The Stacks project

Proof. We prove the equivalence of (1) and (2) by induction on $n$. For $n = 0$ we have $H^0_ T(M) \subset M$ is finite. Hence (1) is true. Since $H^0_ T(M) = \mathop{\mathrm{colim}}\nolimits H^0_{V(J)}(M)$ with $J$ as in (2) we see that (2) is true. Assume that $n > 0$.

Assume (1) is true. Recall that $H^ i_ J(M) = H^ i_{V(J)}(M)$, see Dualizing Complexes, Lemma 47.10.1. Thus $H^ i_ T(M) = \mathop{\mathrm{colim}}\nolimits H^ i_ J(M)$ where the colimit is over ideals $J \subset A$ with $V(J) \subset T$, see Lemma 51.5.3. Since $H^ i_ T(M)$ is finitely generated for $i \leq n$ we can find a $J \subset A$ as in (2) such that $H^ i_ J(M) \to H^ i_ T(M)$ is surjective for $i \leq n$. Thus the finite list of generators are $J$-power torsion elements and we see that (2) holds with $J$ replaced by some power.

Assume we have $J$ as in (2). Let $N = H^0_ T(M)$ and $M' = M/N$. By construction of $R\Gamma _ T$ we find that $H^ i_ T(N) = 0$ for $i > 0$ and $H^0_ T(N) = N$, see Remark 51.5.6. Thus we find that $H^0_ T(M') = 0$ and $H^ i_ T(M') = H^ i_ T(M)$ for $i > 0$. We conclude that we may replace $M$ by $M'$. Thus we may assume that $H^0_ T(M) = 0$. This means that the finite set of associated primes of $M$ are not in $T$. By prime avoidance (Algebra, Lemma 10.15.2) we can find $f \in J$ not contained in any of the associated primes of $M$. Then the long exact local cohomology sequence associated to the short exact sequence

turns into short exact sequences

for $i < n$. We conclude that $J^2$ annihilates $H^ i_ T(M/fM)$ for $i < n$. By induction hypothesis we see that $H^ i_ T(M/fM)$ is finite for $i < n$. Using the short exact sequence once more we see that $H^{i + 1}_ T(M)$ is finite for $i < n$ as desired.

We omit the proof of the equivalence of (2) and (3) in case $T = V(I)$. $\square$

Proof. The implication (1) $\Rightarrow$ (2) is immediate. We prove the converse by induction on $n$. The case $n = 0$ is clear because both (1) and (2) are always true in that case.

Assume $n > 0$ and that (2) is true. Let $N = H^0_ Z(M)$ and $M' = M/N$. By Dualizing Complexes, Lemma 47.11.6 we may replace $M$ by $M'$. Thus we may assume that $H^0_ Z(M) = 0$. This means that $\text{depth}_ I(M) > 0$ (Dualizing Complexes, Lemma 47.11.1). Pick $f \in I$ a nonzerodivisor on $M$ and consider the short exact sequence

which produces a long exact sequence

and similarly after localization. Thus assumption (2) implies that the modules $H^ i_ Z(M/fM)_\mathfrak p$ are finite for $i < n$. Hence by induction assumption $H^ i_ Z(M/fM)$ are finite for $i < n$.

Let $\mathfrak p$ be a prime of $A$ which is associated to $H^ i_ Z(M)$ for some $i \leq n$. Say $\mathfrak p$ is the annihilator of the element $x \in H^ i_ Z(M)$. Then $\mathfrak p \in Z$, hence $f \in \mathfrak p$. Thus $fx = 0$ and hence $x$ comes from an element of $H^{i - 1}_ Z(M/fM)$ by the boundary map $\delta$ in the long exact sequence above. It follows that $\mathfrak p$ is an associated prime of the finite module $\mathop{\mathrm{Im}}(\delta )$. We conclude that $\text{Ass}(H^ i_ Z(M))$ is finite for $i \leq n$, see Algebra, Lemma 10.63.5.

Recall that

by Algebra, Lemma 10.63.19. Since by assumption the modules on the right hand side are finite and $I$-power torsion, we can find integers $e_{\mathfrak p, i} \geq 0$, $i \leq n$, $\mathfrak p \in \text{Ass}(H^ i_ Z(M))$ such that $I^{e_{\mathfrak p, i}}$ annihilates $H^ i_ Z(M)_\mathfrak p$. We conclude that $I^ e$ with $e = \max \{ e_{\mathfrak p, i}\}$ annihilates $H^ i_ Z(M)$ for $i \leq n$. By Lemma 51.7.1 we see that $H^ i_ Z(M)$ is finite for $i \leq n$. $\square$

Proof. Consider the annihilator $\mathfrak a$ of $H^ i_ Z(M)$. Let $\mathfrak p \subset A$ with $\mathfrak p \not\in Z$. By assumption there exists an $f \in I$, $f \not\in \mathfrak p$ and an isomorphism $\varphi : A_ f^{\oplus r} \to M_ f$ of $A_ f$-modules. Clearing denominators (and using that $M$ is of finite presentation) we find maps

with $a_ f = f^ N \varphi$ and $b_ f = f^ N \varphi ^{-1}$ for some $N$. Moreover we may assume that $a \circ b$ and $b \circ a$ are equal to multiplication by $f^{2N}$. Thus we see that $H^ i_ Z(M)$ is annihilated by $f^{2N}J^ n$, i.e., $f^{2N}J^ n \subset \mathfrak a$.

As $U = \mathop{\mathrm{Spec}}(A) \setminus Z$ is quasi-compact we can find finitely many $f_1, \ldots , f_ t$ and $N_1, \ldots , N_ t$ such that $U = \bigcup D(f_ j)$ and $f_ j^{2N_ j}J^ n \subset \mathfrak a$. Then $V(I) = V(f_1, \ldots , f_ t)$ and since $I$ is finitely generated we conclude $I^ M \subset (f_1, \ldots , f_ t)$ for some $M$. All in all we see that $J^ m \subset \mathfrak a$ for $m \gg 0$, for example $m = M (2N_1 + \ldots + 2N_ t) n$ will do. $\square$

Proof. The assumption that $H^ i_ Z(A)$ is finite for $0 \leq i \leq n$ implies there exists an $e \geq 0$ such that $I^ e$ annihilates $H^ i_ Z(A)$ for $0 \leq i \leq n$, see Lemma 51.7.1. Then Lemma 51.7.3 implies that $H^ i_ Z(M)$, $0 \leq i \leq n$ is annihilated by $I^ m$ for some $m = m(M, i)$. We may take the same $m$ for all $0 \leq i \leq n$. Then Lemma 51.7.1 implies that $H^ i_ Z(M)$ is finite for $0 \leq i \leq n$ as desired. $\square$