Let and be rings with 1. Let be a left -module, and let be an -bimodule.

- Define an action of on by . Prove that this action is well defined and makes into a right -module. Deduce that, via this action, is a right -module for any left -module ; this is called the
*dual module*to . - Let be considered as an -bimodule as usual. Show that, under the action defined in part (1), and are isomorphic as right -modules. Deduce that if is a finitely generated free left -module, then is a free right -module of the same rank.
- Show that if is a finitely generated projective left unital -module then its dual module is projective as a right -module.

- Note that, for all and , . So this action is well defined. To see that is a right -module, let and let .
- Note that, for all , . Thus .
- Note that, for all , . Thus .
- Note that, for all , . Thus .
- Note that, for all , , so that .

Thus, under this action, is a right unital -module. In particular, letting and (considering as an -bimodule in the usual way), we have that is a right unital -module.

- We define a mapping as follows: . Note that, for all , , so that is a homomorphism , and thus is well-defined. Now let . Note that for all , . So , so that is a homomorphism of right -modules. Now suppose . Then . In particular, , so that is injective. Now suppose is a left -module homomorphism. Note that, for all , , so that . That is, is surjective. Thus, as right -modules, we have . (Using this action on .)
Next we prove a lemma.

Lemma: Let and be unital rings. Let , , be left unital -modules, and let be an -bimodule. Let be a left -module homomorphism. Note that given by is well-defined. Moreover, is a right -module homomorphism (using the action from part (1)) and if is surjective, then is injective. Proof: Let and let . Note that, for all , . Thus , so that is a right -module homomorphism. Now suppose is surjective. If , then . Because is surjective, it is right cancellable, and we have . Thus is injective.

Now let be a finite index set, and let be a family of left unital -modules. Note that the projections are surjective. Using the lemma, for each we have an injective right unital -module homomorphism . By the universal property of direct sums, we have an injective module homomorphism . Now let . For each , we have the natural injection . Let . Certainly then . That is, is an isomorphism of right -modules, so that . (Using the action defined in (1)).

In particular, if is a free left -module of finite rank, then $latex . Note that this is a chain of right -module isomorphisms. That is, the dual module of is isomorphic to the free

*right*-module having the same rank as . - Suppose is a finitely generated, projective, left unital -module. Then is free for some , and moreover, we may choose so that is finitely generated. Now as right -modules, and using the action described in part (1), we have that is free. Thus the dual module is projective as a right -module.