The Stacks project

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. $\square$

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

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

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

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

This follows from applying the snake lemma to the diagram

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

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

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

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

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

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

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

as before. However, the commutative diagram

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

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

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

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

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

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

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

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

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

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

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. $\square$