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: 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).)
A: 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 $\{0\}$.
