Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In an example I have worked out for my work, I have constructed a category whose objects are graded $R$-modules (where $R$ is a graded ring), and with morphisms the usual morphisms quotient the following class of morphisms:

$\Sigma=\left\lbrace f\in \hom_{\text{gr}R\text{-mod}}\left(A,B\right) \ | \ \ker\left(f\right)_0\neq 0, \ \mathrm{coker}\left(f\right)_0\neq 0\right\rbrace$

(by quotient I mean simply that this class of morphisms are isomorphisms, thus creating an equivalence relation) I am wondering if this category has a better (more canonical) description, or if I can show it is equivalent to some other interesting category.


share|cite|improve this question
Since $\Sigma$ isn't an ideal in the category of graded $R$-modules, I don't think that quotienting by it makes much sense. – Rasmus Aug 13 '10 at 10:38
Do you mean to make the maps in $\Sigma$ all the zero maps (which is quotienting) or do you mean to make the maps in $\Sigma$ isomorphisms (which is localizing)? As $\Sigma$ is not an ideal the first isn't well defined, and as $\Sigma$ can contain the zero map between modules the second doesn't seem to make much sense either. – Jim Feb 27 '13 at 7:29
This is very weird, really: you are making all modules with non-zero zero component isomorphic... Does this really make sense in some context? – Mariano Suárez-Alvarez Jun 30 '13 at 2:54
@MarianoSuárez-Alvarez I was trying to get the altruist badge, so I found the oldest unanswered question and put a bounty on it. It makes sense that this has been unanswered for three years... – Brian Rushton Jul 4 '13 at 3:21
Well, I don't remember. I do remember that I was localizing by these morphisms and that was what I meant by quotienting. Judging by the date, I was probably trying to work out a ncag example. I don't have a clue what it was! – BBischof Jul 6 '13 at 13:20

Your Answer


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

Browse other questions tagged or ask your own question.