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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)

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Are you sure this is the question? What if R=Z and P, Q are <2>, <3>? The localizations are not equal, and I don't think they're isomorphic. – Gadi A May 22 '11 at 16:05
Are they really distinct? an element of $\mathbb{Z}_{(3)}$ is a fraction $\frac{x}{y}$ with y coprime to 3. This element is equal to $\frac{2x}{2y}$, which is an element of $(2)_{(3)}$. – Michalis May 22 '11 at 16:40
Michalis, your excellent response to Gadi's objection is easy to generalize to a proof of the general statement: just use that $P$ is not contained in $Q$. – Georges Elencwajg May 22 '11 at 17:10
@elgeorges: you're right :D I was a bit confused when I posed the question. – Michalis May 22 '11 at 17:15
up vote 6 down vote accepted

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.)

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Oh I shouldve seen that myself! thanks, I will definitely have a look at your notes. – Michalis May 22 '11 at 17:14

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