# Why are injective modules more complicated than projective modules?

For beginners in homological algebra, it is a fact of life that injective modules seems to be more mysterious than projective modules. For example, for finitely generated modules over a noetherian ring, projective resolution can be taken as resolution by free modules of finite rank, but I don't see how one can easily write down injective resolutions.

I'm wondering if there is a deep reason behind this. What makes injective modules so complicated?

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I don't quite agree. Injective envelopes occur more easily than projective covers, and Grothendieck abelian categories have injectives but need not have projectives. Some people find easier to think of projectives because they think of free objects, which behave like vector spaces to some extent. But once you're into the abelian category world, injectives are even easier. – Fernando Muro Jul 27 '12 at 0:55
why is Q more complicated than Z? – roy smith Jul 27 '12 at 1:05
@roy Do you mean why is Q/Z more complicated than Z? I imagine you are thinking that every abelian group has a projective resolution by direct sums of Z, and similarly, every abelian group has an injective resolution by products of Q/Z? – name Jul 27 '12 at 1:36
I myself think $\mathbb Q/\mathbb Z$ is more complicated than $\mathbb Z$. – paul garrett Jul 27 '12 at 1:53
There is, in fact, a classification of sorts for injective modules over a commutative noetherian ring; I believe this is discussed in Maclane's book on homological algebra (but probably in many others as well). The basic idea is this: every $R$-module $M$ has an "injective envelope"; this is an injection of modules $M \hookrightarrow I(M)$ such that every nonzero submodule of $I(M)$ intersects $M$. One can show that $I(M)$ is injective, and that every injective module containing $M$ also contains a (non-unique) copy of $I(M)$. Then an injective module is precisely a direct sum of... – Charles Staats Jul 27 '12 at 15:54

## 2 Answers

Injective modules are of course just projective modules in the opposite category, so it seems to me that the question really is "why is the opposite of a module category more complicated than a module category?" Probably this is because the opposite of a module category is almost never itself a module category (see this MO question). It embeds into a module category by Freyd-Mitchell, but this is quite noncanonical.

For the sake of concreteness, by Pontrjagin duality $\text{Ab}^{op}$ itself is equivalent to the category of compact (Hausdorff) abelian groups (which embeds into $\text{Ab}$ itself but this is not too useful of an embedding for our purposes). An injective abelian group dualizes to a projective object in $\text{Ab}^{op}$, and it is not so straightforward as in a module category to find a projective object here. The simplest nontrivial thing that deserves to be called a free object is the Bohr compactification of $\mathbb{Z}$ (which dualizes to $S^1$ with the discrete topology). The injective abelian group $\mathbb{Q}/\mathbb{Z}$, as a filtered colimit of the groups $\mu_n$, dualizes to a cofiltered limit giving the profinite integers $\hat{\mathbb{Z}}$.

This gives one way to find an injective resolution of an abelian group by following a recipe exactly analogous to the free module recipe, but in $\text{Ab}^{op}$: find a projective resolution of its Pontrjagin dual by products of copies of the Bohr compactification of $\mathbb{Z}$, then dualize it!

Edit: Steven Landsburg makes the following comment below:

But I thought that another part of your answer was that in the categories we often choose to look at (module categories) projectives might look simpler, though in the opposite categories it's the injectives that look simpler. That leaves the question of why it's the projectives that are simpler in the categories we're naturally led to look at.

My revised revised answer is that free objects are projective, and free objects are simpler in the categories we're naturally led to look at.

Let $C$ be a category and let $U : C \to \text{Set}$ be a faithful functor. If $U$ has a left adjoint $F$, then $U$ preserves monomorphisms, so the monomorphisms in $C$ are precisely the maps which are injective on underlying sets. Thus to find projective objects it suffices to find objects with the lifting property. Now, for a set $S$, $F(S)$ clearly has the lifting property (hence is projective) because $S$ has the lifting property in $\text{Set}$ (in other words, is projective).

The categories we're naturally led to look at, such as module categories, are usually concretely defined so are equipped with a faithful functor to $\text{Set}$, and the corresponding objects $F(S)$ usually exist and provide a plentiful supply of projectives in $C$. (Without the functors $U$ and $F$ it is not clear why $C$ should have any projective objects whatsoever, and it might not: for an abelian category example, take $C = \text{FinAb}$.)

On the other hand, the opposite of a category, even one equipped with a faithful functor to $\text{Set}$, doesn't itself come equipped with a faithful functor to $\text{Set}$ (instead it comes equipped with a faithful functor to $\text{Set}^{op}$, which is harder to understand) and finding one with a left adjoint (as in the Pontrjagin duality example above) might be difficult.

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compact abelian topological groups. But I guess finite groups are interesting too... – David Roberts Jul 27 '12 at 1:35
@David: I don't follow. A finite group is also a compact abelian group with the discrete topology. (But thanks for reminding me I forgot to include Hausdorff.) – Qiaochu Yuan Jul 27 '12 at 1:38
A paraphrase: while reversing arrows is abstractly nothing-at-all, for concrete-ish categories (not necessarily refering, as "concrete" sometimes does, to categories whose objects meaningful underlying sets) the concrete-ish interpretation of the dual category is typically significantly different. The simplest case is that projective $\mathbb Z$-modules are easily given by $\mathbb Z$ and sums thereof, while injective $\mathbb Z$ modules are trickier both to exhibit, and trickier to prove are truly injective. I suppose in a different universe things might be more symmetrical? – paul garrett Jul 27 '12 at 1:56
@Steven: part of my answer is that I disagree with the premise of the question (if interpreted as "why are injective objects more complicated than projective objects?") because injectivity and projectivity are dual. Injective objects in $\text{Ab}^{op}$ are exactly as complicated as projective objects in $\text{Ab}$. – Qiaochu Yuan Jul 27 '12 at 2:09
Every object of $Set$ is projective iff you assume the axiom of choice. See also arxiv.org/abs/1111.5180, "Are There Enough Injective Sets?", where they authors say that every non-empty set is injective in the category of ZF-sets. – David Roberts Jul 27 '12 at 3:20

I'll give you my more pedestrian reason. The proofs of existence of injective resolutions require the axiom of choice, in one form or another. Translation: these proofs are not constructive, so there are no general algorithms for producing such objects. This becomes a painful issue in concrete situations. This has similarities with another famous existence result, the Hahn-Banach theorem which postulates the existence of continuous linear functionals with certain properties. It is particularly useful for existence theorems for PDE's. Unfortunately it gives you no guide for finding those solutions.

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A remark about the existence issue of injective resolutions: Everything comes down to prove that $\mathbb{Q}/\mathbb{Z}$ is injective; the rest of the proof is formal. So there is no problem to write down injective resolutions: Rather it requires AC in order to prove that they are indeed injective resolutions. – Martin Brandenburg Jul 27 '12 at 9:02
Does not the existence of projective resolutions also require the axiom of choice? I used to think that one needs AC in order to show that a free module is projective. – Leonid Positselski Jul 27 '12 at 9:26
But one does not need AC if one only needs to know that finitely generated free modules are projective. And the original question mentions Noetherian rings and finitely generated modules. OK. – Leonid Positselski Jul 27 '12 at 9:30
@Martin: Thanks for the clarifying comments. Algebra is really not my thing. In certain concrete topological problems I had to construct extensions of morphisms whose existence was postulated by the injectivity condition and I had to scratch my head. I felt the same when I tried to produce an extension of a linear functional whose existence was postulated by Hahn-Banach. – Liviu Nicolaescu Jul 27 '12 at 10:24
The axiom of choice is also necessary to show that any vector space is free. – Fernando Muro Jul 31 '12 at 16:00