# Properties of injective modules

I am reading A course in Homological Algebra by Hilton and Stammbach. In the first chapter they showed that a $$\Lambda$$-module is projective iff it is a direct summand of a free module. They then defined the categorical dual of projective modules, which are injective modules as follows:

A $$\Lambda$$-module is injective if for every homomorphism $$\alpha:A\to I$$ and every monomorphism $$\mu:A \to B$$ there exists a homomorphism $$\beta: B \to I$$ such that $$\beta \mu = \alpha$$.

Then proceed to show the following characterization for when $$\Lambda$$ is a PID:

Let $$\Lambda$$ be a PID. A $$\Lambda$$-module is injective iff it is divisible.

Now this seems quite concerning to me because the characterization doesn't seem very "dual-like" to projective modules. Two questions natually arise:

1. Does being a divisible module have any categorical relations to being free or being a direct summand?

2. The characterization for injective modules is proved only for PIDs whereas the characterization for projective modules is true for all rings. Is there a generalization to all rings for the injective case, or is there a big-picture reason to why this fails?

Because of my interest in K-theory, I have also two more questions:

1. A special case of projective module is stably free module. Is there a categorical dual to stably-freeness and if so what's its relation to injectivity?

2. Projective modules are used in the construction of the $$K_0$$ group for rings, I'd like to know if injective modules have any significance in the K-theories of rings?

Update: Apparently I was too hasty in asking this question, as the next section of the book provides a better characterization, and that is

A $$\Lambda$$-module $$I$$ is injective iff it is a direct factor (coincides with direct summand in this case) of a cofree module.

This is the kind of result that I was looking for, but the definition of cofree seems even more enigmatic, it is defined to be direct products of $$\Lambda^* = \text{Hom}_\mathbb Z(\Lambda, \mathbb Q / \mathbb Z)$$, where $$\Lambda ^*$$ has the left module structure induced by the right module structure of $$\Lambda$$. I am very puzzled by this $$\mathbb Q / \mathbb Z$$.

I found a thread on MO about cofree modules. Todd explains that free modules does not have a formal dual notion. The definition of cofree with $$\mathbb Q/ \mathbb Z$$ involved is somewhat ad hoc and imprecise. Considering Captain Lama's comment, I will accept that duality in modules aren't perfect.

• I think the key point that explains why duality will not be as nice as you might hope for is: the opposite category of a category of modules is never a category of modules (even though it is an abelian category). Mar 29, 2020 at 8:49
• Just a quick answer to 2. (on top of Captain Lama's relevant comment, which I advise you to think about) : if you want something that is true for all rings and not just PIDs, you can prove : a $\Lambda$-module $A$ is injective if and only for every ideal $I\subset \Lambda$ and map $I\to A$, there is an extension $\Lambda \to A$. For principal ideals $(x)$ this corresponds precisely to saying that $A$ is $x$-divisible, so for PIDs you get the given characterization Mar 29, 2020 at 8:52
• A nice way to state things which does at least feel dual is: Projective iff all short exact sequences with it as last term splits. Injective iff same with "last" replaced by "first". Mar 29, 2020 at 9:32
• Another dualizable definition of projective is that for every morphism $P \rightarrow B$ and epimorphism $E \rightarrow B$ there is some (not necessarily unique) lift $P \rightarrow E$. Mar 29, 2020 at 10:23

This is not really an answer to the question, but is too long for a comment. You mention the module $$\mathbb{Q}/\mathbb{Z}$$, and this actually provides a nice algebraic duality between the injective modules and the flat modules over noetherian rings.
First off, this module is an injective cogenerator, so we have a faithful functor $$(-)^{d}:=\text{Hom}_{\mathbb{Z}}(-,\mathbb{Q}/\mathbb{Z}):R\text{-Mod}\to \text{Mod-}R$$ between the left and right $$R$$-modules (and vice versa). Moreover there are natural isomorphisms $$\text{Ext}_{R}^{j}(M,N^{d})\simeq \text{Tor}_{j}^{R}(M,N)^{d}$$ for all $$R$$-modules $$M$$ and $$N$$ and $$j<\infty$$, and $$\text{Ext}_{R}^{j}(A,B)^{d}\simeq \text{Tor}_{j}^{d}(A,B^{d})$$ for all finitely generated $$R$$-modules $$A$$, all $$R$$-modules $$B$$ and $$j<\infty$$. From these, you can indeed see that if $$M$$ is injective, then $$M^{d}$$ is flat and similarly if $$N$$ is flat then $$N^{d}$$ is injective. Obviously projective modules are also flat.
From this, you can actually recover the cofree situation: if $$M$$ is an injective $$R$$-module, then $$M^{d}$$ is flat and is therefore a direct limit of finitely generated free modules. In particular, there is a pure quotient $$\bigoplus_{I}F_{i}\to M^{d}\to 0$$ with each $$F_{i}$$ a finitely generated free module. Applying $$(-)^{d}$$ to this gives a split sequence $$0\to M^{dd} \to \prod_{I}F_{i}^{d},$$ hence $$M^{dd}$$ is a direct summand of a cofree module as each $$F_{i}^{d}$$ is cofree. Moreover, $$M$$ is a direct summand of $$M^{dd}$$ as it is injective, so is also a direct summand of a cofree module.
In fact, the duality $$(-)^{d}$$ applies to many more classes than flat and injective modules, and is a very useful object. It can also be replaced by any injective cogenerator for $$R$$-Mod.