## Free modules over two sets of the same cardinality are isomorphic

Let be a ring with 1, and let and be sets of the same cardinality. Prove that the free modules and are isomorphic.

We begin with a lemma.

Lemma: Let be a ring with 1 and let be a set map. Note that , where is the free module on and is the inclusion map. By the universal prperty of free modules, the induced map is an -module homomorphism. If is injective, then so is , and if is surjective, then so is . Proof: Suppose is injective, and let . Now . Since is injective, the are distinct. Since is free on , we have for each . Thus , and so . So is injective. Suppose now that is surjective, and let . Since is surjective, we have for some (not necessarily distinct). Then , so that is surjective.

Since in our case is both injective and surjective, is bijective, hence an -module isomorphism.

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