Proving $M = 0 \iff aM = M$ if $a \in R$ is nilpotent and $M$ is an $R$-module for commutative $R$

Question

Aluffi III.5.16 suggests proving the claim in the title.

One direction ($M = 0$ implies $aM = M$) is trivial. Let's focus on the other one (and, since I'm not sure about my proof, I'll go as slow as I can).

So, assume $aM = M$ for some nilpotent $a$. $aM = \{ am \mid m \in M \}$ by definition. What is $a^2M$? By the same definition $a^2M = \{ a^2m \mid m \in M \}$.

But $a^2m = a(am)$, so $\{ a^2m \mid m \in M \} = \{ a(am) \mid m \in M \} = \{ am' | m' \in M \}$ as sets, and the last equality holds because

• $\{ a(am) \mid m \in M \} \subseteq \{ am' | m' \in M \}$, since $\forall m \in M : \exists m' \in M : a (am) = am'$, namely $m' = am$ (because $aM = M$),
• and vice versa: $\{ am' | m' \in M \} \subseteq \{ a(am) \mid m \in M \}$ by a similar argument.

But $\{ am' \mid m' \in M \} = aM$, so we've proven that $a^2M = aM = M$. Doing this $n$ times (where $n : a^n = 0$) gives $0M = M$, but $0M = 0$, and the result immediately follows.

Is this proof correct? If it is, where have I used the commutativity of $R$? I've only used that $a^n$ commutes with $a^m$ for any $m, n$, but this holds for every ring since multiplication associative.

Answer 1

Your proof looks fine to me.

In this case, we do not need $R$ to be commutative for exactly the reasons you state. However, Aluffi mentions that this exercise is meant to be a smaller version of Nakayama’s Lemma which states

The module $M$ is $0$ if and only if $JM = M$ where $J$ is the Jacobson radical of $R$. [Exercise VI.3.8]

In this case, we need $R$ to be commutative so we can talk about the Jacobson radical. This is why Aluffi added this condition on Exercise III.5.16.