TITLE: Prove every prime ideal of a ring is a radical ideal.
QUESTION [9 upvotes]: this is my attempt: Since $R$ is commutative, we let $I$ to be a prime ideal of $R$, the for $a,b\in R$,then the product $ab$ we must have that $a\in I$ or $b \in I$, by definition of a prime ideal. On the other hand,we have: $radI = \{r\in R | r^n\in I $ for some $n\in \mathbb{N}\}$. so the argument goes by induction on $(ab)^n$, for $n=1$ is just the definition of a prime ideal, so we assume that its true for $n$. now for $n+1$, we have the product: $(ab)^{n+1}= (ab)^n(ab)= (a^{n}a)(b^{n}b)$, but if $a\in I$, then $a^n\in I$, similarly if $b\in I$, then $b^n\in I$, therefore we thus have that $radI=I$. Now, i need to give an example that the converse is false, so if $a,b\in radI$, then $a^n\in I$ and $b^n\in I$, so when we consider the product $a^nb^n=(ab)^n$, we thus have that $a^n\in I$ and $b^n\in I$, therefore $I$ cant be a prime ideal. 
My question is if this is the correct proof, especially in the converse.

REPLY [14 votes]: I think a possibly more straightforward proof is as follows. 
Let $P \subset R$ be a prime ideal. To show $P$ is radical, we show that $x^r \in P \implies x\in P$.
Let $x^r \in P$, and assume $r$ is the minimal such natural number. If $r$ is 1, we are done. If $r>1$, because $P$ is prime, either $x\in P$ or $x^{r-1} \in P$. In the first case, we are done; in the second,  this is a contradiction since $r$ was chosen to be minimal. 
A counterexample to the converse is possibly $6\mathbb{Z}\subset \mathbb{Z}$, or any such ideal in $\mathbb{Z}$ generated by a product of distinct primes.