Solution
= Solution
Certainly $A\subseteq\sqrt A$, by taking exponent one. If $x\in\sqrt A$, choose $n\geq1$ with $x^n\in A$. Primality applied to $x\cdot x^{n-1}$ gives $x\in A$ or $x^{n-1}\in A$. In the latter case repeat; induction reduces to $x\in A$. Thus $\sqrt A\subseteq A$ as well, proving
$$
\boxed{\sqrt A=A.}
$$
Every <prime ideal> is therefore a <radical ideal>.