4

Let $Q$ be an ideal of a commutative ring $A$ and $$r(Q) = \{x \in A : x^n \in Q \text{ for some } n >0 \},$$ the radical of $Q$. Suppose that $P$ is a prime ideal of $A$. How to show that $r(P^n) = P$ for all $n>0$?

It is clear that $P \subseteq r(P^n)$ for all $n>0$. We have to show that $r(P^n) \subseteq P$ If $n=1$ and $x \in r(P)$, then there is some $k$ such that $x^k \in P$. Since $P$ is prime, $x \in P$. Therefore $r(P^n) \subseteq P$ is true for $n=1$. How to show $r(P^n) \subseteq P$ for $n>1$? Thank you very much.

Marc van Leeuwen
  • 115,048
LJR
  • 14,520

2 Answers2

3

Prove $I\subseteq J \implies r(I)\subseteq r(J).$ Then $P^n\subseteq P $ gives $r(P^n)\subseteq r(P)=P.$

Ragib Zaman
  • 35,127
1

Let $x\in A\setminus P$. Then since $P$ is a prime ideal $x^k\notin P$ for all $k\in\mathbf N$, and as $P\supseteq P^n$ also $x^k\notin P^n$. This shows that $x\notin r(P^n)$ and therefore $r(P^n)\subseteq P$.

Marc van Leeuwen
  • 115,048