Is there any relation of injective modules to free modules?

Solution 1:

Here is a relevant exercise in Hungerford's Algebra (Chapter, IV, Section 3, #11).

Exercise: If one attempted to dualize the notion of free module over a ring $R$ (and called the object so defined "co-free") the definition would read: An $R$-module $F$ is co-free on a set $X$ if there exists a function $\iota: F \rightarrow X$ such that for any $R$-module $A$ and any function $f: A \rightarrow X$, there exists a unique module homomorphism $\tilde{f}: A \rightarrow F$ such that $\iota \tilde{f} = f$. Show that for any set $X$ with $|X| \geq 2$ no such $R$-module $F$ exists. If $|X| = 1$, then $0$ is the only co-free module.

Thus a straightforward dualization of the notion of free module leads to a property which is too strong to actually exist. There are other, weaker, notions of "co-free module" which are designed to render the analogy projective:free::injective:co-free complete: see e.g. here. However, up to the limits of my own knowledge of the subject, co-free modules do not play a large role in commutative algebra. (They do have a cameo in the construction of "enough injectives": see $\S 3.6.4$ of my commutative algebra notes.)

Solution 2:

Well, like all modules, injective modules are quotients of free modules. But that is not quite what you want.

I suspect that the answer is that there is no good dual notion to "direct summand of free". This is not entirely surprising, given that there is no good notion of "co-free module", and that the duality between injective and projective modules is not really all that good. For one thing: every module has an injective envelope (e.g., Theorem 18.10 in Anderson and Fuller's Rings and Categories of Modules), but not every module has a projective cover.

For an example of a module with no projective cover, recall that a projective cover for $M$ is a pair $(P,p)$, where $P$ is projective, and $p\colon P\to M$ is a superfluous epimorphism; that is, $p$ is onto and if $K=\mathrm{ker}(p)$ and $K+L=P$, then $L=P$. Now take $M$ to be a finite $Z$-module; projective $\mathbb{Z}$-modules are free, and for every nonzero subgroup $H$ of $\mathbb{Z}^k$ there is a proper subgroup $L$ such that $H+L = \mathbb{Z}^k$, so no epimorphism $P\to M$ can be superfluous.