Projective vs Injective Resolutions

Question

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?

Edit:

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?

Answer 1

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.

Answer 2

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.