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Let $R$ be a commutative ring and $M$ a simple $R$-module. Then $\mathfrak{m}=Ann(M)$ is a maximal ideal of $R$. Then it is known that $$ \mathrm{pdim}_{R}(M)=\mathrm{pdim}_{R_{\mathfrak{m}}}(M), $$ where $R_{\mathfrak{m}}$ is the localization of $R$ with respect to $\mathfrak{m}$.

Does anyone know the proof of this simple fact?

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Well, there's no kernel to the localization map $M \rightarrow M \otimes R_\mathfrak{m}$ and whatever resolution gives rise to the projective dimension over the localization is free, because projective is equivalent to free over local rings. I don't see how we pass from that resolution to one no longer over $R$, unfortunately. –  Kevin Carlson Aug 10 '12 at 4:57

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