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

Question

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.

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

score 3 · Accepted Answer · edited Nov 16 '11 at 21:35

3

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

edited Nov 16 '11 at 21:35

t.b.

78,116

answered Jul 15 '11 at 00:16

jspecter

8,152

1

sweet, we don't even need Noetherianness. – B M Jul 15 '11 at 00:22

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

1 Answers1