Prove every prime ideal of a ring is a radical ideal.

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.

If $R$ is a commutative ring and $P$ is a prime ideal of $R$, then $rad(P)=P$. To prove this fact, we first prove by induction over $n$ that if $x \in R$ and $n \in \mathbb N$, $n>0$ then $x^n \in P \Rightarrow x \in P$. If $n=1$, there is nothing to prove. Now suppose $n \ge 1$ and $x^n \in P \Rightarrow x \in P$ for every $x \in R$. If $x^{n+1}=xx^n \in P$ then either $x \in P$ or $x^n \in P$, but in the second case we still have $x \in P$.

Now let's prove the main assertion. If $x \in rad(P)$ then $x^n \in P$ for some $n \in \mathbb N$, $n>0$ by definition of radical. As we saw this imply $x \in P$, so we have $rad(P) \subset P$. The converse inclusion is obvious and holds for every ideal of $R$.

Conversely, if $R$ is a commutative ring and $I$ is an ideal of $R$ such that $rad(I)=I$, it is not necessary for $I$ to be prime. As a counterexample, consider $R= \mathbb Z$, and $I=(6)$. We have $rad(I)=I$ but $I$ is not prime.