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Suppose $R$ is a commutative ring, and $M$ an $R$-module. Is there a nontrivial example of such $M$ where the set of associated primes $\text{Ass}(M)=\varnothing$?

Taking $M=0$ feels kind of cheap, since any annihilator of $x\in M$ is necessarily all of $R$, and hence not prime.

Is there a better example that is still relatively simple, preferably with some explanation about why $\text{Ass}(M)=\varnothing$?

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Any example is going to be somewhat "complicated" because $R$ has to be non-Noetherian. If $R$ is a Noetherian ring, and $M$ is a non-zero $R$-module, then there must be an associated prime ideal. Proof sketch: Consider the set of annihilators of non-zero elements of $M$. This is a non-empty set of ideals of $R$ and therefore has a maximal element, since $R$ is Noetherian. Show that this maximal element is prime. – Ted Feb 21 '12 at 4:51
up vote 5 down vote accepted

Let $k$ be a field and $R=k[X_0,X_1,X_2,..]/ \langle X_0^2,X_1^2,X_2^2,...\rangle=k[x_0,x_1,x_2,..]$. Then $\operatorname{Ass}_R(R)=\emptyset$ .

Edit: a detailed proof
At Lando's request I'll give a complete proof of the above.

1) Let's show that the only prime ideal of $R$ is the obviously maximal ideal $\mathfrak m=\langle x_0,x_1,x_2,...\rangle$.
Indeed, if $\mathfrak p \subset R$ is prime then for all $i\geq 0$ we deduce from $0=x_i^2\in \mathfrak p$ that $x_i\in \mathfrak p$, so that $\mathfrak m=\langle x_0,x_1,x_2,...\rangle \subset \mathfrak p$ and since $\mathfrak m$ is maximal we have $\mathfrak p=\mathfrak m$.

2) It remains to prove that $\mathfrak m$ is not an associated ideal and we will have the desired conclusion $\operatorname{Ass}(R)=\emptyset$ ($R$ is seen as a module over itself).
To say that $\mathfrak m$ is associated means that there exists $0\neq P\in R$ with $\mathfrak m=\operatorname{Ann}(P)$.
Such a $P$ can be written as $P(x_0,...,x_N)$, i.e. can involve only finitely many $x_i$'s.
But then we see that it is not true that $\mathfrak m=\operatorname{Ann}(P)$, because $x_{N+1}\notin \operatorname{Ann}(P(x_0,...,x_N))$: any non-zero term $rx_{i_0}...x_{i_s} \neq 0\; (0\leq i_0\lt\cdots\lt i_s\leq N$ ) of $P(x_0,...,x_N)$ will satisfy $rx_{i_0}...x_{i_s}\cdot x_{N+1} \neq 0$ and so $ P(x_0,...,x_N)\cdot x_{N+1}\neq 0 $.

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Thanks Georges. Can you explain how you indeed know that $\text{Ass}_R(R)=\varnothing$? – Kally Feb 20 '12 at 23:04
@LandoKal: The first step is to determine the prime ideals of $R$. – Ted Feb 20 '12 at 23:32
@Ted Sure, but it's not clear to me what they are. The elements of $R$ are polynomials in $k$ equivalent if their difference is divisible by some $\sum_i k_ix_i^2$. But then what? – Kally Feb 20 '12 at 23:46
Dear Lando, I have added a little hint: the ring is local. I'll add more details tomorrow (it's 02 a.m. !) – Georges Elencwajg Feb 21 '12 at 0:56
Dear Lando, I've had breakfast and then supplied a detailed proof of my assertion. If something is still not clear, do not hesitate to ask: math.stackexchange wants 100 % satisfied customers... – Georges Elencwajg Feb 21 '12 at 8:34

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