Proposition

Let $a$ be an ideal $\ne (1)$ in a ring $A$. Show that $a = r(a) \iff a$ is an intersection of prime ideals.

Solution

Suppose $a = r(a)$. By Proposition 1.14[Atiyah], $r(a)$ is the intersection of the prime ideals containing $a$. Therefore, $a$ is an intersection of prime ideals.

Suppose $a$ is an intersection of prime ideals. Let $\{ P_{\alpha} \}$ be a collection of prime ideals such that $a = \cap_{\alpha} P_{\alpha}$. Then $\forall \alpha, a \subset P_{\alpha}$. Since $r(a)$ is the intersection of all the prime ideals containing $a$, $r(a) \subset \cap_{\alpha} P_{\alpha}$.

Therefore, we have $a \subset r(a) \subset \cap_{\alpha} P_{\alpha} = a$. This implies $r(a) = a$.

The condition that $a \ne (1)$ does not seem necessary if we assume the intersection of no sets is $A$.