# Is there any relation of injective modules to free modules?

Projective modules are direct summands of free modules. As i perceive it, projections and injections are dual notions. Based on that, i was looking whether there is a relation of injective modules to free modules (similar to the natural relation of projective modules to free modules) or to another kind of module that has potentially "dual" properties to that of a free module.

Any insights? Thank you :-)

-
A module $F$ is flat if and only if its dual module $F^{\vee} = \operatorname{Hom}_{R}{(F,R)}$ is injective. In particular $F^{\vee}$ is injective for projective (or free) $F$. –  t.b. Jul 26 '11 at 19:29

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.)

-
Thank you for your nice answer :-) –  Manos Jul 27 '11 at 19:07
I am working on the exercise and will let you know how it's going :-) –  Manos Jul 27 '11 at 19:13
So here is my solution: Assume $F$ is co-free on $X$ and that $|X|\ge 2$. Let $g_1:A \longrightarrow X$ and $g_2:A \longrightarrow X$ be set functions such that $g_1(0)=x_1$ and $g_2(0)=x_2$, with $x_1 \neq x_2$, which is possible since $X$ contains at least two elements. Assuming $F$ co-free, there exist $R$-homomorphisms $\tilde{g}_1 : A \longrightarrow F$ and $\tilde{g}_2 : A \longrightarrow F$ such that $g_1 = \iota \circ \tilde{g}_1$ and $g_2 = \iota \circ \tilde{g}_2$. Since $\tilde{g}_1(0)=0$ and similarly for $\tilde{g}_2$... –  Manos Jul 28 '11 at 16:03
... (since they are homomorphisms) we have $g_1(0)=\iota (\tilde{g}_1(0))$ and $g_2(0)=\iota (\tilde{g}_2(0))$, which yields $x_1=\iota(0)=x_2$, which is a contradiction. What do you think? –  Manos Jul 28 '11 at 16:07
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.