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-necessary-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. 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 consists of the following two steps:
Show that a countably generated projective module over a local ring is free.
The idea of the proof of the theorem was also later used by Hyman Bass to show big projective modules 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.
Proof
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. Now, is a direct summand of ; i.e., for some. Then, by modular law,. Set. Define in the same way. Then, using the early claim, we have: which implies that is countably generated as. This contradicts the maximality of. Proof: Let denote the family of modules that are isomorphic to modules of the form for some finite subset. The assertion is then implied by the following claim:
Given an element, there exists an that contains x and is a direct summand of N.
Indeed, assume the claim is valid. Then choose a sequence in N that is a generating set. Then using the claim, write where. Then we write where. We then decompose with. Note. Repeating this argument, in the end, we have: ; i.e.,. Hence, the proof reduces to proving the claim and the claim is a straightforward consequence of Azumaya's theorem. Proof of the theorem: Let be a projective module over a local ring. Then, by definition, it is a direct summand of some free module. This is in the family in Lemma 1; thus, is a direct sum of countably generated submodules, each a direct summand of F and thus projective. Hence, without loss of generality, we can assume is countably generated. Then Lemma 2 gives the theorem.
Characterization of a local ring
Kaplansky's theorem can be stated in such a way to give a characterization of a local ring. A direct summand is said to be maximal if it has an indecomposable complement. The implication is exactly Kaplansky's theorem and Azumaya's theorem. The converse follows from the following general fact, which is interested itself:
A ring R is local for each nonzero proper direct summand M of, either or.
is by Azumaya's theorem as in the proof of. Conversely, suppose has the above property and that an element x in R is given. Consider the linear map. Set. Then, which is to say splits and the image is a direct summand of. It follows easily from that the assumption that either x or -y is a unit element.
