Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\left(R_{\bullet}^{*}\otimes M_{\bullet}\right)_0=\left(R_{\bullet}^{*}\right)_0\otimes \left(M_{\bullet}\right)_0$?

share|cite|improve this question
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
By the way, the bullet does not help much in the notation :) – Mariano Suárez-Alvarez Aug 1 '10 at 22:51
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
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 $1$.

share|cite|improve this answer
Thanks again friend :D – BBischof Aug 1 '10 at 23:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.