# Rings and Modules in Commutative Algebra

It seems to me that in history of commutative algebra, you come from interest in rings and get more attention of modules by the time.

In my understanding this is because rings build an "ugly" category where you can't look at exact sequences and so you can't use homological algebra, but you can do this for modules and you can watch every ring as a module over itself. So you learn about rings by watching moules. Is this understanding of historical development correct?

We pay some price for working in this way. For getting very nice arguments from homological algebra, we must do some "dirty work" to get from modules back to rings.

As an example, localization and factorication behave well for modules because the localization functor is exact and we can do factorication with exact sequences. But to see that it also works for rings, we must realise by "ugly calculation" that we can do the same mappings as for the moduls also for the rings. This step is often skipped because it's ugly and seems obvious.

But is there a simple argument, why we needn't be frightened of the "ugly part" and why it's "obvious" in the relevant cases?

• It sounds as if you are assuming that homological algebra was around and well understood even before the concept of a ring got studied and that therefore it was used to understand rings. However, I'm not sure if that is true, I would rather think that they developed in parallel (e.g. first understand rings and ideals a little bit, then do some module theory, then try to apply this to rings and ideals, etc.). – Dirk Nov 24 '17 at 7:49
• It's not clear what you're asking, I'm not aware of any "ugly" part to doing factorization with rings. I think when authors skip it they do so because it's considered pretty basic, not because it's ugly. And I certainly don't think anyone is frightened by it. So what exactly are you looking for in an answer? – Jim Nov 27 '17 at 18:32
• As I understand it, it's nice to go from rings to moduls because we can use tools of homological algebra there. But it isn't clear for me, why we can be sure, that everything works when we go back from moduls to rings. If two rings are isomorphic as moduls, why should they be isomorphic as rings? – user302982 Nov 27 '17 at 18:46
• They shouldn't. $k[x]$ and $k[x, y]$ are isomorphic as $k$-modules. And in general what are you considering rings as modules over? As a module over one of the rings? I can also make $k[x, y]$ into a $k[x]$ module that's isomorphic to $k[x]$, because I'm just gonna throw away the multiplication of $k[x, y]$ and use that it's isomorphic to $k[x]$ as a vector space. – Jim Nov 30 '17 at 0:36