The Stacks project

Proof. We will prove this by verifying hypotheses (a), (b), and (c) of Lemma 52.16.10.

Since $\mathcal{F}_ n$ is locally free over $\mathcal{O}_ U/f^ n\mathcal{O}_ U$ we see that we have short exact sequences $0 \to \mathcal{F}_ n \to \mathcal{F}_{n + 1} \to \mathcal{F}_1 \to 0$ for all $n$. Thus condition (b) holds by Cohomology, Lemma 20.36.2.

By induction on $n$ and the short exact sequences $0 \to A/f^ n \to A/f^{n + 1} \to A/f \to 0$ we see that the associated primes of $A/f^ nA$ agree with the associated primes of $A/fA$. Since the associated points of $\mathcal{F}_ n$ correspond to the associated primes of $A/f^ nA$ not equal to $\mathfrak m$ by assumption (3), we conclude that $M_ n = H^0(U, \mathcal{F}_ n)$ is a finite $A$-module by (5) and Local Cohomology, Proposition 51.8.7. Thus hypothesis (c) holds.

To finish the proof it suffices to show that there exists an $n > 1$ such that the image of

has finite length as an $A$-module. Namely, this will imply hypothesis (a) by Cohomology, Lemma 20.36.5. The image is independent of $n$ for $n$ large enough by Lemma 52.5.2. Let $\omega _ A^\bullet $ be a normalized dualizing complex for $A$. By the local duality theorem and Matlis duality (Dualizing Complexes, Lemma 47.18.4 and Proposition 47.7.8) our claim is equivalent to: the image of

has finite length for $n \gg 1$. The modules in question are finite $A$-modules supported at $V(f)$. Thus it suffices to show that this map is zero after localization at a prime $\mathfrak q$ containing $f$ and different from $\mathfrak m$. Let $\omega _{A_\mathfrak q}^\bullet $ be a normalized dualizing complex on $A_\mathfrak q$ and recall that $\omega _{A_\mathfrak q}^\bullet = (\omega _ A^\bullet )_\mathfrak q[\dim (A/\mathfrak q)]$ by Dualizing Complexes, Lemma 47.17.3. Using the local structure of $\mathcal{F}_ n$ given in (4) we find that it suffices to show the vanishing of

for $n$ large enough. If $\dim (A/\mathfrak q) > 3$, then this is immediate from Local Cohomology, Lemma 51.9.4. For the other cases we will use the long exact sequence

If $\dim (A/\mathfrak q) = 2$, then $H^0(\omega _{A_\mathfrak q}^\bullet ) = 0$ because $\text{depth}(A_\mathfrak q) \geq 1$ as $f$ is a nonzerodivisor. Thus the long exact sequence shows the condition is that

is zero. Now $H^{-1}(\omega ^\bullet _\mathfrak q)$ is a finite module supported in the primes $\mathfrak p \subset A_\mathfrak q$ such that $\text{depth}(A_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) \leq 1$. Since $\dim ((A/\mathfrak p)_\mathfrak q) = \dim (A/\mathfrak p) - 2$ condition (6) tells us these primes are contained in $V(f)$. Thus the desired vanishing for $n$ large enough. Finally, if $\dim (A/\mathfrak q) = 1$, then condition (5) combined with the fact that $f$ is a nonzerodivisor insures that $A_\mathfrak q$ has depth at least $2$. Hence $H^0(\omega _{A_\mathfrak q}^\bullet ) = H^{-1}(\omega _{A_\mathfrak q}^\bullet ) = 0$ and the long exact sequence shows the claim is equivalent to the vanishing of

Now $H^{-2}(\omega ^\bullet _\mathfrak q)$ is a finite module supported in the primes $\mathfrak p \subset A_\mathfrak q$ such that $\text{depth}(A_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) \leq 2$. By condition (6) all of these primes are contained in $V(f)$. Thus the desired vanishing for $n$ large enough. $\square$

Proof. Let $\omega _ A^\bullet $ be a normalized dualizing complex. Then $\delta ^ Y_ Z = \delta $ is the dimension function associated with this dualizing complex. Observe that $\mathop{\mathrm{Ext}}\nolimits ^{-i}_ A(M_ n, \omega _ A^\bullet )$ is a finite $A$-module annihilated by $I^ n$. Fix $0 \leq i \leq s$. Below we will find $n_1 > n_0 > 0$ such that if we set

then the kernels of the maps

are finite length $A$-modules and the cokernels $Q_ n$ form a system such that $\mathop{\mathrm{Im}}(Q_{n'} \to Q_{n''})$ has finite length for $n'' \gg n' \gg n_1$. This is equivalent to the statement that the system $\{ \mathop{\mathrm{Ext}}\nolimits ^{-i}_ A(M_ n, \omega _ A^\bullet )\} _{n \geq 1}$ is essentially constant in the quotient of the category of finite $A$-modules modulo the Serre subcategory of finite length $A$-modules. By the local duality theorem (Dualizing Complexes, Lemma 47.18.4) and Matlis duality (Dualizing Complexes, Proposition 47.7.8) we conclude that there are maps

as in the statement of the lemma.

Pick $f \in \mathfrak m$. Let $B = A_ f^\wedge $ be the $I$-adic completion of the localization $A_ f$. Recall that $\omega _{A_ f}^\bullet = \omega _ A^\bullet \otimes _ A A_ f$ and $\omega _ B^\bullet = \omega _ A^\bullet \otimes _ A B$ are dualizing complexes (Dualizing Complexes, Lemma 47.15.6 and 47.22.3). Let $M$ be the finite $B$-module $\mathop{\mathrm{lim}}\nolimits M_{n, f}$ (compare with discussion in Cohomology of Schemes, Lemma 30.23.1). Then

Since $\mathfrak m$ can be generated by finitely many $f \in \mathfrak m$ it suffices to show that for each $f$ the system

is essentially constant. Some details omitted.

Let $\mathfrak q \subset IB$ be a prime ideal. Then $\mathfrak q$ corresponds to a point $y \in U \cap Y$. Observe that $\delta (\mathfrak q) = \dim (\overline{\{ y\} })$ is also the value of the dimension function associated to $\omega _ B^\bullet $ (we omit the details; use that $\omega _ B^\bullet $ is gotten from $\omega _ A^\bullet $ by tensoring up with $B$). Assumption (3) guarantees via Lemma 52.19.2 that Lemma 52.10.4 applies to $B_\mathfrak q, IB_\mathfrak q, \mathfrak qB_\mathfrak q, M_\mathfrak q$ with $s$ replaced by $s - \delta (y)$. We obtain that

and this module is annihilated by a power of $I$. By Lemma 52.5.3 we find that the inverse systems $H^{i - \delta (\mathfrak q)}_{\mathfrak qB_\mathfrak q}((M/I^ nM)_\mathfrak q)$ are essentially constant with value $H^{i - \delta (\mathfrak q)}_{\mathfrak qB_\mathfrak q}(M_\mathfrak q)$. Since $(\omega _ B^\bullet )_\mathfrak q[-\delta (\mathfrak q)]$ is a normalized dualizing complex on $B_\mathfrak q$ the local duality theorem shows that the system

is essentially constant with value $\mathop{\mathrm{Ext}}\nolimits ^{-i}_ B(M, \omega _ B^\bullet )_\mathfrak q$.

To finish the proof we globalize as in the proof of Lemma 52.10.6; the argument here is easier because we know the value of our system already. Namely, consider the maps

for varying $n$. By the above, for every $\mathfrak q$ we can find an $n$ such that $\alpha _ n$ is surjective after localization at $\mathfrak q$. Since $B$ is Noetherian and $\mathop{\mathrm{Ext}}\nolimits ^{-i}_ B(M, \omega _ B^\bullet )$ a finite module, we can find an $n$ such that $\alpha _ n$ is surjective. For any $n$ such that $\alpha _ n$ is surjective, given a prime $\mathfrak q \in V(IB)$ we can find an $n' > n$ such that $\mathop{\mathrm{Ker}}(\alpha _ n)$ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^{-i}(M/I^{n'}M, \omega _ B^\bullet )$ at least after localizing at $\mathfrak q$. Since $\mathop{\mathrm{Ker}}(\alpha _ n)$ is a finite $A$-module and since supports of sections are quasi-compact, we can find an $n'$ such that $\mathop{\mathrm{Ker}}(\alpha _ n)$ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^{-i}(M/I^{n'}M, \omega _ B^\bullet )$. In this way we see that $\mathop{\mathrm{Ext}}\nolimits ^{-i}(M/I^ n M, \omega _ B^\bullet )$ is essentially constant with value $\mathop{\mathrm{Ext}}\nolimits ^{-i}(M, \omega _ B^\bullet )$. This finishes the proof. $\square$

Proof. Recall that $\textit{Coh}(U, I\mathcal{O}_ U)$ is an abelian category, see Cohomology of Schemes, Lemma 30.23.2. Over affine opens of $U$ the object $(\mathcal{F}_ n)$ corresponds to a finite module over a Noetherian ring (Cohomology of Schemes, Lemma 30.23.1). Thus the kernels of the maps $f^ N : (\mathcal{F}_ n) \to (\mathcal{F}_ n)$ stabilize for $N$ large enough. By Lemmas 52.17.1 and 52.16.3 in order to prove the lemma we may replace $(\mathcal{F}_ n)$ by the image of such a map. Thus we may assume $f$ is injective on $(\mathcal{F}_ n)$. After this replacement the equivalent conditions of Lemma 52.3.1 hold for the inverse system $(\mathcal{F}_ n)$ on $U$. We will use this without further mention in the rest of the proof.

We will check hypotheses (a), (b), and (c) of Lemma 52.16.10. Hypothesis (b) holds by Cohomology, Lemma 20.36.2.

Pick a inverse system of modules $\{ M_ n\} $ as in Lemma 52.16.2. We may assume $H^0_\mathfrak m(M_ n) = 0$ by replacing $M_ n$ by $M_ n/H^0_\mathfrak m(M_ n)$ if necessary. Then we obtain short exact sequences

for all $n$. Let $E$ be an injective hull of the residue field of $A$. By Lemma 52.20.3 and our current assumption (4) we can choose, an integer $m \geq 0$, finite $A$-modules $N_1$ and $N_2$ annihilated by $f^ c$ for some $c \geq 0$ and compatible systems of maps

for $n \geq m$ with the properties stated in the lemma.

We know that $M = \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}_ n)$ is an $A$-module whose limit topology is the $f$-adic topology. Thus, given $n$, the module $M/f^ nM$ is a subquotient of $H^0(U, \mathcal{F}_ N)$ for some $N \gg n$. Looking at the information obtained above we see that $f^ cM/f^ nM$ is a finite $A$-module. Since $f$ is a nonzerodivisor on $M$ we conclude that $M/f^{n - c}M$ is a finite $A$-module. In this way we see that hypothesis (c) of Lemma 52.16.10 holds.

Next, we study the module

For $n \geq m$ let $K_ n$ be the kernel of the map $H^2_\mathfrak m(M_ n) \to \mathop{\mathrm{Hom}}\nolimits _ A(N_2, E)$. Set $K = \mathop{\mathrm{lim}}\nolimits K_ n$. We obtain an exact sequence

By the above the limit topology on $Ob = \mathop{\mathrm{lim}}\nolimits H^2_\mathfrak m(M_ n)$ is the $f$-adic topology. Since $N_2$ is annihilated by $f^ c$ we conclude the same is true for the limit topology on $K = \mathop{\mathrm{lim}}\nolimits K_ n$. Thus $K/fK$ is a subquotient of $K_ n$ for $n \gg 1$. However, since $\{ K_ n\} $ is pro-isomorphic to a inverse system of finite length $A$-modules (by the conclusion of Lemma 52.20.3) we conclude that $K/fK$ is a subquotient of a finite length $A$-module. It follows that $K$ is a finite $A$-module, see Algebra, Lemma 10.96.12. (In fact, we even see that $\dim (\text{Supp}(K)) = 1$ but we will not need this.)

Given $n \geq 1$ consider the boundary map

(the second map is an isomorphism) coming from the short exact sequences

For each $n$ set

where $m$ is as above. Observe that $\{ P_ n\} $ is an inverse system and that the map $f : \mathcal{F}_ n \to \mathcal{F}_{n + 1}$ on global sections maps $P_ n$ into $P_{n + 1}$. If $p \in P_ n$, then $\delta _ n(p) \in K \subset Ob$ because $\delta _ n(p)$ maps to zero in $H^1(U, f^ n\mathcal{F}_{n + m}) = H^2_\mathfrak m(M_ m)$ and the composition of $\delta _ n$ and $Ob \to \mathop{\mathrm{Hom}}\nolimits _ A(N_2, E)$ factors through $H^2_\mathfrak m(M_ m)$ by our choice of $m$. Hence

is a finite graded $A[T]$-module where $T$ acts via multiplication by $f$. Namely, it is a graded submodule of $K[T]$ and $K$ is finite over $A$. Arguing as in the proof of Cohomology, Lemma 20.35.1¹ we find that the inverse system $\{ P_ n\} $ satisfies ML. Since $\{ P_ n\} $ is pro-isomorphic to $\{ H^0(U, \mathcal{F}_ n)\} $ we conclude that $\{ H^0(U, \mathcal{F}_ n)\} $ has ML. Thus hypothesis (a) of Lemma 52.16.10 holds and the proof is complete. $\square$

Proof. We will use Lemma 52.19.3 without further mention. In particular, we see the condition is necessary. Conversely, suppose the condition is true. Note that $\delta ^ Y_ Z(y) = \dim (\overline{\{ y\} })$ by Lemma 52.18.1. Let us write $\delta $ for this function. Let $y \in U \cap Y$. If $\delta (y) > 2$, then the inequality of Lemma 52.19.3 holds. Finally, suppose $\delta (y) = 2$. We have to show that

Choose a specialization $y \leadsto y'$ with $\delta (y') = 1$. Then there is a ring map $\mathcal{O}_{X, y'}^\wedge \to \mathcal{O}_{X, y}^\wedge $ which identifies the target with the completion of the localization of $\mathcal{O}_{X, y'}^\wedge $ at a prime $\mathfrak q$ with $\dim (\mathcal{O}_{X, y'}^\wedge /\mathfrak q) = 1$. Moreover, we then obtain

Let $\mathfrak p' \subset \mathcal{O}_{X, y'}^\wedge $ be the image of $\mathfrak p$. By Local Cohomology, Lemma 51.11.3 we have

the last equality because the specialization is immediate. Thus the lemma is prove by the assumed inequality for $y', \mathfrak p'$. $\square$

Proof. We will use the criterion of Lemma 52.20.5. Let $y \in U \cap Y$ with $\dim (\overline{\{ y\} } = 1$ and let $\mathfrak p$ be a prime $\mathfrak p \subset \mathcal{O}_{X, y}^\wedge $ with $\mathfrak p \not\in V(I\mathcal{O}_{X, y}^\wedge )$. Condition (4) shows that $\text{depth}((\mathcal{F}_ y^\wedge )_\mathfrak p) = \text{depth}((\mathcal{O}_{X, y}^\wedge )_\mathfrak p)$. Thus we have to prove

Let $\mathfrak p_0 \subset A$ be the image of $\mathfrak p$. Let $\mathfrak q \subset A$ be the prime corresponding to $y$. By Local Cohomology, Lemma 51.11.3 we have

If (5)(a) holds, then we get that this is

Note that in any case $\dim (A/\mathfrak p_0) \geq 2$. Hence if we get $2$ for the minimum, then we are done. If not we get

because every component of $\mathop{\mathrm{Spec}}(A)$ passing through $\mathfrak p_0$ has dimension $\geq 4$. If (5)(b) holds, then we win immediately. $\square$