# The intersection of the associated primes of a reduced ring

Let $$R$$ be a commutative ring with identity. Recall that a prime ideal is called associated prime ideal whenever it is the annihilator of a nonzero element. Also a ring is called reduced whenever has no nonzero nilpotent.

Why is zero the intersection of all associated prime ideals of a reduced ring?

• a nonzero element.... of $R$? Associated primes are usually defined with respect to a given module $M$. Are we supposed to assume $M=R$ or something? Commented Jan 11, 2019 at 17:04
• Yes. A prime ideal is a associated prime ideal of an $R$-module, if there exists a non zero element of that module such that the prime ideal is its annihilator. Commented Jan 11, 2019 at 17:35
• I already knew everything after "yes", and it is not relevant to my question. Does your "yes" mean you are confirming that you are talking about a nonzero element of $R$ or do you mean something else (all associated primes for all $R$ modules?) Commented Jan 11, 2019 at 17:44
• About non zero elements of $R$. Commented Jan 11, 2019 at 18:31

Let $$R$$ be a commutative ring. We denote by $${\rm Ass}(R)$$ the set of associated primes of $$R$$ and by $${\rm Ass}^f(R)$$ the set of weakly associated primes of $$R$$.

(1) By Bourbaki, AC.IV.1 Exercice 17 b), we have $$\bigcap{\rm Ass}^f(R)={\rm Nil}(R)$$. Thus, if $$R$$ is reduced, then $$\bigcap{\rm Ass}^f(R)=0$$.

(2) If $$R$$ is noetherian, then by Bourbaki, AC.IV.1 Exercice 17 g) we have $${\rm Ass}^f(R)={\rm Ass}(R)$$, and therefore $$\bigcap{\rm Ass}(R)={\rm Nil}(R)$$ by (1). Thus, if $$R$$ is reduced and noetherian, then $$\bigcap{\rm Ass}(R)=0$$.

(3) There exists a non-noetherian reduced ring $$R$$ such that $$\bigcap{\rm Ass}(R)=0$$. For this it suffices to exhibit a non-noetherian reduced ring $$R$$ such that $${\rm Ass}^f(R)={\rm Ass}(R)$$. Such an example (with $$R$$ even a domain) is given in P.-J. Cahen, Ascending chain conditions and associated primes, Commutative ring theory (Fès, 1992), 41-46, Lecture Notes in Pure and Appl. Math. 153, Dekker, New York, 1994.

(4) There exists a reduced ring $$R$$ such that $$\bigcap{\rm Ass}(R)\neq 0$$. For this, any nonzero reduced ring $$R$$ with $${\rm Ass}(R)=\emptyset$$ will do.

(Thanks to user25867 for pointing out how to improve statement (1).)

• I think the nilradical equals the intersection of all weakly associated primes for any commutative ring, no need to assume that R is reduced. And this is the way one should read Bernard's answer. Commented Jan 18, 2019 at 11:21
• @user26857: Thanks for pointing this out! Commented Jan 18, 2019 at 12:05

The intersection of all associated prime ideals is the nilradical of $$R$$, i.e. the set of nilpotent elements. If the ring is reduced, this set is $$\{0\}$$.

• I've seen that the nilradical is this intersection for Noetherian rings but does it hold in general? Commented Jan 11, 2019 at 14:28
• I believe this is true with correct notion of associated primes (weakly associated prime ideals – which are the same if the ring is noetherian). Commented Jan 11, 2019 at 14:33
• The intersection of all minimal prime ideals is the nilradical of $R$. I want to know why it is true for associated primes in reduced rings, with the mentioned definition? Commented Jan 11, 2019 at 14:33
• That's because one shows $\operatorname{Spec} A$ and $\operatorname{Ass} A$ have the same minimal elements. Commented Jan 12, 2019 at 11:13
• As far as I remember, you can replace associated with weakly associated. Commented Jan 12, 2019 at 12:50