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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;[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 "[reminiscence] 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.

Lemma 1 — [6] Let ${\displaystyle {\mathfrak {F}}}$  denote the family of modules that are direct sums of some countably generated submodules (here modules can be those over a ring, a group or even a set of endomorphisms). If ${\displaystyle M}$  is in ${\displaystyle {\mathfrak {F}}}$ , then each direct summand of ${\displaystyle M}$  is also in ${\displaystyle {\mathfrak {F}}}$ .

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 ${\displaystyle M}$ ); i.e., ${\displaystyle N_{J}\oplus M'=M}$  for some ${\displaystyle M'}$ . Then, by modular law, ${\displaystyle N_{J'}=N_{J}\oplus (M'\cap N_{J'})}$ . Set ${\displaystyle {\widetilde {N_{J}}}=M'\cap N_{J'}}$ . Define ${\displaystyle {\widetilde {L_{J}}}}$  in the same way. Then, using the early claim, we have:

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

which implies that

${\displaystyle {\widetilde {N_{J}}}\oplus {\widetilde {L_{J}}}\simeq M_{J'}/M_{J}\simeq M_{J'-J}}$

is countably generated as ${\displaystyle J'-J\subset I'}$ . This contradicts the maximality of ${\displaystyle (J,B,C)}$ . ${\displaystyle \square }$

Lemma 2 — If ${\displaystyle M_{i},i\in I}$  are countably generated modules with local endomorphism rings and if ${\displaystyle N}$  is a countably generated module that is a direct summand of ${\displaystyle \bigoplus _{i\in I}M_{i}}$ , then ${\displaystyle N}$  is isomorphic to ${\displaystyle \bigoplus _{i\in I'}M_{i}}$  for some at most countable subset ${\displaystyle I'\subset I}$ .

Proof:[7] Let ${\displaystyle {\mathcal {G}}}$  denote the family of modules that are isomorphic to modules of the form ${\displaystyle \bigoplus _{i\in F}M_{i}}$  for some finite subset ${\displaystyle F\subset I}$ . The assertion is then implied by the following claim:

• Given an element ${\displaystyle x\in N}$ , there exists an ${\displaystyle H\in {\mathcal {G}}}$  that contains x and is a direct summand of N.

Indeed, assume the claim is valid. Then choose a sequence ${\displaystyle x_{1},x_{2},\dots }$  in N that is a generating set. Then using the claim, write ${\displaystyle N=H_{1}\oplus N_{1}}$  where ${\displaystyle x_{1}\in H_{1}\in {\mathcal {G}}}$ . Then we write ${\displaystyle x_{2}=y+z}$  where ${\displaystyle y\in H_{1},z\in N_{1}}$ . We then decompose ${\displaystyle N_{1}=H_{2}\oplus N_{2}}$  with ${\displaystyle z\in H_{2}\in {\mathcal {G}}}$ . Note ${\displaystyle \{x_{1},x_{2}\}\subset H_{1}\oplus H_{2}}$ . Repeating this argument, in the end, we have: ${\textstyle \{x_{1},x_{2},\dots \}\subset \bigoplus _{0}^{\infty }H_{n}}$ ; i.e., ${\textstyle N=\bigoplus _{0}^{\infty }H_{n}}$ . 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). ${\displaystyle \square }$

Proof of the theorem: Let ${\displaystyle N}$  be a projective module over a local ring. Then, by definition, it is a direct summand of some free module ${\displaystyle F}$ . This ${\displaystyle F}$  is in the family ${\displaystyle {\mathfrak {F}}}$  in Lemma 1; thus, ${\displaystyle N}$  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 ${\displaystyle N}$  is countably generated. Then Lemma 2 gives the theorem. ${\displaystyle \square }$

## 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.

Theorem — [8] Let R be a ring. Then the following are equivalent.

1. R is a local ring.
2. Every projective module over R is free and has an indecomposable decomposition ${\displaystyle M=\bigoplus _{i\in I}M_{i}}$  such that for each maximal direct summand L of M, there is a decomposition ${\displaystyle M={\Big (}\bigoplus _{j\in J}M_{j}{\Big )}\bigoplus L}$  for some subset ${\displaystyle J\subset I}$ .

The implication ${\displaystyle 1.\Rightarrow 2.}$  is exactly (usual) Kaplansky's theorem and Azumaya's theorem. The converse ${\displaystyle 2.\Rightarrow 1.}$  follows from the following general fact, which is interesting itself:

• A ring R is local ${\displaystyle \Leftrightarrow }$  for each nonzero proper direct summand M of ${\displaystyle R^{2}=R\times R}$ , either ${\displaystyle R^{2}=(0\times R)\oplus M}$  or ${\displaystyle R^{2}=(R\times 0)\oplus M}$ .

${\displaystyle (\Rightarrow )}$  is by Azumaya's theorem as in the proof of ${\displaystyle 1.\Rightarrow 2.}$ . Conversely, suppose ${\displaystyle R^{2}}$  has the above property and that an element x in R is given. Consider the linear map ${\displaystyle \sigma :R^{2}\to R,\,\sigma (a,b)=a-b}$ . Set ${\displaystyle y=x-1}$ . Then ${\displaystyle \sigma (x,y)=1}$ , which is to say ${\displaystyle \eta :R\to R^{2},a\mapsto (ax,ay)}$  splits and the image ${\displaystyle M}$  is a direct summand of ${\displaystyle R^{2}}$ . It follows easily from that the assumption that either x or -y is a unit element. ${\displaystyle \square }$

## Notes

1. ^ a b Anderson & Fuller 1992, Corollary 26.7.
2. ^ Anderson & Fuller 1992, Proposition 15.15.
3. ^ Matsumura 1989, Theorem 2.5.
4. ^ Lam 2000, Part 1. § 1.
5. ^ Bass 1963
6. ^ Anderson & Fuller 1992, Theorem 26.1.
7. ^ Anderson & Fuller 1992, Proof of Theorem 26.5.
8. ^ Anderson & Fuller 1992, Exercise 26.3.

## References

• Anderson, Frank W.; Fuller, Kent R. (1992), Rings and categories of modules, Graduate Texts in Mathematics, vol. 13 (2 ed.), New York: Springer-Verlag, pp. x+376, doi:10.1007/978-1-4612-4418-9, ISBN 0-387-97845-3, MR 1245487
• Bass, Hyman (February 28, 1963). "Big projective modules are free". Illinois Journal of Mathematics. 7 (1). University of Illinois at Champagne-Urbana: 24–31. doi:10.1215/ijm/1255637479.
• Kaplansky, Irving (1958), "Projective modules", Ann. of Math., 2, 68 (2): 372–377, doi:10.2307/1970252, hdl:10338.dmlcz/101124, JSTOR 1970252, MR 0100017
• Lam, T.Y. (2000). "Bass's work in ring theory and projective modules". arXiv:math/0002217. MR1732042
• Matsumura, Hideyuki (1989), Commutative Ring Theory, Cambridge Studies in Advanced Mathematics (2nd ed.), Cambridge University Press, ISBN 978-0-521-36764-6