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.

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    $\begingroup$ 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). $\endgroup$ Mar 29, 2020 at 8:49
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    $\begingroup$ 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 $\endgroup$ Mar 29, 2020 at 8:52
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    $\begingroup$ 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". $\endgroup$ Mar 29, 2020 at 9:32
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    $\begingroup$ 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 $P \rightarrow E$. $\endgroup$ Mar 29, 2020 at 10:23

1 Answer 1


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.


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