Kaplansky's theorem on projective modules explained

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 (

  1. 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:

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.

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.,

M=NL

. Using the assumption, we write

M=oplusiMi

where each

Mi

is a countably generated submodule. For each subset

A\subsetI

, we write

MA=oplusiMi,NA=

the image of

MA

under the projection

M\toN\hookrightarrowM

and

LA

the same way. Now, consider the set of all triples (

J

,

B

,

C

) consisting of a subset

J\subsetI

and subsets

B,C\subsetak{F}

such that

MJ=NJLJ

and

NJ,LJ

are the direct sums of the modules in

B,C

. We give this set a partial ordering such that

(J,B,C)\le(J',B',C')

if and only if

J\subsetJ'

,

B\subsetB',C\subsetC'

. By Zorn's lemma, the set contains a maximal element

(J,B,C)

. We shall show that

J=I

; i.e.,

N=NJ=oplusN'N'\inak{F}

. Suppose otherwise. Then we can inductively construct a sequence of at most countable subsets

I1\subsetI2\subset\subsetI

such that

I1\not\subsetJ

and for each integer

n\ge1

,
M
In

\subset

N
In

+

L
In

\subset

M
In+1
.Let

I'=

infty
cup
0

In

and

J'=J\cupI'

. We claim:

MJ'=NJ'LJ'.

The inclusion

\subset

is trivial. Conversely,

NJ'

is the image of

NJ+LJ+MI'\subsetNJ+MI'

and so

NJ'\subsetMJ'

. The same is also true for

LJ'

. Hence, the claim is valid.

Now,

NJ

is a direct summand of

M

(since it is a summand of

MJ

, which is a summand of

M

); i.e.,

NJM'=M

for some

M'

. Then, by modular law,

NJ'=NJ(M'\capNJ')

. Set

\widetilde{NJ}=M'\capNJ'

. Define

\widetilde{LJ}

in the same way. Then, using the early claim, we have:

MJ'=MJ\widetilde{NJ}\widetilde{LJ},

which implies that

\widetilde{NJ}\widetilde{LJ}\simeqMJ'/MJ\simeqMJ'

is countably generated as

J'-J\subsetI'

. This contradicts the maximality of

(J,B,C)

.

\square

Proof: Let

l{G}

denote the family of modules that are isomorphic to modules of the form

oplusiMi

for some finite subset

F\subsetI

. The assertion is then implied by the following claim:

x\inN

, there exists an

H\inl{G}

that contains x and is a direct summand of N.Indeed, assume the claim is valid. Then choose a sequence

x1,x2,...

in N that is a generating set. Then using the claim, write

N=H1N1

where

x1\inH1\inl{G}

. Then we write

x2=y+z

where

y\inH1,z\inN1

. We then decompose

N1=H2N2

with

z\inH2\inl{G}

. Note

\{x1,x2\}\subsetH1H2

. Repeating this argument, in the end, we have: \ \subset \bigoplus_0^\infty H_n; i.e., N = \bigoplus_0^\infty H_n. Hence, the proof reduces to proving the claim and the claim is a straightforward consequence of Azumaya's theorem (see the linked article for the argument).

\square

Proof of the theorem: Let

N

be a projective module over a local ring. Then, by definition, it is a direct summand of some free module

F

. This

F

is in the family

ak{F}

in Lemma 1; thus,

N

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

N

is countably generated. Then Lemma 2 gives the theorem.

\square

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

1.2.

is exactly (usual) Kaplansky's theorem and Azumaya's theorem. The converse

2.1.

follows from the following general fact, which is interesting itself:

\Leftrightarrow

for each nonzero proper direct summand M of

R2=R x R

, either

R2=(0 x R)M

or

R2=(R x 0)M

.

()

is by Azumaya's theorem as in the proof of

1.2.

. Conversely, suppose

R2

has the above property and that an element x in R is given. Consider the linear map

\sigma:R2\toR,\sigma(a,b)=a-b

. Set

y=x-1

. Then

\sigma(x,y)=1

, which is to say

η:R\toR2,a\mapsto(ax,ay)

splits and the image

M

is a direct summand of

R2

. It follows easily from that the assumption that either x or -y is a unit element.

\square

See also

References