I'd like to show the following claim:
The radical of a primary ideal $\mathfrak q$, $r(\mathfrak q)$, is the smallest prime ideal containing $\mathfrak q$.
Can you tell me if my proof is correct:
I'll use the following two facts:
(i) There is a bijection between ideals $J$ of $R$ containing $I$ and ideals $\overline{J}$ of $R/I$.
(ii) The radical of an ideal $r(I)$ equals the nilradical $n(R/I)$. (To see this, let $x \in r(I)$. Then $x^n \in I$ and hence $\overline{x}^n = 0$ in $R/I$. On the other hand, let $\overline{x}^n = 0$ then $x^n \in I$ and hence $x \in r(I)$.)
Now for the claim: By (ii), we have $r(\mathfrak q) = n(R/\mathfrak q)$. We know that $n(R/\mathfrak q) = \bigcap_{I \in R/\mathfrak q; I \text{ prime}} I$. Going back to $R$ and using (i) we hence see that $r(\mathfrak q) = \bigcap_{I \subset R; \mathfrak q \subset I } I$. At first I thought that I could write "$\mathfrak q \subset I; I \text{ prime}$" here but I think that might be wrong. (Do you know an example where that's wrong?)
Now we have established that $r(\mathfrak q)$ is the smallest ideal containing $\mathfrak q$ so to finish the proof we want to show that $r(\mathfrak q)$ is prime:
Let $xy \in r(\mathfrak q)$. Then $x^n y^n \in \mathfrak q$ for some $n$. Since $\mathfrak q$ is primary we hence know that for some $m$ we have $x^m$ or $y^m$ in $ \mathfrak q$. Hence $x \in r(\mathfrak q)$ or $y \in r(\mathfrak q)$.