The Stacks project

Lemma 115.11.2. Let $X$ be a Noetherian scheme. Let $Z_0 \subset X$ be an irreducible closed subset with generic point $\xi $. Let $\mathcal{P}$ be a property of coherent sheaves on $X$ such that

  1. For any short exact sequence of coherent sheaves if two out of three of them have property $\mathcal{P}$ then so does the third.

  2. If $\mathcal{P}$ holds for a direct sum of coherent sheaves then it holds for both.

  3. For every integral closed subscheme $Z \subset Z_0 \subset X$, $Z \not= Z_0$ and every quasi-coherent sheaf of ideals $\mathcal{I} \subset \mathcal{O}_ Z$ we have $\mathcal{P}$ for $(Z \to X)_*\mathcal{I}$.

  4. There exists some coherent sheaf $\mathcal{G}$ on $X$ such that

    1. $\text{Supp}(\mathcal{G}) = Z_0$,

    2. $\mathcal{G}_\xi $ is annihilated by $\mathfrak m_\xi $, and

    3. property $\mathcal{P}$ holds for $\mathcal{G}$.

Then property $\mathcal{P}$ holds for every coherent sheaf $\mathcal{F}$ on $X$ whose support is contained in $Z_0$.

Proof. The proof is a variant on the proof of Cohomology of Schemes, Lemma 30.12.5. In exactly the same manner as in that proof we see that any coherent sheaf whose support is strictly contained in $Z_0$ has property $\mathcal{P}$.

Consider a coherent sheaf $\mathcal{G}$ as in (3). By Cohomology of Schemes, Lemma 30.12.2 there exists a sheaf of ideals $\mathcal{I}$ on $Z_0$ and a short exact sequence

\[ 0 \to \left((Z_0 \to X)_*\mathcal{I}\right)^{\oplus r} \to \mathcal{G} \to \mathcal{Q} \to 0 \]

where the support of $\mathcal{Q}$ is strictly contained in $Z_0$. In particular $r > 0$ and $\mathcal{I}$ is nonzero because the support of $\mathcal{G}$ is equal to $Z$. Since $\mathcal{Q}$ has property $\mathcal{P}$ we conclude that also $\left((Z_0 \to X)_*\mathcal{I}\right)^{\oplus r}$ has property $\mathcal{P}$. By (2) we deduce property $\mathcal{P}$ for $(Z_0 \to X)_*\mathcal{I}$. Slotting this into the proof of Cohomology of Schemes, Lemma 30.12.5 at the appropriate point gives the lemma. Some details omitted. $\square$

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