Kaplansky's theorem on projective modules

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-necessary-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 of 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 of 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 interested 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. ${\displaystyle \square }$

See also

• Krull–Schmidt category

Notes

1. Anderson & Fuller 1992, Corollary 26.7.
2. Anderson & Fuller 1992, Proposition 15.15.
3. Matsumura, Theorem 2.5. harvnb error: no target: CITEREFMatsumura (help)
4. Lam, Part 1. § 1. harvnb error: no target: CITEREFLam (help)
5. Bass 1963 harvnb error: no target: CITEREFBass1963 (help)
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, 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
• H. Bass: Big projective modules are free, Illinois J. Math. 7(1963), 24-31.
• 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
• Y. Lam, Bass’s work in ring theory and projective modules [MR 1732042]
• Matsumura, Hideyuki (1989), Commutative Ring Theory, Cambridge Studies in Advanced Mathematics (2nd ed.), Cambridge University Press, ISBN 978-0-521-36764-6