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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.2011-05-22
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    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 $(2)_{(3)}$.2011-05-22
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    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 $Q$.2011-05-22
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    @elgeorges: you're right :D I was a bit confused when I posed the question.2011-05-22

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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.2011-05-22