12
$\begingroup$

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.

Thanks!

  • 0
    Since $\Sigma$ isn't an ideal in the category of graded $R$-modules, I don't think that quotienting by it makes much sense.2010-08-13
  • 2
    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.2013-02-27
  • 6
    This is very weird, really: you are making all modules with non-zero zero component isomorphic... Does this *really* make sense in some context?2013-06-30
  • 0
    @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...2013-07-04
  • 0
    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!2013-07-06

0 Answers 0