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.

Commutative rings over which every module $M$ satisfies $\mathrm{Ass}(R/\mathrm{Ann}\; M) \subseteq \mathrm{Ass}(M)$ might be related


1 Answer 1


I agree that this condition is complicated, and I do not have any insight into the structure of such rings. However, I can think of one class of (not necessarily noetherian) rings with this property.

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 $\mathbb{Q}[C(p^\infty)]$.


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