# Commutative rings with unity over which every non-zero module has an associated prime

Let $R$ be a commutative ring with unity such that every non-zero module over $R$ has an associated prime. Then is it true that $R$ is Noetherian ? If not, then can we say something about any possible structure of $R$ at least ?

COMMENTS: Since for every proper ideal $I$ of $R$, and non-zero $x+I \in R/I$, $\mathrm{ann}_R(x+I)=(I:x)$, so if every non-zero module over $R$ has an associated prime, then for every proper ideal $I$ of $R$, $\exists x \notin I$ such that $(I:x)$ is a prime ideal, and conversely, if this holds, then for every proper ideal $I$ of $R$, $\mathrm{Ass}(R/I)$ is non-empty, and then since for every non-zero $R$-module $M$, we have $\mathrm{Ass}(R/\mathrm{ann}(m)) =\mathrm{Ass}(Rm)\subseteq \mathrm{Ass}(M)$, for every non-zero $m$ in $M$, so every non-zero $R$-module would have an associated prime. So we have proved :

For a commutative ring (with unity) $R$, TFAE :

1) for every proper ideal $I$ of $R$, $\exists x \notin I$ such that $(I:x)$ is a prime ideal.

2) $R/I$ has an associated prime for every proper ideal $I$ of $R$.

3) Every non-zero $R$-module has an associated prime.

But condition (1) still seems very complicated. Noetherian rings of course satisfy condition (1), but I don't know what other type of rings satisfy that condition.

A commutative ring is $$R$$ semi-artinian if every nonzero $$R$$-module contains a simple submodule. If $$R$$ is semi-artinian, then every module $$M \neq 0$$ has an element $$x \in M$$ that generates a simple module, so that $$\operatorname{ann}_R(x)$$ is a maximal ideal. Notice also that a semi-artinian ring is noetherian if and only if it is artinian; for the nontrivial direction, one begins with a semi-artinian and noetherian ring $$R$$ and then produces an ascending chain of finite-length submodules that eventually terminates at $$R$$.
While I am familiar with the semi-artinian property from the literature on noncommutative rings, I was able to find an example of a commutative semi-artinian from the first page of the paper Semi-artinian group rings by Annie Vogel. It claims that the rational group algebra $$\mathbb{Q}[C(p^\infty)]$$ of a Prüfer $$p$$-group $$C(p^\infty) \cong \frac{1}{p}\mathbb{Z}/\mathbb{Z}$$ is semi-artinian. (Disclaimer: I cannot view the rest of the pages of that paper.)
However, $$\mathbb{Q}[C(p^\infty)]$$ is not noetherian. Indeed, this abelian group has an ascending chain of cyclic subgroups $$C(p) \leq C(p^2) \leq C(p^3) \leq \cdots$$ such that each $$C(p^\infty)/C(p^n) \cong C(p^\infty)$$; thus we have surjective group homomorphisms $$\phi_n \colon C(p^\infty) \twoheadrightarrow C(p^\infty)$$ with kernel $$C(p^n)$$. By the universal property of the group algebra, each $$\phi_n$$ induces a unique algebra homomrphism $$\psi_n \colon \mathbb{Q}[C(p^\infty)] \to \mathbb{Q}[C(p^\infty)]$$ that restricts to $$\phi_n$$ on $$C(p^\infty)$$. The kernel $$K_n$$ of $$\psi_n$$ will be generated by $$\{1-h \mid h \in C(p^n)\}$$. Thus the $$K_n$$ will form an infinite ascending chain of ideals in $$\mathbb{Q}[C(p^\infty)]$$.