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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?

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  • $\begingroup$ 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? $\endgroup$
    – rschwieb
    Commented Jan 11, 2019 at 17:04
  • $\begingroup$ 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. $\endgroup$
    – Deroty
    Commented Jan 11, 2019 at 17:35
  • $\begingroup$ 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?) $\endgroup$
    – rschwieb
    Commented Jan 11, 2019 at 17:44
  • $\begingroup$ About non zero elements of $R $. $\endgroup$
    – Deroty
    Commented Jan 11, 2019 at 18:31

2 Answers 2

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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).)

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    $\begingroup$ 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. $\endgroup$
    – user26857
    Commented Jan 18, 2019 at 11:21
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    $\begingroup$ @user26857: Thanks for pointing this out! $\endgroup$ Commented Jan 18, 2019 at 12:05
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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 $\{0\}$.

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

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