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Concept
Kaplansky's theorem on projective modules
Summary
In abstract algebra, Kaplansky's theorem on projective modules, first proven by Irving Kaplansky, states that a projective module over a local ring is free; where a not-necessarily-commutative ring is called local if for each element x, either x or 1 − x is a unit element. The theorem can also be formulated so to characterize a local ring (#Characterization of a local ring).
For a finite projective module over a commutative local ring, the theorem is an easy consequence of Nakayama's lemma. For the general case, the proof (both the original as well as later one) consists of the following two steps:
Observe that a projective module over an arbitrary ring is a direct sum of countably generated projective modules.
Show that a countably generated projective module over a local ring is free (by a "[reminiscence] of the proof of Nakayama's lemma").
The idea of the proof of the theorem was also later used by Hyman Bass to show big projective modules (under some mild conditions) are free. According to , Kaplansky's theorem "is very likely the inspiration for a major portion of the results" in the theory of semiperfect rings.
The proof of the theorem is based on two lemmas, both of which concern decompositions of modules and are of independent general interest.
Proof: Let N be a direct summand; i.e., . Using the assumption, we write where each is a countably generated submodule. For each subset , we write the image of under the projection and the same way. Now, consider the set of all triples (, , ) consisting of a subset and subsets such that and are the direct sums of the modules in . We give this set a partial ordering such that if and only if , . By Zorn's lemma, the set contains a maximal element . We shall show that ; i.e., . Suppose otherwise. Then we can inductively construct a sequence of at most countable subsets such that and for each integer ,
Let and . We claim:
The inclusion is trivial. Conversely, is the image of and so . The same is also true for . Hence, the claim is valid.
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