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Since not all facts about projective and injective modules are not dual, i was wondering what similarities and differences are there between the information we get from projective resolutions and the information we get from injective resolutions of modules over commutative rings?


More specifically: i have a particular interest in the analysis of the complexity of the structure of a module over a commutative ring. Resolutions give information about that complexity. From that perspective, what is the relation between injective and projective resolutions? Are they equivalent? Do they complement each other?

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Categorically speaking, they are precisely the duals of injective modules... but I know what you mean if you're saying that not all facts about them are dual. So: I like this question. –  rschwieb Feb 22 '13 at 19:26
Instead of asking "which is more important," which is not a very objective question, would you consider revising it to "What similarities and differences are there between the info we get from proj resolutions and the info we get from inj resolutions?" I think that's what you're getting at, really. –  rschwieb Feb 22 '13 at 19:35
@rschwieb: I am interested in your first comment and i think i will make it a question. Tell me if you encourage: "since injective and projective modules are categorically dual, but not all facts about them are dual, then what is being lost by viewing the modules categorically?" –  Manos Feb 22 '13 at 19:44
I think the difference arises in our minds when we try to connect the behavior of the category of $R$ modules directly to the ring $R$ itself. Morita equivalence compares rings "by module category," and you probably know that there are some properties which are not Morita invariant. Another thing that is "not dual" about projs and injs is that all modules have an injective hull, but not every module has a projective cover. The (ring!) property that makes that happen is perfectness, but again that is a ring property, not a property of modules themselves. –  rschwieb Feb 22 '13 at 20:10
@Manos Yes, every module has a free resolution and an injective resolution. Not every module has a projective cover, but every module does have a flat cover. Rings are always free modules over themselves, but not all of them are injective modules over themselves. These are the asymmetries I always keep in mind when thinking of injectives and projectives. –  rschwieb Feb 22 '13 at 20:49

2 Answers 2

up vote 8 down vote accepted

Projective and injective are dual notions but facts about them aren't dual since the dual of a module category is almost never a module category. However, they do give similar information when "averaged" over all modules. Given a right $R$-module $M$ the projective dimension of $M$ is the least $n$ such that there is a projective resolution $0\to P_n \to ... \to P_0\to M\to 0$ and there is the evident notion of injective dimension. It happens that the supremum over all right $R$-modules of their projective dimension is the same as the supremum over all right $R$-modules of their injective dimension; this is called the "right global dimension"; you can replace "right" by "left" everywhere. If $R$ is right-noetherian then this concept also corresponds to flat dimension.

The natural place to look for this type of information is homological dimension theory, which will also show you some of the differences between projective and injective dimension. Typically some assumptions on the base ring will give you restrictions on either the projective or injective dimension (e.g. "grade zero lemma"), which could count as a kind of complexity.

Some nice treatments are Barbara Osofsky's "Homological Dimensions of Modules" and Kaplansky's "Commutative Rings", and the chapter four in Weibel's book on homological algebra. You should also check out MathOverflow Question 34704.

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Very cool!${}{}{}$ –  rschwieb Feb 22 '13 at 22:07
In order to support your answer I'd like to add this link. –  user26857 Feb 22 '13 at 22:43
+1 This is useful, thanks. –  Manos Feb 22 '13 at 22:53

To add to Jason’s answer, the Ext groups of an $ R $-module $ A $ can be computed using both injective and projective resolutions. For example, let $ P^{\bullet} $ be a projective resolution of $ A $: $$ \cdots \longrightarrow P^{2} \longrightarrow P^{1} \longrightarrow P^{0} \longrightarrow A \longrightarrow 0. $$ Let $ B $ be another $ R $-module. We then have the following (not necessarily exact) sequence: $$ 0 \longrightarrow {\text{Hom}_{R}}(P^{0},B) \longrightarrow {\text{Hom}_{R}}(P^{1},B) \longrightarrow {\text{Hom}_{R}}(P^{2},B) \longrightarrow \cdots. $$ As you know, computing the homology groups of the sequence above yields the sequence $ ({\text{Ext}_{R}^{n}}(A,B))_{n \in \mathbb{N}_{0}} $ of Ext groups.

Let us start with an injective resolution $ I^{\bullet} $ of $ B $ instead: $$ 0 \longrightarrow B \longrightarrow I^{0} \longrightarrow I^{1} \longrightarrow I^{2} \longrightarrow \cdots. $$ This leads to the following (not necessarily exact) sequence: $$ 0 \longrightarrow {\text{Hom}_{R}}(A,I^{0}) \longrightarrow {\text{Hom}_{R}}(A,I^{1}) \longrightarrow {\text{Hom}_{R}}(A,I^{2}) \longrightarrow \cdots. $$ Computing the homology groups of this sequence also yields $ ({\text{Ext}_{R}^{n}}(A,B))_{n \in \mathbb{N}_{0}} $ up to isomorphism.

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I don't see how the same thing could be said of Tor since in both variables the tensor functor is right-exact. –  Jason Polak Feb 22 '13 at 22:38
Yes, you’re right! I overlooked that. Thanks! –  Haskell Curry Feb 22 '13 at 22:55
There is a general balancing result for the derived functors of bifunctors. It can be applied both to $\mathrm{Ext} : C \times C^{op} \to C^{op}$ and to $\mathrm{Tor} : C \times C \to C$. Both these functors are right exact in each variable. Remember that projective objects in $C^{op}$ are injective objects in $C$. With respect to this there is really no difference between Tor and Ext! –  Martin Brandenburg Feb 22 '13 at 23:58

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