Corollary 2

Let $A$ be a Noetherian ring, $M$ an $A$-module and $a$ an element of $A$. For the homothety on $M$ with ratio $a$ to be injective, it is necessary and sufficient that $a$ belong to no prime ideal associated with $M$.

If $a$ belongs to a prime ideal $p \in \operatorname{Ass}(M)$, then $p = \operatorname{Ann}(x)$, where $x \in M, x \neq 0$; whence $ax = 0$ and the homothety with ratio $a$ is not injective. Conversely, if $ax = 0$ for some $x \in M$ such that $x \# 0$, then $Ax \# \{0\}$, whence $\operatorname{Ass}(x) \neq \varnothing$ (Corollary 1). Let $p \in \operatorname{Ass}(Ax)$; then obviously $p \in \operatorname{Ass}(M)$ and $p = \operatorname{Ann}(bx)$, where $b \in A$; whence $a \in p$, since $abx = 0$.