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 $T = V(I)$. $\square$