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I am looking for an alternate proof of the fact that in a commutative ring $R$ with a prime ideal $P$, the ideal $P^{(n)}=P^n R_P\cap R$ is primary. I understand once we localize, $P^n R_P$ is a power of the maximal ideal in $R_P$ and hence primary and hence the contraction to $R$ is primary.

How would one prove this by first principles i.e. using the definition of primary ideals? Suppose $xy\in P^{(n)}$. Then there exists $t\notin P$ such that $txy\in P^n$. How do I show either $x\in P^{(n)}$ or $y\in P$?

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The notation is confusing me. Do you just want the symbolic power $P^{(n)} = (P^nR_P) \cap R$? – Dylan Moreland Mar 13 '12 at 5:01
@DylanMoreland: Sorry I somehow missed the $n$ which made the notation confusing and in the second instance I had just copy pasted from the first. – Wayne S Mar 13 '12 at 5:03
Dear Wayne, You are trying to show $x \in P^{(n)}$ or $y \in P$ (not $x \in P^n$ or $y \in P$). So suppose that $y\not\in P,$ and consider the statement $txy \in P^n$. Can you see how to conclude that $x \in P^{(n)}$? Regards, – Matt E Mar 13 '12 at 5:46
@Matt E: Sorry for another typo. But yes that's what I was trying. $y\notin P$ gets me $tx\in P$. Can't seem to do better than that. I would like to get $tx\in P^n$, but don't see how. – Wayne S Mar 13 '12 at 5:50
@Matt E: Ah, I see it. $ty \notin P$, so $(ty)x\in P^n$ gets $x\in P^{(n)}$. – Wayne S Mar 13 '12 at 5:51

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