52.25 Coherent triples

Let $(A, \mathfrak m)$ be a Noetherian local ring. Let $f \in \mathfrak m$ be a nonzerodivisor. Set $X = \mathop{\mathrm{Spec}}(A)$, $X_0 = \mathop{\mathrm{Spec}}(A/fA)$, $U = X \setminus V(\mathfrak m)$, and $U_0 = U \cap X_0$. We say $(\mathcal{F}, \mathcal{F}_0, \alpha )$ is a coherent triple if we have

1. $\mathcal{F}$ is a coherent $\mathcal{O}_ U$-module such that $f : \mathcal{F} \to \mathcal{F}$ is injective,

2. $\mathcal{F}_0$ is a coherent $\mathcal{O}_{X_0}$-module,

3. $\alpha : \mathcal{F}/f\mathcal{F} \to \mathcal{F}_0|_{U_0}$ is an isomorphism.

There is an obvious notion of a morphism of coherent triples which turns the collection of all coherent triples into a category.

The category of coherent triples is additive but not abelian. However, it is clear what a short exact sequence of coherent triples is.

Given two coherent triples $(\mathcal{F}, \mathcal{F}_0, \alpha )$ and $(\mathcal{G}, \mathcal{G}_0, \beta )$ it may not be the case that $(\mathcal{F} \otimes _{\mathcal{O}_ U} \mathcal{G}, \mathcal{F}_0 \otimes _{\mathcal{O}_{X_0}} \mathcal{G}_0, \alpha \otimes \beta )$ is a coherent triple1. However, if the stalks $\mathcal{G}_ x$ are free for all $x \in U_0$, then this does hold.

We will say the coherent triple $(\mathcal{G}, \mathcal{G}_0, \beta )$ is locally free, resp. invertible if $\mathcal{G}$ and $\mathcal{G}_0$ are locally free, resp. invertible modules. In this case tensoring with $(\mathcal{G}, \mathcal{G}_0, \beta )$ makes sense (see above) and turns short exact sequences of coherent triples into short exact sequences of coherent triples.

Lemma 52.25.1. For any coherent triple $(\mathcal{F}, \mathcal{F}_0, \alpha )$ there exists a coherent $\mathcal{O}_ X$-module $\mathcal{F}'$ such that $f : \mathcal{F}' \to \mathcal{F}'$ is injective, an isomorphism $\alpha ' : \mathcal{F}'|_ U \to \mathcal{F}$, and a map $\alpha '_0 : \mathcal{F}'/f\mathcal{F}' \to \mathcal{F}_0$ such that $\alpha \circ (\alpha ' \bmod f) = \alpha '_0|_{U_0}$.

Proof. Choose a finite $A$-module $M$ such that $\mathcal{F}$ is the restriction to $U$ of the coherent $\mathcal{O}_ X$-module associated to $M$, see Local Cohomology, Lemma 51.8.2. Since $\mathcal{F}$ is $f$-torsion free, we may replace $M$ by its quotient by $f$-power torsion. On the other hand, let $M_0 = \Gamma (X_0, \mathcal{F}_0)$ so that $\mathcal{F}_0$ is the coherent $\mathcal{O}_{X_0}$-module associated to the finite $A/fA$-module $M_0$. By Cohomology of Schemes, Lemma 30.10.5 there exists an $n$ such that the isomorphism $\alpha _0$ corresponds to an $A/fA$-module homomorphism $\mathfrak m^ n M/fM \to M_0$ (whose kernel and cokernel are annihilated by a power of $\mathfrak m$, but we don't need this). Thus if we take $M' = \mathfrak m^ n M$ and we let $\mathcal{F}'$ be the coherent $\mathcal{O}_ X$-module associated to $M'$, then the lemma is clear. $\square$

Let $(\mathcal{F}, \mathcal{F}_0, \alpha )$ be a coherent triple. Choose $\mathcal{F}', \alpha ', \alpha '_0$ as in Lemma 52.25.1. Set

52.25.1.1
$$\label{algebraization-equation-chi-triple} \chi (\mathcal{F}, \mathcal{F}_0, \alpha ) = \text{length}_ A(\mathop{\mathrm{Coker}}(\alpha '_0)) - \text{length}_ A(\mathop{\mathrm{Ker}}(\alpha '_0))$$

The expression on the right makes sense as $\alpha '_0$ is an isomorphism over $U_0$ and hence its kernel and coherent are coherent modules supported on $\{ \mathfrak m\}$ which therefore have finite length (Algebra, Lemma 10.62.3).

Lemma 52.25.2. The quantity $\chi (\mathcal{F}, \mathcal{F}_0, \alpha )$ in (52.25.1.1) does not depend on the choice of $\mathcal{F}', \alpha ', \alpha '_0$ as in Lemma 52.25.1.

Proof. Let $\mathcal{F}', \alpha ', \alpha '_0$ and $\mathcal{F}'', \alpha '', \alpha ''_0$ be two such choices. For $n > 0$ set $\mathcal{F}'_ n = \mathfrak m^ n \mathcal{F}'$. By Cohomology of Schemes, Lemma 30.10.5 for some $n$ there exists an $\mathcal{O}_ X$-module map $\mathcal{F}'_ n \to \mathcal{F}''$ agreeing with the identification $\mathcal{F}''|_ U = \mathcal{F}'|_ U$ determined by $\alpha '$ and $\alpha ''$. Then the diagram

$\xymatrix{ \mathcal{F}'_ n/f\mathcal{F}'_ n \ar[r] \ar[d] & \mathcal{F}'/f\mathcal{F}' \ar[d]^{\alpha _0'} \\ \mathcal{F}''/f\mathcal{F}'' \ar[r]^{\alpha _0''} & \mathcal{F}_0 }$

is commutative after restricting to $U_0$. Hence by Cohomology of Schemes, Lemma 30.10.5 it is commutative after restricting to $\mathfrak m^ l(\mathcal{F}'_ n/f\mathcal{F}'_ n)$ for some $l > 0$. Since $\mathcal{F}'_{n + l}/f\mathcal{F}'_{n + l} \to \mathcal{F}'_ n/f\mathcal{F}'_ n$ factors through $\mathfrak m^ l(\mathcal{F}'_ n/f\mathcal{F}'_ n)$ we see that after replacing $n$ by $n + l$ the diagram is commutative. In other words, we have found a third choice $\mathcal{F}''', \alpha ''', \alpha '''_0$ such that there are maps $\mathcal{F}''' \to \mathcal{F}''$ and $\mathcal{F}''' \to \mathcal{F}'$ over $X$ compatible with the maps over $U$ and $X_0$. This reduces us to the case discussed in the next paragraph.

Assume we have a map $\mathcal{F}'' \to \mathcal{F}'$ over $X$ compatible with $\alpha ', \alpha ''$ over $U$ and with $\alpha '_0, \alpha ''_0$ over $X_0$. Observe that $\mathcal{F}'' \to \mathcal{F}'$ is injective as it is an isomorphism over $U$ and since $f : \mathcal{F}'' \to \mathcal{F}''$ is injective. Clearly $\mathcal{F}'/\mathcal{F}''$ is supported on $\{ \mathfrak m\}$ hence has finite length. We have the maps of coherent $\mathcal{O}_{X_0}$-modules

$\mathcal{F}''/f\mathcal{F}'' \to \mathcal{F}'/f\mathcal{F}' \xrightarrow {\alpha '_0} \mathcal{F}_0$

whose composition is $\alpha ''_0$ and which are isomorphisms over $U_0$. Elementary homological algebra gives a $6$-term exact sequence

$\begin{matrix} 0 \to \mathop{\mathrm{Ker}}(\mathcal{F}''/f\mathcal{F}'' \to \mathcal{F}'/f\mathcal{F}') \to \mathop{\mathrm{Ker}}(\alpha ''_0) \to \mathop{\mathrm{Ker}}(\alpha '_0) \to \\ \mathop{\mathrm{Coker}}(\mathcal{F}''/f\mathcal{F}'' \to \mathcal{F}'/f\mathcal{F}') \to \mathop{\mathrm{Coker}}(\alpha ''_0) \to \mathop{\mathrm{Coker}}(\alpha '_0) \to 0 \end{matrix}$

By additivity of lengths (Algebra, Lemma 10.52.3) we find that it suffices to show that

$\text{length}_ A( \mathop{\mathrm{Coker}}(\mathcal{F}''/f\mathcal{F}'' \to \mathcal{F}'/f\mathcal{F}')) - \text{length}_ A( \mathop{\mathrm{Ker}}(\mathcal{F}''/f\mathcal{F}'' \to \mathcal{F}'/f\mathcal{F}')) = 0$

This follows from applying the snake lemma to the diagram

$\xymatrix{ 0 \ar[r] & \mathcal{F}'' \ar[r]_ f \ar[d] & \mathcal{F}'' \ar[r] \ar[d] & \mathcal{F}''/f\mathcal{F}'' \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \mathcal{F}' \ar[r]^ f & \mathcal{F}' \ar[r] & \mathcal{F}'/f\mathcal{F}' \ar[r] & 0 }$

and the fact that $\mathcal{F}'/\mathcal{F}''$ has finite length. $\square$

Lemma 52.25.3. We have $\chi (\mathcal{G}, \mathcal{G}_0, \beta ) = \chi (\mathcal{F}, \mathcal{F}_0, \alpha ) + \chi (\mathcal{H}, \mathcal{H}_0, \gamma )$ if

$0 \to (\mathcal{F}, \mathcal{F}_0, \alpha ) \to (\mathcal{G}, \mathcal{G}_0, \beta ) \to (\mathcal{H}, \mathcal{H}_0, \gamma ) \to 0$

is a short exact sequence of coherent triples.

Proof. Choose $\mathcal{G}', \beta ', \beta '_0$ as in Lemma 52.25.1 for the triple $(\mathcal{G}, \mathcal{G}_0, \beta )$. Denote $j : U \to X$ the inclusion morphism. Let $\mathcal{F}' \subset \mathcal{G}'$ be the kernel of the composition

$\mathcal{G}' \xrightarrow {\beta '} j_*\mathcal{G} \to j_*\mathcal{H}$

Observe that $\mathcal{H}' = \mathcal{G}'/\mathcal{F}'$ is a coherent subsheaf of $j_*\mathcal{H}$ and hence $f : \mathcal{H}' \to \mathcal{H}'$ is injective. Hence by the snake lemma we obtain a short exact sequence

$0 \to \mathcal{F}'/f\mathcal{F}' \to \mathcal{G}'/f\mathcal{G}' \to \mathcal{H}'/f\mathcal{H}' \to 0$

We have isomorphisms $\alpha ' : \mathcal{F}'|_ U \to \mathcal{F}$, $\beta ' : \mathcal{G}'|_ U \to \mathcal{G}$, and $\gamma ' : \mathcal{H}'|_ U \to \mathcal{H}$ by construction. To finish the proof we'll need to construct maps $\alpha '_0 : \mathcal{F}'/f\mathcal{F}' \to \mathcal{F}_0$ and $\gamma '_0 : \mathcal{H}'/f\mathcal{H}' \to \mathcal{H}_0$ as in Lemma 52.25.1 and fitting into a commutative diagram

$\xymatrix{ 0 \ar[r] & \mathcal{F}'/f\mathcal{F}' \ar[r] \ar@{..>}[d]^{\alpha '_0} & \mathcal{G}'/f\mathcal{G}' \ar[r] \ar[d]^{\beta '_0} & \mathcal{H}'/f\mathcal{H}' \ar[r] \ar@{..>}[d]^{\gamma '_0} & 0 \\ 0 \ar[r] & \mathcal{F}_0 \ar[r] & \mathcal{G}_0 \ar[r] & \mathcal{H}_0 \ar[r] & 0 }$

However, this may not be possible with our initial choice of $\mathcal{G}'$. From the displayed diagram we see the obstruction is exactly the composition

$\delta : \mathcal{F}'/f\mathcal{F}' \to \mathcal{G}'/f\mathcal{G}' \xrightarrow {\beta '_0} \mathcal{G}_0 \to \mathcal{H}_0$

Note that the restriction of $\delta$ to $U_0$ is zero by our choice of $\mathcal{F}'$ and $\mathcal{H}'$. Hence by Cohomology of Schemes, Lemma 30.10.5 there exists an $k > 0$ such that $\delta$ vanishes on $\mathfrak m^ k \cdot (\mathcal{F}'/f\mathcal{F}')$. For $n > k$ set $\mathcal{G}'_ n = \mathfrak m^ n \mathcal{G}'$, $\mathcal{F}'_ n = \mathcal{G}'_ n \cap \mathcal{F}'$, and $\mathcal{H}'_ n = \mathcal{G}'_ n/\mathcal{F}'_ n$. Observe that $\beta '_0$ can be composed with $\mathcal{G}'_ n/f\mathcal{G}'_ n \to \mathcal{G}'/f\mathcal{G}'$ to give a map $\beta '_{n, 0} : \mathcal{G}'_ n/f\mathcal{G}'_ n \to \mathcal{G}_0$ as in Lemma 52.25.1. By Artin-Rees (Algebra, Lemma 10.51.2) we may choose $n$ such that $\mathcal{F}'_ n \subset \mathfrak m^ k \mathcal{F}'$. As above the maps $f : \mathcal{F}'_ n \to \mathcal{F}'_ n$, $f : \mathcal{G}'_ n \to \mathcal{G}'_ n$, and $f : \mathcal{H}'_ n \to \mathcal{H}'_ n$ are injective and as above using the snake lemma we obtain a short exact sequence

$0 \to \mathcal{F}'_ n/f\mathcal{F}'_ n \to \mathcal{G}'_ n/f\mathcal{G}'_ n \to \mathcal{H}'_ n/f\mathcal{H}'_ n \to 0$

As above we have isomorphisms $\alpha '_ n : \mathcal{F}'_ n|_ U \to \mathcal{F}$, $\beta '_ n : \mathcal{G}'_ n|_ U \to \mathcal{G}$, and $\gamma '_ n : \mathcal{H}'_ n|_ U \to \mathcal{H}$. We consider the obstruction

$\delta _ n : \mathcal{F}'_ n/f\mathcal{F}'_ n \to \mathcal{G}'_ n/f\mathcal{G}'_ n \xrightarrow {\beta '_{n, 0}} \mathcal{G}_0 \to \mathcal{H}_0$

as before. However, the commutative diagram

$\xymatrix{ \mathcal{F}'_ n/f\mathcal{F}'_ n \ar[r] \ar[d] & \mathcal{G}'_ n/f\mathcal{G}'_ n \ar[r]_{\beta '_{n, 0}} \ar[d] & \mathcal{G}_0 \ar[r] \ar[d] & \mathcal{H}_0 \ar[d] \\ \mathcal{F}'/f\mathcal{F}' \ar[r] & \mathcal{G}'/f\mathcal{G}' \ar[r]^{\beta '_0} & \mathcal{G}_0 \ar[r] & \mathcal{H}_0 }$

our choice of $n$ and our observation about $\delta$ show that $\delta _ n = 0$. This produces the desired maps $\alpha '_{n, 0} : \mathcal{F}'_ n/f\mathcal{F}'_ n \to \mathcal{F}_0$, and $\gamma '_{n, 0} : \mathcal{H}'_ n/f\mathcal{H}'_ n \to \mathcal{H}_0$. OK, so we may use $\mathcal{F}'_ n, \alpha '_ n, \alpha '_{n, 0}$, $\mathcal{G}'_ n, \beta '_ n, \beta '_{n, 0}$, and $\mathcal{H}'_ n, \gamma '_ n, \gamma '_{n, 0}$ to compute $\chi (\mathcal{F}, \mathcal{F}_0, \alpha )$, $\chi (\mathcal{G}, \mathcal{G}_0, \beta )$, and $\chi (\mathcal{H}, \mathcal{H}_0, \gamma )$. Now finally the lemma follows from an application of the snake lemma to

$\xymatrix{ 0 \ar[r] & \mathcal{F}'_ n/f\mathcal{F}'_ n \ar[r] \ar[d] & \mathcal{G}'_ n/f\mathcal{G}'_ n \ar[r] \ar[d] & \mathcal{H}'_ n/f\mathcal{H}'_ n \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \mathcal{F}_0 \ar[r] & \mathcal{G}_0 \ar[r] & \mathcal{H}_0 \ar[r] & 0 }$

and additivity of lengths (Algebra, Lemma 10.52.3). $\square$

Proposition 52.25.4. Let $(\mathcal{F}, \mathcal{F}_0, \alpha )$ be a coherent triple. Let $(\mathcal{L}, \mathcal{L}_0, \lambda )$ be an invertible coherent triple. Then the function

$\mathbf{Z} \longrightarrow \mathbf{Z},\quad n \longmapsto \chi ((\mathcal{F}, \mathcal{F}_0, \alpha ) \otimes (\mathcal{L}, \mathcal{L}_0, \lambda )^{\otimes n})$

is a polynomial of degree $\leq \dim (\text{Supp}(\mathcal{F}))$.

More precisely, if $\mathcal{F} = 0$, then the function is constant. If $\mathcal{F}$ has finite support in $U$, then the function is constant. If the support of $\mathcal{F}$ in $U$ has dimension $1$, i.e., the closure of the support of $\mathcal{F}$ in $X$ has dimension $2$, then the function is linear, etc.

Proof. We will prove this by induction on the dimension of the support of $\mathcal{F}$.

The base case is when $\mathcal{F} = 0$. Then either $\mathcal{F}_0$ is zero or its support is $\{ \mathfrak m\}$. In this case we have

$(\mathcal{F}, \mathcal{F}_0, \alpha ) \otimes (\mathcal{L}, \mathcal{L}_0, \lambda )^{\otimes n} = (0, \mathcal{F}_0 \otimes \mathcal{L}_0^{\otimes n}, 0) \cong (0, \mathcal{F}_0, 0)$

Thus the function of the lemma is constant with value equal to the length of $\mathcal{F}_0$.

Induction step. Assume the support of $\mathcal{F}$ is nonempty. Let $\mathcal{G}_0 \subset \mathcal{F}_0$ denote the submodule of sections supported on $\{ \mathfrak m\}$. Then we get a short exact sequence

$0 \to (0, \mathcal{G}_0, 0) \to (\mathcal{F}, \mathcal{F}_0, \alpha ) \to (\mathcal{F}, \mathcal{F}_0/\mathcal{G}_0, \alpha ) \to 0$

This sequence remains exact if we tensor by the invertible coherent triple $(\mathcal{L}, \mathcal{L}_0, \lambda )$, see discussion above. Thus by additivity of $\chi$ (Lemma 52.25.3) and the base case explained above, it suffices to prove the induction step for $(\mathcal{F}, \mathcal{F}_0/\mathcal{G}_0, \alpha )$. In this way we see that we may assume $\mathfrak m$ is not an associated point of $\mathcal{F}_0$.

Let $T = \text{Ass}(\mathcal{F}) \cup \text{Ass}(\mathcal{F}/f\mathcal{F})$. Since $U$ is quasi-affine, we can find $s \in \Gamma (U, \mathcal{L})$ which does not vanish at any $u \in T$, see Properties, Lemma 28.29.7. After multiplying $s$ by a suitable element of $\mathfrak m$ we may assume $\lambda (s \bmod f) = s_0|_{U_0}$ for some $s_0 \in \Gamma (X_0, \mathcal{L}_0)$; details omitted. We obtain a morphism

$(s, s_0) : (\mathcal{O}_ U, \mathcal{O}_{X_0}, 1) \longrightarrow (\mathcal{L}, \mathcal{L}_0, \lambda )$

in the category of coherent triples. Let $\mathcal{G} = \mathop{\mathrm{Coker}}(s : \mathcal{F} \to \mathcal{F} \otimes \mathcal{L})$ and $\mathcal{G}_0 = \mathop{\mathrm{Coker}}(s_0 : \mathcal{F}_0 \to \mathcal{F}_0 \otimes \mathcal{L}_0)$. Observe that $s_0 : \mathcal{F}_0 \to \mathcal{F}_0 \otimes \mathcal{L}_0$ is injective as it is injective on $U_0$ by our choice of $s$ and as $\mathfrak m$ isn't an associated point of $\mathcal{F}_0$. It follows that there exists an isomorphism $\beta : \mathcal{G}/f\mathcal{G} \to \mathcal{G}_0|_{U_0}$ such that we obtain a short exact sequence

$0 \to (\mathcal{F}, \mathcal{F}_0, \alpha ) \to (\mathcal{F}, \mathcal{F}_0, \alpha ) \otimes (\mathcal{L}, \mathcal{L}_0, \lambda ) \to (\mathcal{G}, \mathcal{G}_0, \beta ) \to 0$

By induction on the dimension of the support we know the proposition holds for the coherent triple $(\mathcal{G}, \mathcal{G}_0, \beta )$. Using the additivity of Lemma 52.25.3 we see that

$n \longmapsto \chi ((\mathcal{F}, \mathcal{F}_0, \alpha ) \otimes (\mathcal{L}, \mathcal{L}_0, \lambda )^{\otimes n + 1}) - \chi ((\mathcal{F}, \mathcal{F}_0, \alpha ) \otimes (\mathcal{L}, \mathcal{L}_0, \lambda )^{\otimes n})$

is a polynomial. We conclude by a variant of Algebra, Lemma 10.58.5 for functions defined for all integers (details omitted). $\square$

Lemma 52.25.5. Assume $\text{depth}(A) \geq 3$ or equivalently $\text{depth}(A/fA) \geq 2$. Let $(\mathcal{L}, \mathcal{L}_0, \lambda )$ be an invertible coherent triple. Then

$\chi (\mathcal{L}, \mathcal{L}_0, \lambda ) = \text{length}_ A \mathop{\mathrm{Coker}}(\Gamma (U, \mathcal{L}) \to \Gamma (U_0, \mathcal{L}_0))$

and in particular this is $\geq 0$. Moreover, $\chi (\mathcal{L}, \mathcal{L}_0, \lambda ) = 0$ if and only if $\mathcal{L} \cong \mathcal{O}_ U$.

Proof. The equivalence of the depth conditions follows from Algebra, Lemma 10.72.7. By the depth condition we see that $\Gamma (U, \mathcal{O}_ U) = A$ and $\Gamma (U_0, \mathcal{O}_{U_0}) = A/fA$, see Dualizing Complexes, Lemma 47.11.1 and Local Cohomology, Lemma 51.8.2. Using Local Cohomology, Lemma 51.12.2 we find that $M = \Gamma (U, \mathcal{L})$ is a finite $A$-module. This in turn implies $\text{depth}(M) \geq 2$ for example by part (4) of Local Cohomology, Lemma 51.8.2 or by Divisors, Lemma 31.6.6. Also, we have $\mathcal{L}_0 \cong \mathcal{O}_{X_0}$ as $X_0$ is a local scheme. Hence we also see that $M_0 = \Gamma (X_0, \mathcal{L}_0) = \Gamma (U_0, \mathcal{L}_0|_{U_0})$ and that this module is isomorphic to $A/fA$.

By the above $\mathcal{F}' = \widetilde{M}$ is a coherent $\mathcal{O}_ X$-module whose restriction to $U$ is isomorphic to $\mathcal{L}$. The isomorphism $\lambda : \mathcal{L}/f\mathcal{L} \to \mathcal{L}_0|_{U_0}$ determines a map $M/fM \to M_0$ on global sections which is an isomorphism over $U_0$. Since $\text{depth}(M) \geq 2$ we see that $H^0_\mathfrak m(M/fM) = 0$ and it follows that $M/fM \to M_0$ is injective. Thus by definition

$\chi (\mathcal{L}, \mathcal{L}_0, \lambda ) = \text{length}_ A \mathop{\mathrm{Coker}}(M/fM \to M_0)$

which gives the first statement of the lemma.

Finally, if this length is $0$, then $M \to M_0$ is surjective. Hence we can find $s \in M = \Gamma (U, \mathcal{L})$ mapping to a trivializing section of $\mathcal{L}_0$. Consider the finite $A$-modules $K$, $Q$ defined by the exact sequence

$0 \to K \to A \xrightarrow {s} M \to Q \to 0$

The supports of $K$ and $Q$ do not meet $U_0$ because $s$ is nonzero at points of $U_0$. Using Algebra, Lemma 10.72.6 we see that $\text{depth}(K) \geq 2$ (observe that $As \subset M$ has $\text{depth} \geq 1$ as a submodule of $M$). Thus the support of $K$ if nonempty has dimension $\geq 2$ by Algebra, Lemma 10.72.3. This contradicts $\text{Supp}(M) \cap V(f) \subset \{ \mathfrak m\}$ unless $K = 0$. When $K = 0$ we find that $\text{depth}(Q) \geq 2$ and we conclude $Q = 0$ as before. Hence $A \cong M$ and $\mathcal{L}$ is trivial. $\square$

[1] Namely, it isn't necessarily the case that $f$ is injective on $\mathcal{F} \otimes _{\mathcal{O}_ U} \mathcal{G}$.

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