Exercise A0.III.5.16

Proposition

Let $R$ be a ring, $M$ an $R$-module, and let $a \in R$ be a nilpotent element. Then $M = 0 \iff aM = M.$

Proof

Suppose $M = 0$. Then its only element is $0$ and $a0 = 0$.

Suppose conversely that $aM = M$. Since $a$ is nilpotent, there exists $n \in \mathbb{N}$ such that $a^n = 0$. Putting this all together, we get

$$M = a^n M = 0M = 0.$$

$\blacksquare$

NOTE

The book includes an extra hypothesis requiring $R$ to be commutative. It is not necessary for the proof.