Example of a non-zero module that has no associated primes

Question

I'm currently reading up on Associated Primes and localization. I came across the following theorem. Let $M$ be an $R$ module.

If $M = 0$ then $Ass(M)$ is empty. The converse is true if $R$ is a Noetherian ring. I understood the proof of this result.

Now I'm trying to come up with an example of $M$ being a non-zero $R$ module with $R$ NOT Noetherian but $Ass(M)$ is empty. Is there an easy example to show that the converse fails to hold when $R$ is not Noetherian? Thank you so much for any assistance! :)

Answer 1

Let $O$ be the valuation ring of a non-discrete valuation $v: K^\times\rightarrow\mathbb{R}$ of some field $K$. Let $x\in O$ be a non-unit and consider the $O$-module $O/xO$. The ring $O$ has only one non-zero prime ideal, namely its maximal ideal $p=\{y\in O : v(y)>0\}$; it is not finitely generated, since $v$ is non-discrete. The annihilator of an element $y+xO$ of $M$ by definition equals $\{z\in O : v(z)+v(y)\geq v(x)\}$ -- note that $w\in xO$ if and only if $v(w)\geq v(x)$. Hence the annihilator equals $wO$, where $w\in O$ is any element satisfying $v(w)=v(x)-v(y)$. In particular this annihilator is different from $p$, hence no annihilator ideal is a prime ideal.

Answer 2

An easy example is $R=M=C(\mathbb{R}, \mathbb{R})$ the ring of continuous functions on the real line.

Let $0\neq f\in M$. There exist two distinct points $a, b\in \mathbb{R}$ such that $f(a), f(b)$ are both nonzero (there exists at least one point as $f\neq 0$ and by continuity there are lots of these points nearby). Now we can construct $g, h\in R$ such that $g(a)=1=h(b)$, but $gh=0$ (you can for example do this piecewise linear, or if you cannot be bothered to get the constant terms right just use Urysohn's lemma). In particular $g$ and $h$ are both not in the annihilator of $f$, but $gh$ is. Therefore, the annihilator of $f$ is not a prime ideal.