The Stacks project

Proof. The sequences in the diagram are exact in the middle and the first arrow is injective. Thus the final statement follows from an easy diagram chase. For the rest of the proof we assume the left and right downward arrows are isomorphisms. A diagram chase shows that the middle downward arrow is injective. All that remains is to show that it is surjective.

We may choose finite $A$-modules $M$ and $M'$ such that $\mathcal{F}$ and $\mathcal{F}'$ are the restriction of $\widetilde{M}$ and $\widetilde{M}'$ to $U$, see Local Cohomology, Lemma 51.8.2. After replacing $M'$ by $\mathfrak a^ n M'$ for some $n \geq 0$ we may assume that $\mathcal{F}' \to \mathcal{F}$ corresponds to a module map $M' \to M$, see Cohomology of Schemes, Lemma 30.10.5. After replacing $M'$ by the image of $M' \to M$ and setting $M'' = M/M'$ we see that our short exact sequence corresponds to the restriction of the short exact sequence of coherent modules associated to the short exact sequence $0 \to M' \to M \to M'' \to 0$ of $A$-modules.

Let $\hat s \in \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F})$ with image $\hat s'' \in \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}''/I^ n\mathcal{F}'')$. By assumption we find $V \in \mathcal{V}$ and a section $s'' \in \mathcal{F}''(V)$ mapping to $\hat s''$. Let $J \subset A$ be an ideal such that $V(J) = \mathop{\mathrm{Spec}}(A) \setminus V$. By Cohomology of Schemes, Lemma 30.10.5 after replacing $J$ by a power, we may assume there is an $A$-linear map $\varphi : J \to M''$ corresponding to $s''$. We fix this choice of $J$; in the rest of the proof we will replace $V$ by a smaller $V$ in $\mathcal{V}$, i.e, we will have $V \cap V(J) = \emptyset $.

Choose a presentation $A^{\oplus m} \to A^{\oplus n} \to J \to 0$. Denote $g_1, \ldots , g_ n \in J$ the images of the basis vectors of $A^{\oplus n}$, so that $J = (g_1, \ldots , g_ n)$. Let $A^{\oplus m} \to A^{\oplus n}$ be given by the matrix $(a_{ji})$ so that $\sum a_{ji} g_ i = 0$, $j = 1, \ldots , m$. Since $M \to M''$ is surjective, for each $i$ we can choose $m_ i \in M$ mapping to $\varphi (g_ i) \in M''$. Then the element $g_ i \hat s - m_ i$ of $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F})$ lies in the submodule $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}'/I^ n\mathcal{F}')$. By assumption after shrinking $V$ we may assume there are $s'_ i \in \mathcal{F}'(V)$, $i = 1, \ldots , n$ with $s'_ i$ mapping to $g_ i \hat s - m_ i$. Set $s_ i = s'_ i + m_ i$ in $\mathcal{F}(V)$. Note that $\sum a_{ji} s_ i$ maps to $\sum a_{ji}g_ i\hat s = 0$ by the map

Since this map is injective (see above), we may after shrinking $V$ assume that $\sum a_{ji}s_ i = 0$ in $\mathcal{F}(V)$ for all $j = 1, \ldots , m$. Then it follows that we obtain an $A$-module map $J \to \mathcal{F}(V)$ sending $g_ i$ to $s_ i$. By the universal property of $\widetilde{J}$ this $A$-module map corresponds to an $\mathcal{O}_ V$-module map $\widetilde{J}|_ V \to \mathcal{F}$. However, since $V(J) \cap V = \emptyset $ we have $\widetilde{J}|_ V = \mathcal{O}_ V$. Thus we have produced a section $s \in \mathcal{F}(V)$. We omit the computation that shows that $s$ maps to $\hat s$ by the map displayed above. $\square$

Proof. Choose a finite $A$-module $M$ such that $\mathcal{F}$ is the restriction to $U$ of the coherent module associated to $M$, see Local Cohomology, Lemma 51.8.2. Set $d = \text{cd}(A, I)$. Let $\mathfrak p$ be a prime of $A$ not contained in $V(I)$ and let $\mathfrak q \in V(\mathfrak p) \cap V(\mathfrak a)$. Then either $\mathfrak p$ is not an associated prime of $M$ and hence $\text{depth}(M_\mathfrak p) \geq 1$ or we have $\dim ((A/\mathfrak p)_\mathfrak q) > d + 1$ by (2). Thus the hypotheses of Lemma 52.8.5 are satisfied for $s = 1$ and $d$; here we use condition (3). Thus we find there exists an ideal $J_0 \subset \mathfrak a$ with $V(J_0) \cap V(I) = V(\mathfrak a)$ such that for any $J \subset J_0$ with $V(J) \cap V(I) = V(\mathfrak a)$ the maps

are isomorphisms for $i = 0, 1$. Consider the morphisms of exact triangles

where $V = \mathop{\mathrm{Spec}}(A) \setminus V(J)$. Recall that $R\Gamma _\mathfrak a(M)^\wedge \to R\Gamma _ J(M)^\wedge $ is an isomorphism (because $\mathfrak a$, $\mathfrak a + I$, and $J + I$ cut out the same closed subscheme, for example see proof of Lemma 52.8.5). Hence $R\Gamma (U, \mathcal{F})^\wedge = R\Gamma (V, \mathcal{F})^\wedge $. This produces a commutative diagram

with exact rows and isomorphisms for the lower vertical arrows. Hence we obtain an isomorphism $\Gamma (V, \mathcal{F}) \to H^0(R\Gamma (U, \mathcal{F})^\wedge )$. By Lemmas 52.6.20 and 52.7.2 we have

and we find $H^0(R\Gamma (U, \mathcal{F})^\wedge ) = \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F})$ by Cohomology, Lemma 20.37.1. $\square$

Proof. Let $T \subset U$ be the set of points $x$ with $\overline{\{ x\} } \cap V(I) \subset V(\mathfrak a)$. Let $\mathcal{F} \to \mathcal{F}'$ be the surjection of coherent modules on $U$ constructed in Local Cohomology, Lemma 51.15.1. Since $\mathcal{F} \to \mathcal{F}'$ is an isomorphism over an open $V \subset U$ containing $U \cap V(I)$ it suffices to prove the lemma with $\mathcal{F}$ replaced by $\mathcal{F}'$. Hence we may and do assume for $x \in U$ with $\overline{\{ x\} } \cap V(I) \subset V(\mathfrak a)$ we have $\text{depth}(\mathcal{F}_ x) \geq 1$.

Let $\mathcal{V}$ be the set of open subschemes $V \subset U$ containing $U \cap V(I)$ ordered by reverse inclusion. This is a directed set. We first claim that

is injective for any $V \in \mathcal{F}$ (and in particular the map of the lemma is injective). Namely, an associated point $x$ of $\mathcal{F}$ must have $\overline{\{ x\} } \cap U \cap Y \not= \emptyset $ by the previous paragraph. If $y \in \overline{\{ x\} } \cap U \cap Y$ then $\mathcal{F}_ x$ is a localization of $\mathcal{F}_ y$ and $\mathcal{F}_ y \subset \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ y/I^ n \mathcal{F}_ y$ by Krull's intersection theorem (Algebra, Lemma 10.51.4). This proves the claim as a section $s \in \mathcal{F}(V)$ in the kernel would have to have empty support, hence would have to be zero.

Choose a finite $A$-module $M$ such that $\mathcal{F}$ is the restriction of $\widetilde{M}$ to $U$, see Local Cohomology, Lemma 51.8.2. We may and do assume that $H^0_\mathfrak a(M) = 0$. Let $\text{Ass}(M) \setminus V(I) = \{ \mathfrak p_1, \ldots , \mathfrak p_ n\} $. We will prove the lemma by induction on $n$. After reordering we may assume that $\mathfrak p_ n$ is a minimal element of the set $\{ \mathfrak p_1, \ldots , \mathfrak p_ n\} $ with respect to inclusion, i.e, $\mathfrak p_ n$ is a generic point of the support of $M$. Set

and $M'' = M/M'$. Let $\mathcal{F}'$ and $\mathcal{F}''$ be the coherent $\mathcal{O}_ U$-modules corresponding to $M'$ and $M''$. Dualizing Complexes, Lemma 47.11.6 implies that $M''$ has only one associated prime, namely $\mathfrak p_ n$. Hence $\mathcal{F}''$ has only one associated point and we see that condition (3)(a) of Lemma 52.12.2 holds; thus the map $\mathop{\mathrm{colim}}\nolimits H^0(V, \mathcal{F}'') \to \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}''/I^ n\mathcal{F}'')$ is an isomorphism. On the other hand, since $\mathfrak p_ n \not\in V(\mathfrak p_1 \ldots \mathfrak p_{n - 1} I)$ we see that $\mathfrak p_ n$ is not an associated prime of $M'$. Hence the induction hypothesis applies to $M'$; note that since $\mathcal{F}' \subset \mathcal{F}$ the condition $\text{depth}(\mathcal{F}'_ x) \geq 1$ at points $x$ with $\overline{\{ x\} } \cap V(I) \subset V(\mathfrak a)$ holds, see Algebra, Lemma 10.72.6. Thus the map $\mathop{\mathrm{colim}}\nolimits H^0(V, \mathcal{F}') \to \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}'/I^ n\mathcal{F}')$ is an isomorphism too. We conclude by Lemma 52.12.1. $\square$

Proof. Let $\hat s \in \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F})$. By Proposition 52.12.3 we find that $\hat s$ is the image of an element $s \in \mathcal{F}(V)$ for some $V \subset U$ open containing $U \cap V(I)$. However, condition (3) shows that $\text{depth}(\mathcal{F}_ x) \geq 2$ for all $x \in U \setminus V$ and hence we find that $\mathcal{F}(V) = \mathcal{F}(U)$ by Divisors, Lemma 31.5.11 and the proof is complete. $\square$

Proof. Set $\mathcal{F} = \widetilde{M}|_ U$. The finiteness of $H^1_\mathfrak a(M/fM)$ implies that $H^0(U, \mathcal{F}/f\mathcal{F})$ is finite, see Local Cohomology, Lemma 51.8.2. By Cohomology, Lemma 20.36.3 (which applies as $f$ is a nonzerodivisor on $\mathcal{F}$) we see that $N = \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/f^ n\mathcal{F})$ is a finite $A$-module, is $f$-torsion free, and $N/fN \subset H^0(U, \mathcal{F}/f\mathcal{F})$. On the other hand, we have a map $M \to N$ and a compatible map

For $g \in \mathfrak a$ we see that $(M/fM)_ g$ maps isomorphically to $H^0(U \cap D(f), \mathcal{F}/f\mathcal{F})$ since $\mathcal{F}/f\mathcal{F}$ is the restriction of $\widetilde{M/fM}$ to $U$. We conclude that $M_ g \to N_ g$ induces an isomorphism

Since $f$ is a nonzerodivisor on both $N$ and $M$ we conclude that $M_ g \to N_ g$ induces an isomorphism on $f$-adic completions which in turn implies $M_ g \to N_ g$ is an isomorphism in an open neighbourhood of $V(f) \cap D(g)$. Since $g \in \mathfrak a$ was arbitrary, we conclude that $M$ and $N$ determine isomorphic coherent modules over an open $V$ as in the statement of the lemma. This finishes the proof. $\square$

First proof. Recall that $A$ is universally catenary and with Gorenstein formal fibres, see Dualizing Complexes, Lemmas 47.23.2 and 47.17.4. Thus we may consider the map $\mathcal{F} \to \mathcal{F}'$ constructed in Local Cohomology, Lemma 51.15.3 for the closed subset $V(f) \cap U$ of $U$. Observe that

1. The kernel and cokernel of $\mathcal{F} \to \mathcal{F}'$ are supported on $V(f) \cap U$.

2. The module $\mathcal{F}'$ is $f$-torsion free as its stalks have depth $\geq 1$ for all points of $V(f) \cap U$, i.e., $\mathcal{F}'$ has no associated points in $V(f) \cap U$.

3. If $y \in V(f) \cap U$ is an associated point of $\mathcal{F}'/f\mathcal{F}'$, then $\text{depth}(\mathcal{F}'_ y) = 1$ and hence (by the construction of $\mathcal{F}'$) there is an immediate specialization $x \leadsto y$ with $x \not\in V(f)$ an associated point of $\mathcal{F}$. It follows that $y$ cannot have an immediate specialization in $\mathop{\mathrm{Spec}}(A)$ to a point $z \in V(\mathfrak a)$ by our assumption (2).

4. It follows from (3) that $H^0(U, \mathcal{F}'/f\mathcal{F}')$ is a finite $A$-module, see Local Cohomology, Lemma 51.12.1.

These observations will allow us to finish the proof.

First, we claim the lemma holds for $\mathcal{F}'$. Namely, choose a finite $A$-module $M'$ such that $\mathcal{F}'$ is the restriction to $U$ of the coherent module associated to $M'$, see Local Cohomology, Lemma 51.8.2. Since $\mathcal{F}'$ is $f$-torsion free, we may assume $M'$ is $f$-torsion free as well. Observation (4) above shows that $H^1_\mathfrak a(M')$ is a finite $A$-module, see Local Cohomology, Lemma 51.8.2. Thus the claim by Lemma 52.12.5.

Second, we observe that the lemma holds trivially for any coherent $\mathcal{O}_ U$-module supported on $V(f) \cap U$. Let $\mathcal{K}$, resp. $\mathcal{G}$, resp. $\mathcal{Q}$ be the kernel, resp. image, resp. cokernel of the map $\mathcal{F} \to \mathcal{F}'$. The short exact sequence $0 \to \mathcal{G} \to \mathcal{F}' \to \mathcal{Q} \to 0$ and Lemma 52.12.1 show that the result holds for $\mathcal{G}$. Then we do this again with the short exact sequence $0 \to \mathcal{K} \to \mathcal{F} \to \mathcal{G} \to 0$ to finish the proof. $\square$

Second proof. The proposition is a special case of Proposition 52.12.3. $\square$

Proof. We may apply Lemma 52.3.2 to $U$ and $\mathcal{F} = \widetilde{M}|_ U$ because $\mathcal{F}$ is a Noetherian object in the category of coherent $\mathcal{O}_ U$-modules. Since $H^1(U, \mathcal{F}) = H^2_\mathfrak a(M)$ (Local Cohomology, Lemma 51.8.2) is annihilated by a power of $f$, we see that its $f$-adic Tate module is zero. Hence the lemma shows $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/f^ n \mathcal{F})$ is equal to the usual $f$-adic completion of $H^0(U, \mathcal{F})$. Consider the short exact sequence

of Local Cohomology, Lemma 51.8.2. Since $M/H^0_\mathfrak a(M)$ is a finite $A$-module, it is complete, see Algebra, Lemma 10.97.1. Since $H^1_\mathfrak a(M)$ is killed by a power of $f$, we conclude from Algebra, Lemma 10.96.4 that $H^0(U, \mathcal{F})$ is complete as well. This finishes the proof. $\square$