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;^[1] where a not-necessarily-commutative ring is called local if for each element x, either x or 1 − x is a unit element.^[2] 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.^[3] 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 " of the proof of Nakayama's lemma"^[4]).

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.^[5] According to (Anderson & Fuller 1992), Kaplansky's theorem "is very likely the inspiration for a major portion of the results" in the theory of semiperfect rings.^[1]

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., ${\displaystyle M=N\oplus L}$. Using the assumption, we write ${\displaystyle M=\bigoplus _{i\in I}M_{i}}$ where each ${\displaystyle M_{i}}$ is a countably generated submodule. For each subset ${\displaystyle A\subset I}$, we write ${\displaystyle M_{A}=\bigoplus _{i\in A}M_{i},N_{A}=}$ the image of ${\displaystyle M_{A}}$ under the projection ${\displaystyle M\to N\hookrightarrow M}$ and ${\displaystyle L_{A}}$ the same way. Now, consider the set of all triples (${\displaystyle J}$, ${\displaystyle B}$, ${\displaystyle C}$) consisting of a subset ${\displaystyle J\subset I}$ and subsets ${\displaystyle B,C\subset {\mathfrak {F}}}$ such that ${\displaystyle M_{J}=N_{J}\oplus L_{J}}$ and ${\displaystyle N_{J},L_{J}}$ are the direct sums of the modules in ${\displaystyle B,C}$. We give this set a partial ordering such that ${\displaystyle (J,B,C)\leq (J',B',C')}$ if and only if ${\displaystyle J\subset J'}$, ${\displaystyle B\subset B',C\subset C'}$. By Zorn's lemma, the set contains a maximal element ${\displaystyle (J,B,C)}$. We shall show that ${\displaystyle J=I}$; i.e., ${\displaystyle N=N_{J}=\bigoplus _{N'\in B}N'\in {\mathfrak {F}}}$. Suppose otherwise. Then we can inductively construct a sequence of at most countable subsets ${\displaystyle I_{1}\subset I_{2}\subset \cdots \subset I}$ such that ${\displaystyle I_{1}\not \subset J}$ and for each integer ${\displaystyle n\geq 1}$,

${\displaystyle M_{I_{n}}\subset N_{I_{n}}+L_{I_{n}}\subset M_{I_{n+1}}}$.

Let ${\displaystyle I'=\bigcup _{0}^{\infty }I_{n}}$ and ${\displaystyle J'=J\cup I'}$. We claim:

${\displaystyle M_{J'}=N_{J'}\oplus L_{J'}.}$

The inclusion ${\displaystyle \subset }$ is trivial. Conversely, ${\displaystyle N_{J'}}$ is the image of ${\displaystyle N_{J}+L_{J}+M_{I'}\subset N_{J}+M_{I'}}$ and so ${\displaystyle N_{J'}\subset M_{J'}}$. The same is also true for ${\displaystyle L_{J'}}$. Hence, the claim is valid.

Now, ${\displaystyle N_{J}}$ is a direct summand of ${\displaystyle M}$ (since it is a summand of ${\displaystyle M_{J}}$, which is a summand of $$Zdroj:https://en.wikipedia.org?pojem=Kaplansky's_theorem_on_projective_modules
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