**Proof.**
We will prove part (1) by induction on the dimension of the support of $M$. The statement holds if $M = 0$, thus we may and do assume $M$ is not zero.

Base case of the induction. If $\dim (\text{Supp}(M)) = 0$, then the support of $M$ is $\{ \mathfrak m\} $ and we see that $H^0_\mathfrak m(M) = M$ and $H^ i_\mathfrak m(M) = 0$ for $i > 0$ as is clear from the construction of local cohomology, see Dualizing Complexes, Section 47.9. Since $M$ has finite length (Algebra, Lemma 10.52.8) it has the descending chain condition.

Induction step. Assume $\dim (\text{Supp}(M)) > 0$. By the base case the finite module $H^0_\mathfrak m(M) \subset M$ has the descending chain condition. By Dualizing Complexes, Lemma 47.11.6 we may replace $M$ by $M/H^0_\mathfrak m(M)$. Then $H^0_\mathfrak m(M) = 0$, i.e., $M$ has depth $\geq 1$, see Dualizing Complexes, Lemma 47.11.1. Choose $x \in \mathfrak m$ such that $x : M \to M$ is injective. By Algebra, Lemma 10.63.10 we have $\dim (\text{Supp}(M/xM)) = \dim (\text{Supp}(M)) - 1$ and the induction hypothesis applies. Pick an index $i$ and consider the exact sequence

\[ H^{i - 1}_\mathfrak m(M/xM) \to H^ i_\mathfrak m(M) \xrightarrow {x} H^ i_\mathfrak m(M) \]

coming from the short exact sequence $0 \to M \xrightarrow {x} M \to M/xM \to 0$. It follows that the $x$-torsion $H^ i_\mathfrak m(M)[x]$ is a quotient of a module with the descending chain condition, and hence has the descending chain condition itself. Hence the $\mathfrak m$-torsion submodule $H^ i_\mathfrak m(M)[\mathfrak m]$ has the descending chain condition (and hence is finite dimensional over $A/\mathfrak m$). Thus we conclude that the $\mathfrak m$-power torsion module $H^ i_\mathfrak m(M)$ has the descending chain condition by Dualizing Complexes, Lemma 47.7.7.

Part (2) follows from (1) via Local Cohomology, Lemma 51.8.2.
$\square$

## Comments (2)

Comment #6796 by Johan on

Comment #6943 by Johan on