The Stacks project

50.15 Improving coherent modules

Similar constructions can be found in [EGA] and more recently in [Kollar-local-global-hulls] and [Kollar-variants].

Lemma 50.15.1. 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

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

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

  3. $\text{depth}_{\mathcal{O}_{X, x}}(\mathcal{F}'_ x) \geq 1$ 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 30.3.1 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 50.5. 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 $\mathfrak p \in T$. $\square$

Lemma 50.15.2. Let $j : U \to X$ be an open immersion of Noetherian schemes. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. Assume $\mathcal{F}' = j_*(\mathcal{F}|_ U)$ is coherent. Then $\mathcal{F} \to \mathcal{F}'$ is the unique map of coherent $\mathcal{O}_ X$-modules such that

  1. $\mathcal{F}|_ U \to \mathcal{F}'|_ U$ is an isomorphism,

  2. $\text{depth}_{\mathcal{O}_{X, x}}(\mathcal{F}'_ x) \geq 2$ for $x \in X$, $x \not\in U$.

If $f : Y \to X$ is a flat morphism with $Y$ Noetherian, then $f^*\mathcal{F} \to f^*\mathcal{F}'$ is the corresponding map for $f^{-1}(U) \subset Y$.

Proof. We have $\text{depth}_{\mathcal{O}_{X, x}}(\mathcal{F}'_ x) \geq 2$ by Divisors, Lemma 30.6.6 part (3). The uniqueness of $\mathcal{F} \to \mathcal{F}'$ follows from Divisors, Lemma 30.5.11. The compatibility with flat pullbacks follows from flat base change, see Cohomology of Schemes, Lemma 29.5.2. $\square$

Lemma 50.15.3. Let $X$ be a Noetherian scheme. Let $Z \subset X$ be a closed subscheme. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. Assume $X$ is universally catenary and the formal fibres of local rings have $(S_1)$. Then there exists a unique map $\mathcal{F} \to \mathcal{F}''$ of coherent $\mathcal{O}_ X$-modules such that

  1. $\mathcal{F}_ x \to \mathcal{F}''_ x$ is an isomorphism for $x \in X \setminus Z$,

  2. $\mathcal{F}_ x \to \mathcal{F}''_ x$ is surjective and $\text{depth}_{\mathcal{O}_{X, x}}(\mathcal{F}''_ x) = 1$ for $x \in Z$ such that there exists an immediate specialization $x' \leadsto x$ with $x' \not\in Z$ and $x' \in \text{Ass}(\mathcal{F})$,

  3. $\text{depth}_{\mathcal{O}_{X, x}}(\mathcal{F}''_ x) \geq 2$ for the remaining $x \in Z$.

If $f : Y \to X$ is a Cohen-Macaulay morphism with $Y$ Noetherian, then $f^*\mathcal{F} \to f^*\mathcal{F}''$ satisfies the same properties with respect to $f^{-1}(Z) \subset Y$.

Proof. Let $\mathcal{F} \to \mathcal{F}'$ be the map constructed in Lemma 50.15.1 for the subset $Z$ of $X$. Recall that $\mathcal{F}'$ is the quotient of $\mathcal{F}$ by the subsheaf of sections supported on $Z$.

We first prove uniqueness. Let $\mathcal{F} \to \mathcal{F}''$ be as in the lemma. We get a factorization $\mathcal{F} \to \mathcal{F}' \to \mathcal{F}''$ since $\text{Ass}(\mathcal{F}'') \cap Z = \emptyset $ by conditions (2) and (3). Let $U \subset X$ be a maximal open subscheme such that $\mathcal{F}'|_ U \to \mathcal{F}''|_ U$ is an isomorphism. We see that $U$ contains all the points as in (2). Then by Divisors, Lemma 30.5.11 we conclude that $\mathcal{F}'' = j_*(\mathcal{F}'|_ U)$. In this way we get uniqueness (small detail: if we have two of these $\mathcal{F}''$ then we take the intersection of the opens $U$ we get from either).

Proof of existence. Recall that $\text{Ass}(\mathcal{F}') = \{ x_1, \ldots , x_ n\} $ is finite and $x_ i \not\in Z$. Let $Y_ i$ be the closure of $\{ x_ i\} $. Let $Z_{i, j}$ be the irreducible components of $Z \cap Y_ i$. Observe that $\text{Supp}(\mathcal{F}') \cap Z = \bigcup Z_{i, j}$. Let $z_{i, j} \in Z_{i, j}$ be the generic point. Let

\[ d_{i, j} = \dim (\mathcal{O}_{\overline{\{ x_ i\} }, z_{i, j}}) \]

If $d_{i, j} = 1$, then $z_{i, j}$ is one of the points as in (2). Thus we do not need to modify $\mathcal{F}'$ at these points. Furthermore, still assuming $d_{i, j} = 1$, using Lemma 50.9.2 we can find an open neighbourhood $z_{i, j} \in V_{i, j} \subset X$ such that $\text{depth}_{\mathcal{O}_{X, z}}(\mathcal{F}'_ z) \geq 2$ for $z \in Z_{i, j} \cap V_{i, j}$, $z \not= z_{i, j}$. Set

\[ Z' = X \setminus \left( X \setminus Z \cup \bigcup \nolimits _{d_{i, j} = 1} V_{i, j}) \right) \]

Denote $j' : X \setminus Z' \to X$. By our choice of $Z'$ the assumptions of Lemma 50.8.9 are satisfied. We conclude by setting $\mathcal{F}'' = j'_*(\mathcal{F}'|_{X \setminus Z'})$ and applying Lemma 50.15.2.

The final statement follows from the formula for the change in depth along a flat local homomorphism, see Algebra, Lemma 10.157.1 and the assumption on the fibres of $f$ inherent in $f$ being Cohen-Macaulay. Details omitted. $\square$

Lemma 50.15.4. Let $X$ be a Noetherian scheme which locally has a dualizing complex. Let $T' \subset X$ be a subset stable under specialization. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. Assume that if $x \leadsto x'$ is an immediate specialization of points in $X$ with $x' \in T'$ and $x \not\in T'$, then $\text{depth}(\mathcal{F}_ x) \geq 1$. Then there exists a unique map $\mathcal{F} \to \mathcal{F}''$ of coherent $\mathcal{O}_ X$-modules such that

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

  2. $\text{depth}_{\mathcal{O}_{X, x}}(\mathcal{F}''_ x) \geq 2$ for $x \in T'$.

If $f : Y \to X$ is a Cohen-Macaulay morphism with $Y$ Noetherian, then $f^*\mathcal{F} \to f^*\mathcal{F}''$ satisfies the same properties with respect to $f^{-1}(T') \subset Y$.

Proof. Let $\mathcal{F} \to \mathcal{F}'$ be the quotient of $\mathcal{F}$ constructed in Lemma 50.15.1 using $T'$. Recall that $\mathcal{F}'$ is the quotient of $\mathcal{F}$ by the subsheaf of sections supported on $T'$.

Proof of uniqueness. Let $\mathcal{F} \to \mathcal{F}''$ be as in the lemma. We get a factorization $\mathcal{F} \to \mathcal{F}' \to \mathcal{F}''$ since $\text{Ass}(\mathcal{F}'') \cap T' = \emptyset $ by condition (2). Let $U \subset X$ be a maximal open subscheme such that $\mathcal{F}'|_ U \to \mathcal{F}''|_ U$ is an isomorphism. We see that $U$ contains all the points of $T'$. Then by Divisors, Lemma 30.5.11 we conclude that $\mathcal{F}'' = j_*(\mathcal{F}'|_ U)$. In this way we get uniqueness (small detail: if we have two of these $\mathcal{F}''$ then we take the intersection of the opens $U$ we get from either).

Proof of existence. We will define

\[ \mathcal{F}'' = \mathop{\mathrm{colim}}\nolimits j_*(\mathcal{F}'|_ V) \]

where $j : V \to X$ runs over the open subschemes such that $X \setminus V \subset T'$. Observe that the colimit is filtered as $T'$ is stable under specialization. Each of the maps $\mathcal{F}' \to j_*(\mathcal{F}'|_ V)$ is injective as $\text{Ass}(\mathcal{F}')$ is disjoint from $T'$. Thus $\mathcal{F}' \to \mathcal{F}''$ is injective.

Suppose $X = \mathop{\mathrm{Spec}}(A)$ is affine and $\mathcal{F}$ corresponds to the finite $A$-module $M$. Then $\mathcal{F}'$ corresponds to $M' = M / H^0_{T'}(M)$, see proof of Lemma 50.15.1. Applying Lemmas 50.2.2 and 50.5.3 we see that $\mathcal{F}''$ corresponds to an $A$-module $M''$ which fits into the short exact sequence

\[ 0 \to M' \to M'' \to H^1_{T'}(M') \to 0 \]

By Proposition 50.11.1 and our condition on immediate specializations in the statement of the lemma we see that $M''$ is a finite $A$-module. In this way we see that $\mathcal{F}''$ is coherent.

The final statement follows from the formula for the change in depth along a flat local homomorphism, see Algebra, Lemma 10.157.1 and the assumption on the fibres of $f$ inherent in $f$ being Cohen-Macaulay. Details omitted. $\square$

Lemma 50.15.5. Let $X$ be a Noetherian scheme which locally has a dualizing complex. Let $T' \subset T \subset X$ be subsets stable under specialization such that if $x \leadsto x'$ is an immediate specialization of points in $X$ and $x' \in T'$, then $x \in T$. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. Then there exists a unique map $\mathcal{F} \to \mathcal{F}''$ of coherent $\mathcal{O}_ X$-modules such that

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

  2. $\mathcal{F}_ x \to \mathcal{F}''_ x$ is surjective and $\text{depth}_{\mathcal{O}_{X, x}}(\mathcal{F}''_ x) \geq 1$ for $x \in T$, $x \not\in T'$, and

  3. $\text{depth}_{\mathcal{O}_{X, x}}(\mathcal{F}''_ x) \geq 2$ for $x \in T'$.

If $f : Y \to X$ is a Cohen-Macaulay morphism with $Y$ Noetherian, then $f^*\mathcal{F} \to f^*\mathcal{F}''$ satisfies the same properties with respect to $f^{-1}(T') \subset f^{-1}(T) \subset Y$.

Proof. First, let $\mathcal{F} \to \mathcal{F}'$ be the quotient of $\mathcal{F}$ constructed in Lemma 50.15.1 using $T$. Second, let $\mathcal{F}' \to \mathcal{F}''$ be the unique map of coherent modules construction in Lemma 50.15.4 using $T'$. Then $\mathcal{F} \to \mathcal{F}''$ is as desired. $\square$


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