# Should one interested in commutative algebra learn the “noncommutative complications” of homological algebra?

I have found that in most books on homological algebra, the author(s) work in the general setting of noncommutative rings. This leads to several irritating complications. For example, the tensor product M⊗RNM\otimes_R N only makes sense if MM is a right RR-module and NN is a left RR-module.

Personally I am completely uninterested in noncommutative algebra. Because of this, whenever I have studied homological algebra in the past I simply ignored these details and mentally assumed all rings are commutative.

However, it has occurred to me that perhaps I may be crippling myself by doing this. There are certainly many places in mathematics where considering a more general setting provides richer information about the specific setting one is interested in. For example, allowing nilpotent coordinate rings in algebraic geometry allows one to construct useful maps to or from the integral domains that one is interested in, which can give otherwise unavailable information.

Generally speaking, I want to ask for some examples where being aware of the intricacies of noncommutative homological algebra is important for one interested in the study of commutative rings.

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That the tensor product only makes sense for modules of the correct variances is hardly an intricacy! I’d suggest you just learn homological algebra and stop worrying…
– Mariano Suárez-Álvarez♦
Oct 21 at 2:46

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Perhaps I should have clarified: The intricacy for me in this setting comes from the fact that simply considering the category of left RR-modules is not enough in the noncommutative setting, the way that one can simply consider the category of RR-modules in the commutative setting. This is because important constructions such as the tensor product of two left RR-modules may not be defined.
– Ethan MacBrugh
Oct 21 at 2:50

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It is enough or not depending on what you need, as everything. That dependence makes your question an essentially unanswererable one, unless you either have a specific objective â€”which you should explain so as to judge if the non commutative theory is neededor notâ€” or you are looking for a general purpose answer, which is then of course yes, just go learn homological algebra.
– Mariano Suárez-Álvarez♦
Oct 21 at 3:59

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A simple example is Galois cohomology, which is homological algebra of noncommutative rings (the group rings of Galois groups of field extensions) The objectives are usually commutative, yet you obviously need to know how to deal with non commutative stuff to do that!
– Mariano Suárez-Álvarez♦
Oct 21 at 4:03

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If you need to quickly learn some tools to apply them to commutative algebra problems, then in principle you may assume that RR is commutative, that all modules are left (or maybe right, if you like), and so on. Actually, a great introduction to homological algebra is an appendix to Eisenbud’s textbook on commutative algebra.

But then there are a couple of important points:

Distinguishing left and right modules and writing the correct formulas is not hard at all and may even lead to cleaner exposition of some topics. It is also nice to know where the commutativity of the base ring really matters in homological algebra (answer: almost nowhere in the very basic constructions and arguments).
You may be uninterested in noncommutative algebra, but there are some situations where noncommutative rings arise very naturally. For instance, group (co)homology is very important, and there you basically deal with modules over the group ring Z[G]\mathbb{Z} [G] for some GG:
Hn(G,A)=ExtnZ[G](Z,A),Hn(G,A)=TorZ[G]n(Z,A).H^n(G, A) = \operatorname{Ext}^n_{\mathbb{Z} [G]} (\mathbb{Z},A), \quad H_n (G,A) = \operatorname{Tor}^{\mathbb{Z} [G]}_n (\mathbb{Z}, A).
The ring Z[G]\mathbb{Z} [G] is noncommutative, if GG is. Of course, it is not very noncommutative, since it can be canonically identified with its opposite ring via g↦g−1g \mapsto g^{-1}. But still, you should be careful about left and right to avoid silly mistakes.
The same thing happens with Lie algebra (co)homology, as you basically work with modules over the universal enveloping algebra U(g)U (\mathfrak{g}):
Hn(g;M)=ExtnU(g)(R,M),Hn(g;M)=TorU(g)n(R,M).H^n (\mathfrak{g};M) = \operatorname{Ext}_{U (\mathfrak{g})}^n (R,M), \quad H_n (\mathfrak{g};M) = \operatorname{Tor}^{U (\mathfrak{g})}_n (R,M).
I believe other people can come up with more examples of noncommutative rings arising very naturally.
Besides, don’t forget that homological algebra is about arbitrary abelian categories (their derived categories, etc.), not just the ones coming from RR-modules… In many fundamental arguments, there won’t be any modules at all.

This is just a humble personal opinion that comes from my experience of teaching homological algebra: I just realized that if I wanted to talk about group or Lie algebra cohomology (to give some applications!), I had to deal with noncommutative rings from the very beginning, and also that noncommutativity of RR was not really affecting any basic arguments / definitions.