Lemma 1

For every ideal a of A, the set S of elements $1 + a$, where $a \in a$, is a multiplicative subset of A and the set $a'$ of elements of $S^{-1}A$ of the form $a/s$, where $a \in a$ and $s \in S$, is an ideal contained in the Jacobson radical of $S^{-1}A$.

The first assertion is obvious, as well as the fact that $a'$ is an ideal of $S^{-1}A$. On the other hand, $(1/1) + (a/s) = (s + a)/s$ and $s + a \in S$ for all $s \in S$ and $a \in a$ by definition of S; hence $(1/1) + (a/s)$ is invertible in $S^{-1}A$ for all $a/s \in a'$, which completes the proof of the lemma (Algebra, Chapter VIII, § 6, no. 3, Theorem 1).

This being so, if we set $N = S^{-1}M$, clearly N is a finitely generated $S^{-1}A$-module; if $aM = M$, then $a'N = N$ and it follows that $N = 0$ by Nakayama's Lemma (Algebra, Chapter VIII, § 6, no. 3, Corollary to Proposition 6); the corollary then follows from Corollary 2.