}$ is the coherent module associated to the finite $A$-module $M$. Set $M' = M / H^0_ T(M)$ with $H^0_ T(M)$ as in Section 51.5. Then $M_\mathfrak p = M'_\mathfrak p$ for $\mathfrak p \not\in T$ which proves (1). On the other hand, we have $H^0_ T(M) = \mathop{\mathrm{colim}}\nolimits H^0_ Z(M)$ where $Z$ runs over the closed subsets of $X$ contained in $T$. Thus by Dualizing Complexes, Lemmas 47.11.6 we have $H^0_ T(M') = 0$, i.e., no associated prime of $M'$ is in $T$. Therefore $\text{depth}(M'_\mathfrak p) \geq 1$ for $\mathfrak p \in T$.
$\square$

## Comments (0)