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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 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... –  BBischof Aug 1 '10 at 22:36
    
Wrap the whole formula in backticks ` (including the dollar signs) –  Mariano Suárez-Alvarez Aug 1 '10 at 22:36
    
@Mariano You must be some kind of sorcerer... –  BBischof Aug 1 '10 at 22:46
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By the way, the bullet does not help much in the notation :) –  Mariano Suárez-Alvarez Aug 1 '10 at 22:51
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Suppose $R=k[t]$ with its usual grading, and $M=R(1)$ is free of rank one generated in degree $1$. –  Mariano Suárez-Alvarez Aug 1 '10 at 22:58
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1 Answer

up vote 1 down vote accepted

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

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Thanks again friend :D –  BBischof Aug 1 '10 at 23:14
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