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.
If $(R,m)$ is Noetherian, $P$ a prime ideal s.t. $R/P$ is 1 dim, then if $x\in m-P$, then rad$(x,P)=m$
0
$\begingroup$
commutative-algebra
-
1@Dylan: Thanks. I have made the change. – 2011-07-15
1 Answers
3
Hint: $\operatorname{Rad}{(x,P)} = \displaystyle\bigcap_{I \in \mathrm{Spec R}, I \supset (x,P)} I.$
-
1sweet, we don't even need Noetherianness. – 2011-07-15