Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\text{Ass}(M)=\varnothing$?

share|improve this question
    
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

1 Answer 1

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 $.

share|improve this answer
    
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.