Remark 3

Suppose q is contained in the radical of A. Then, by Nakayama's lemma, the sequence (q^nM) is stationary if and only if one has q^nM = 0 for n sufficiently large. It then follows from part a) of the corollary that one has d_q(M) \leq 0 if and only if M is of finite length, and that one then has e_q(M) = \operatorname{long}_A(M).