0
$\begingroup$

Let $R$ be a one-dimensional noetherian domain and $I$ a nonzero ideal of $R$. If $P$ is a prime ideal containing $I$, why is $P$ the only prime ideal containing $J=R \cap IR_P$ (where $R_P$ is the localization at $P$)?

I see that $P$ contains $J$, but can't see why some other prime $Q$ containing $I$ can't also contain $J$. I know $QR_P=R_P$.

I know $J=\{ r \in R \mid \exists s\notin P,\ \ \ sr\in I\}$.

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

This follows from the following two facts.
1) $R \cap IR_P$ is $P$-primary.
2) $R$ is 1-dimensional.

Claim: $P$ is the only prime ideal containing $J$.
Proof: Since $J$ is $P$-primary, any prime containing $J$ will contain $\sqrt{J} = P$. Since $R$ is 1-dimensional and $P \neq 0 $, $P$ is a maximal ideal. Thus, $P$ is the only prime containing $J$.

$\endgroup$
2
  • $\begingroup$ ok but why is $J$ primary? $\endgroup$
    – usr0192
    Commented Apr 28, 2020 at 22:50
  • $\begingroup$ The contraction of a primary ideal is primary. $\endgroup$
    – Youngsu
    Commented Apr 28, 2020 at 22:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .