# If $(R,m)$ is Noetherian, $P$ a prime ideal s.t. $R/P$ is 1 dim, then if $x\in m-P$, then rad$(x,P)=m$

If $(R,m)$ is Noetherian, $P$ a prime ideal s.t. $R/P$ is 1 dim, then if $x\in R-P$, then rad$(x,P)=m$. I am looking to prove this statement but I am at a loss how to start. It's part of a proof that I am reading, but this statement appears without any further elaboration.

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I'm afraid I don't know/remember what $\text{rad}(x,P)$ stands for, could someone remind me? –  Zev Chonoles Jul 14 '11 at 23:44
@Zev: It is the radical of the ideal generated by $x$ and $P$. –  B M Jul 14 '11 at 23:51
Should you exclude $x \notin \mathfrak{m}$? –  Dylan Moreland Jul 14 '11 at 23:58
@Dylan: Thanks. I have made the change. –  B M Jul 15 '11 at 0:15

Hint: $\operatorname{Rad}{(x,P)} = \displaystyle\bigcap_{I \in \mathrm{Spec R}, I \supset (x,P)} I.$

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sweet, we don't even need Noetherianness. –  B M Jul 15 '11 at 0:22