5
$\begingroup$

Note that if we have a ring $R$, we can talk about $R$-modules, and if we have a ring homomorphism $R\to S$, there is a map from $R$-modules to $S$-modules given by $-\otimes_RS$ (just assume everything is a bi-module, or adjust things or whatever). So is there some functor $(-)-$modules$:\underline{Rng}\to\mathcal{C}$ where $\mathcal{C}$ is some kind of category whose objects are categories of modules or something? Does this question even make sense??

Thanks! Jon

  • 2
    You are associating to each (commutative, to make it simpler) ring $R$ a category, $R$-$\mathcal{M}od$, and to each morphism $R\to S$ a functor $R-\mathcal{M}od\to S-\mathcal{M}od$; you would need to check that it behaves well with respect to composition, but (assuming suitable set-theoretic axioms if your category theory is based on set theory) you could say that $\Box-\mathcal{M}od$ is a functor from the category of commutative rings to the category whose objects are categories of modules over commutative rings and whose morphisms are functors between those categories.2011-10-10
  • 1
    I would be concerned a bit about set-theoretic issues here, because there is [technically no "category of all categories"](http://en.wikipedia.org/wiki/Category_of_small_categories) that could be the target of this functor (perhaps you need to utter the phrase "[Grothendieck universe](http://en.wikipedia.org/wiki/Grothendieck_universe)", this appears to help sometimes), and I am not sure things get much better if one restricts to [abelian categories](http://en.wikipedia.org/wiki/Mitchell%27s_embedding_theorem). (I am unfortunately entirely ignorant about these things though)2011-10-10
  • 1
    Thanks guys. Yeah, I'm afraid I don't entirely understand the set-theoretic issues here myself. I guess I often see people say things like "Grothendieck Universe" and then ignore such issues, so perhaps I'll do the same thing. I guess we'd have to have a functor then from a small category into a large category or something?2011-10-10
  • 0
    Recently someone explained to me the "descent" problem in terms of $R$ and $S$ modules. I'm not familiar with "descent" problems and was wondering if there is some nice generalization categorically, i.e. functors inducing adjoint relations and then some relationship to (co)monadicity.2011-10-10
  • 2
    You should note that $R$-Mod is a category and not a set and the tensor product $S\otimes_R -$ is not uniquely defined but defined only up to a unique isomorphism. So what you looking at is not a really a functor $Ring \to Cat$ but a lax 2-functor from the category of rings to the 2-category of categories or, the category of modules $Mod$ of couples $(R, M)$ of a ring $R$ and an $R$-module $M$ which, in the terminology of SGA1, is fibered over the category of rings by the functor $p: Mod \to Ring$, $(R,M) \mapsto R$.2011-10-10
  • 0
    Wow thanks that's really interesting. I know next to nothing about lax 2-functors. Is this a subject of interest to people, or some special case of an interesting situation?2011-10-10
  • 0
    Modulo various set-theoretic issues, it is possible to make it into a strict 1-functor by giving an explicit construction of $S \otimes_R M$ for each $R$-module $M$ and each ring homomorphism $R \to S$. However, user8882's comments are probably more in the spirit of category theory and algebraic geometry than what I'm saying.2011-10-11
  • 0
    It has also been suggested to me by John Lind that we can consider the category $\mathscr{Mod}$ as a bi-category. However, he said that it is not in fact a 2-category, as we don't have strict associativity on the 2-cells. But I'm not sure. However, this has all been very enlightening.2011-10-12
  • 0
    @JBeardz Maybe you want to write up what you learned from the comments to an answer, so that the question gets removed from the unanswered tab? If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-09
  • 0
    @YBL Can you consider making your comments into an answer, since you seem to have a good perspective on the problem?2013-07-26
  • 0
    Sorry all, I've got a lot of unanswered questions out there, with answers more or less in the comments. I'd love for other people to turn these comments into answers, but it's sort of an ongoing project of mine to try to do this, at some point, but I worry it may not be until I finish with writing my thesis.2013-07-26

0 Answers 0