Intersection of prime ideals

Solution 1:

Here's my attempt. If you didn't want an actual solution then don't read this. Please correct me if I made any mistakes! I'm using the definition $\mbox{rad}(I) = \{ r \in R \,|\,\,\, r^k \in I \text{ for some } k \in \mathbb N \}$.

Let $P$ be a prime ideal containing $I$. If $r \in R$ is such that $r^k \in I$, then $r^k \in P$, so $r \in P$ since $P$ is prime. Thus $\mbox{rad}(I) \subset \bigcap_{P \supset I} P$.

Conversely, if $r \notin \mbox{rad}(I)$, then $r^k \notin I$ for any $k$, so $S = \{1, r, r^2, \ldots \}$ is a multiplicatively closed set disjoint from $I$. By a basic theorem on prime ideals, we have that $R \smallsetminus S$ contains a prime ideal $P_r$ containing I. Since $r \notin P_r$, we have $r \notin \bigcap_{P \supset I} P$. Therefore $\mbox{rad}(I) = \bigcap_{P \supset I} P$.

The problem posted is the special case where $I = \mbox{rad}(I)$.

Solution 2:

Hint: Show that the radical of the zero ideal (the nilradical) is the intersection of all prime ideals of $R.$ Then apply this to $R/I$ and use the fourth isomorphism theorem to obtain your desired result.