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 (
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.
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=N ⊕ L
M=oplusiMi
Mi
A\subsetI
MA=oplusiMi,NA=
MA
M\toN\hookrightarrowM
LA
J
B
C
J\subsetI
B,C\subsetak{F}
MJ=NJ ⊕ LJ
NJ,LJ
B,C
(J,B,C)\le(J',B',C')
J\subsetJ'
B\subsetB',C\subsetC'
(J,B,C)
J=I
N=NJ=oplusN'N'\inak{F}
I1\subsetI2\subset … \subsetI
I1\not\subsetJ
n\ge1
M | |
In |
\subset
N | |
In |
+
L | |
In |
\subset
M | |
In+1 |
I'=
infty | |
cup | |
0 |
In
J'=J\cupI'
MJ'=NJ' ⊕ LJ'.
\subset
NJ'
NJ+LJ+MI'\subsetNJ+MI'
NJ'\subsetMJ'
LJ'
Now,
NJ
M
MJ
M
NJ ⊕ M'=M
M'
NJ'=NJ ⊕ (M'\capNJ')
\widetilde{NJ}=M'\capNJ'
\widetilde{LJ}
MJ'=MJ ⊕ \widetilde{NJ} ⊕ \widetilde{LJ},
\widetilde{NJ} ⊕ \widetilde{LJ}\simeqMJ'/MJ\simeqMJ'
J'-J\subsetI'
(J,B,C)
\square
Proof: Let
l{G}
oplusiMi
F\subsetI
x\inN
H\inl{G}
x1,x2,...
N=H1 ⊕ N1
x1\inH1\inl{G}
x2=y+z
y\inH1,z\inN1
N1=H2 ⊕ N2
z\inH2\inl{G}
\{x1,x2\}\subsetH1 ⊕ H2
\square
Proof of the theorem: Let
N
F
F
ak{F}
N
N
\square
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.
2. ⇒ 1.
\Leftrightarrow
R2=R x R
R2=(0 x R) ⊕ M
R2=(R x 0) ⊕ M
( ⇒ )
1. ⇒ 2.
R2
\sigma:R2\toR,\sigma(a,b)=a-b
y=x-1
\sigma(x,y)=1
η:R\toR2,a\mapsto(ax,ay)
M
R2
\square