Proposition 3

Let $A$ be a torsion-free commutative group, $A$ a graded Noetherian ring of type $A$, $p$ a graded ideal of $A$ and $M$ a graded $A$-module of type $A$ not reduced to $0$. Suppose that for every homogeneous element $a$ of $p$ the homothety of ratio $a$ on $M$ is almost nilpotent and that for every homogeneous element $b$ of $A - p$ the homothety of ratio $b$ on $M$ is injective. Then $p$ is prim and the submodule $\{0\}$ of $M$ is $p$-primary.

It suffices to show that $\operatorname{Ass}(M) = \{p\}$ (§ 2, no. 1, Proposition 1). Let $q$ be a prime ideal associated with $M$; it is a graded ideal and it is the annihilator of a homogeneous element $x \neq 0$ of $M$ (no. 1, Proposition 1). For every homogeneous element $a$ of $q$, $ax = 0$ and hence the homothety of ratio $a$ on $M$ is not injective, whence $a \in p$. Conversely, let $b$ be a homogeneous element of $p$; there exists an integer $n > 0$ such that $b^n x = 0$, whence $b^n \in \operatorname{Ann}(x) = q$ and, as $q$ is prime, $b \in q$. As $p$ and $q$ are generated by their respective homogeneous element, $p = q$, which proves that $\operatorname{Ass}(M) \subset \{p\}$. As $M \neq \{0\}, \operatorname{Ass}(M) \# \varnothing$ (§ 1, no. 1, Corollary 1 to Proposition 2), whence $\operatorname{Ass}(M) = \{p\}$.