Let be a ring with 1. Let be a left unital -module, and let be a nonempty family of left unital -modules. Prove that, as abelian groups, and . Prove also that if is commutative, these pairs are -module isomorphic.
First we show that as abelian groups. Recall that for each , we have the canonical injection . Define for each the map by ; certainly each is well defined. By the universal property of direct products of abelian groups, there exists a unique group homomorphism such that , where denotes the th natural projection from a direct product. We claim that this is a group isomorphism.
Suppose , so . Then for each , so that for each . Thus for all . That is, applied to the th component of any element in is zero, for all . So , and thus . So is injective.
Now suppose . Define by ; this map is well defined since only finitely many terms of are nonzero. Moreover, it is clear that is an -module homomorphism, so in fact . Note that for all , . Thus for all , and so . Thus is surjective.
So is an isomorphism of abelian groups. Suppose now that is commutative, so that both and are naturally left -modules. For all and all module homomorphisms , we have . Thus is an isomorphism of left -modules.
Next we show that as abelian groups. For each , define by , where denotes the th canonical projection from a direct product. By the universal property of direct products, we have a unique group homomorphism such that for all . We claim that this is an isomorphism.
Suppose . So , and thus for all . Thus for all , so for all . That is, every coordinate of any element in the image of is 0. So . Thus , and so is injective.
Now let . Define by . Certainly is a module homomorphism. Note that, for all , . So for all , and we have . So is surjective, and thus an isomorphism of groups.
Finally, suppose is commutative. If and is a module homomorphism, then . Thus in this case is an isomorphism of -modules.