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I guarantee there is an easy reference on this, but for some reason I cannot find it. If you can point me to a reference or just write a short proof for me, I would be very appreciative.

Given a graded ring $R_{\bullet}$ and a localization $R_{\bullet}^{*}$. We also have a graded $R_{\bullet}$-mod, $M_{\bullet}$.

So what I want to know; is $\left(R_{\bullet}^{*}\otimes M_{\bullet}\right)_0=\left(R_{\bullet}^{*}\right)_0\otimes \left(M_{\bullet}\right)_0$?

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    I cant get the dumb tex to work...2010-08-01
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    Wrap the whole formula in backticks \` (including the dollar signs)2010-08-01
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    @Mariano You must be some kind of sorcerer...2010-08-01
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    By the way, the bullet does not help much in the notation :)2010-08-01
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    Oops that was a typo, it should not be a module over the localization2010-08-01
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    Suppose $R=k[t]$ with its usual grading, and $M=R(1)$ is free of rank one generated in degree $1$.2010-08-01
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    Thank you, I guess there wont be a reference... I suppose you should post an answer.2010-08-01

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For a counterexample, take $R=k[t]$ with its usual grading and $M=R(1)$, the free module of rank one generated in degree $1$.