Localisation of an ideal

This need to be fairly very easy, yet in some way I can not locate the evidence. Allow $P\neq Q$ be 2 topmost perfects in the commutative ring $R$. After that $P_Q=R_Q$.

($P_Q$ is the localisation of the R - component $P$ at $Q$ and also $R_Q$ is the localisation of R at Q)

3
2022-06-07 14:32:00
Source Share
Answers: 1

Since $P$ and also $Q$ stand out topmost perfects, $P$ is not had in $Q$ and also hence there exists $x \in P \cap (R \setminus Q)$. This component comes to be a device in the localization, so the local perfect has a device and also is hence the whole local ring $R_Q$.

This is a grandfather clause of standard outcomes on pressing onward and also drawing back perfects under a localization map: see as an example $\S 7.2$ of my commutative algebra notes for even more information. (Or see any kind of various other commutative algebra message, certainly.)

7
2022-06-07 14:53:29
Source