This should be quite easy, but somehow I can't find the proof. Let $P\neq Q$ be two maximal ideals in the commutative ring $R$. Then $P_Q=R_Q$.
($P_Q$ is the localisation of the R-module $P$ at $Q$ and $R_Q$ is the localisation of R at Q)
This should be quite easy, but somehow I can't find the proof. Let $P\neq Q$ be two maximal ideals in the commutative ring $R$. Then $P_Q=R_Q$.
($P_Q$ is the localisation of the R-module $P$ at $Q$ and $R_Q$ is the localisation of R at Q)
Since $P$ and $Q$ are distinct maximal ideals, $P$ is not contained in $Q$ and thus there exists $x \in P \cap (R \setminus Q)$. This element becomes a unit in the localization, so the localized ideal contains a unit and is thus the entire localized ring $R_Q$.
This is a special case of basic results on pushing forward and pulling back ideals under a localization map: see e.g. $\S 7.2$ of my commutative algebra notes for more details. (Or see any other commutative algebra text, of course.)