Math. Log. Quart. 51, No. 2, 187 – 190 (2005) / DOI 10.1002/malq.200410018 / www.mlq-journal.org

If vector spaces are projective modules then multiple choice holds

Paul Howard∗

Department of Mathematics, Eastern Michigan University, Ypsilanti, MI 48197, U. S. A.

Received 23 June 2004, accepted 21 October 2004Published online 31 January 2005

Key words Axiom of multiple choice, vector space, projective module.MSC (2000) 03E30, 03E25, 03E35

We show that the assertion that every vector space is a projective module implies the axiom of multiple choiceand that the reverse implication does not hold in set theory weakened to permit the existence of atoms.

c© 2005 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

1 Introduction

The usual proofs of the assertion that every vector space is a projective module (see, for example, [2]) use thevector space basis theorem (every vector space has a basis). A. Blass [1] has shown that the vector space basistheorem is equivalent to the axiom of choice (AC) in Zermelo-Fraenkel Set Theory (ZF) and so it is natural toask whether the use of the axiom of choice in the proof that every vector space is a projective module can beeliminated or if some form ’of choice is essential. Our primary purpose is to answer this question by showingthat the assertion “vector spaces are projective modules” implies the axiom of multiple choice (MC) in Zermelo-Fraenkel Set Theory weakened to permit the existence of atoms (ZFU) and that the reverse implication does nothold in this weaker theory. Since MC implies AC in ZF, it follows that the assertion “vector spaces are projectivemodules” implies AC in ZF.

2 Definitions and preliminary results

We refer the reader to [2] for definitions of the basic terms that will be used in the paper, specifically R-module,where R is a ring with 1 ([2, p. 318]) (we note that if R is a field, then an R-module is a vector space over R),R-module homomorphism ([2, p. 326]), direct sum of R-modules ([2, p. 334]), free R-module ([2, p. 335]), andexact sequence ([2, p. 359 ff.]). If M and N are R-modules, we also follow [2] in letting HomR(M,N) be theset of all R-module homomorphisms from M to N . We note that HomR(M,N) is an Abelian group, where forψ and ϕ in HomR(M,N) we define (ψ + ϕ)(m) = ψ(m) + ϕ(m) for all m ∈M . For convenience we give thedefinition of the notion of short exact sequence.

Definition 2.1 If A, B and C are R-modules (respectively groups) and ψ : A −→ B and ϕ : B −→ C

are R-module homomorphisms (respectively group homomorphisms) then 0 −→ Aψ−→ B

ϕ−→ C −→ 0 isa short exact sequence if ψ is one-to-one, ϕ is onto C and ran(ψ) = ker(ϕ) (where ran(ψ) denotes the set{ψ(a) : a ∈ dom(ψ)})

We now proceed to the definition of “projective” for modules as given in [2]. There the following propositionis proved ([2, Proposition 30, p. 369]).

Proposition 2.2 Let R be a ring and P an R-module. Then the following are equivalent:

(1) For any R modules L, M and N , if 0 −→ Lψ−→ M

ϕ−→ N −→ 0 is a short exact sequence, then

0 −→ HomR(P,L)ψ′−→ HomR(P,M)

ϕ′−→ HomR(P,N) −→ 0 is also a short exact sequence (here ψ′

∗ e-mail: phoward@emunix.emich.edu

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188 P. Howard: If vector spaces are projective modules then multiple choice holds

denotes the homomorphism from HomR(P,L) to HomR(P,M) defined by ψ′(f) = ψ ◦ f for f ∈ HomR(P,L),where ◦ is function composition, and ϕ′ is defined similarly).

(2) For any R modules M and N , if ϕ : M −→ N and f : P −→ N are R-module homomorphisms, thenthere is an F ∈ HomR(P,M) for which ϕ ◦ F = f .

(3) If P is a quotient of the R-module M , then P is isomorphic to a direct summand of M .

(4) P is a direct summand of a free R-module.

An R-module is defined (in [2]) to be projective if it satisfies any of the conditions (1) through (4) in Proposi-tion 2.2.

Although the context for the proof of Proposition 2.2 is set theory with the axiom of choice, the proof of theequivalence of (1), (2) and (3) does not depend on the axiom of choice. Nor does the proof that (1), (2) and (3)imply (4). However the proof that (4) implies (1), (2) and (3) does use the axiom of ’choice and, as we shall showshortly, this use of choice cannot be eliminated. Therefore, since the context of this paper is set theory withoutthe axiom of choice, we will have two definitions of projective.

Definition 2.3 If R is a ring and P is an R-module, then

1. P is a projective R-module if condition (2) (or (1) or (3)) of Proposition 2.2 holds.

2. P is a projective∗ R-module if condition (4) holds.

We shall consider five statements all of which are known to be consequences of the axiom of choice (AC).They are

MC (The axiom of multiple choice: For every set X of non-empty sets, there is a function withdomain X such that for all y ∈ X , f(y) is a non-empty, finite subset of y. (Form 67 in [4].)

VBT The vector space basis theorem: Every vector space over a field has a basis. (Form 66 in [4].)

SMC The strong multiple choice axiom: For every prime natural number r and every set X of non-empty sets, there is a function f such that for all y ∈ X , f(y) is a finite non-empty subset of yand r is not a divisor of |f(y)|.

VPM Every vector space is a projective module.

VPM∗ Every vector space is a projective∗ module.

It is known that in ZF, MC and VBT are each equivalent to the axiom of choice. In the weaker theory ZFU(ZF weakened to permit the existence of atoms) the following are known: AC implies VBT, VBT implies MC,the latter implication is not reversible, SMC implies MC and this implication is not reversible, and MC is not atheorem of ZFU (see [4], [5], [3], and [6]) We also note that the proof of the assertion that every vector space isa projective module (say in [2]) is a proof in ZFU that VBT implies VPM∗.

3 The Results

Theorem 3.1 In ZFU (without the axiom of choice) VBT + SMC imply VPM.

P r o o f. Assume that R is a field and that P is a vector space over R. We will prove condition (2) of Propo-sition 2.2. Assume that M and N are vector spaces over R and assume that ϕ and f are in HomR(M,N) andHomR(P,N), respectively. Let B be a basis for P and, for each b ∈ B, let

yb = ϕ−1[{f(b)}] = {m ∈M : ϕ(m) = f(b)}.

Since B is a basis for P , given a function G : B −→ M there is a unique homomorphismG∗ ∈ HomR(P,M)extending G. We would like to construct a function G : B −→ M such that for each b ∈ B, G(b) ∈ yb. It isstraightforward to prove that for such a G, the homomorphism F = G∗ satisfies ϕ ◦ F = f and we leave theproof to the reader.

To complete the proof of Theorem 3.1 it suffices to construct a choice function for the set X = {yb : b ∈ B}.Using SMC, let c be a function with domain X such that for all yb ∈ X , c(yb) is a ’finite subset of yb. If thecharacteristic of the field R is r we assume that for all yb ∈ X , r is not a divisor of |c(yb)|. Temporarily fix

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Math. Log. Quart. 51, No. 2 (2005) / www.mlq-journal.org 189

b ∈ B and let k = |c(yb)| and assume that yb = {t1, t2, . . . , tk}. Let kR = 1R + 1R + · · ·+ 1R, where there arek terms in the sum on the right hand side and 1R is the identity in R. By our assumptions on |c(yb)|, kR �= 0Rand therefore has an inverse k−1

R in R. Let G(b) = k−1R t1 + k−1

R t2 + · · · + k−1R tk. To show that G(b) ∈ yb we

calculate

ϕ(G(b))=ϕ(k−1R t1 + k−1

R t2 + · · · + k−1R tk) = k−1

R ϕ(t1) + k−1R ϕ(t2) + · · · + k−1

R ϕ(tk)= k−1

R b + k−1R b+ · · · + k−1

R b = kR(k−1R b) = b.

This completes the proof of the theorem.

Theorem 3.2 In ZFU (without AC), VPM implies MC.

P r o o f. Assume the hypothesis and assume that X is a collection of non-empty sets with the property thatfor all y1, y2 ∈ X , if y1 �= y2,then y1 ∩ y2 = ∅. It suffices to prove that X has a multiple choice function.

Let M , N and P be the following vector spaces over R:

M = {f :⋃X −→ R : {t ∈ ⋃

X : f(t) �= 0} is finite},N =P = {g : X −→ R : {y ∈ X : g(y) �= 0} is finite}.

Note thatM is (isomorphic to) the free R-module on⋃X andN and P are each isomorphic to the free R-module

on X . Define ϕ : M −→ N by ϕ(β)(y) =∑

t∈y β(t) and let f : P −→ N be the identity may 1P on P . Thenϕ ∈ HomR(M,N) and f ∈ HomR(P,N). By VPM, P is projective and therefore there is a homomorphism Ffrom P to M such that ϕ ◦ F = f .

We first note that for any g ∈ P and any y ∈ X ,

(ϕ ◦ F (g))(y) = (ϕ(F (g)))(y) =∑t∈y(F (g))(t).(1)

For any z ∈ X , let gz ∈ P be defined by

gz(y) =

{1 if y = z,

0 otherwise.

(If we identify P with the free module on X , then gz is just z.) Then since ϕ ◦ F = f = 1P , we have for allz ∈ X , (ϕ ◦F )(gz) = gz and therefore, for all z ∈ X , (ϕ ◦F )(gz)(z) = gz(z) = 1. Combining this with (1) weconclude that

∑t∈z(F (gz))(t) = 1. It follows that the set Wz = {t ∈ z : (F (gz))(t) �= 0} is non-empty. Since

F (gz) ∈ M , the set Wz is also finite. Therefore if we define the function h on X by h(z) = Wz we obtain therequired multiple choice function.

Since the vector space P used in the proof of Theorem 3.2 is a free R-module, we obtain as a corollary to theproof

Corollary 3.3 The assertion that every projective∗ R-module is a projective R-module implies MC (in thetheory ZFU) and therefore this assertion is not provable in ZFU.

If r is a prime integer and we repeat the proof of Theorem 3.2 using the field Zr of integers modulo r weobtain a slightly stronger result.

Corollary 3.4 In Zermelo-Fraenkel set theory weakened to permit the existence of atoms and without theaxiom of choice VPM implies SMC.

P r o o f. Let X be a pairwise disjoint set of non-empty sets and assume that r is prime. Replace the field R byin the proof of Theorem 3.2 by Zr. That is, let M , N and P be the following vector spaces over Zr :

M = {f :⋃X −→ Zr : {t ∈ ⋃

X : f(t) �= 0} is finite},N =P = {g : X −→ Zr : {y ∈ X : g(y) �= 0} is finite}.

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190 P. Howard: If vector spaces are projective modules then multiple choice holds

The homomorphisms ϕ and f are defined as before and as before by VPM we obtain a homomorphism F fromP to M such that ϕ ◦ F = f . For each z ∈ X we define gz as we did in the proof of Theorem 3.2 and repeatingthe argument there we conclude that

∑t∈z(F (gz))(t) = 1 (the sum being taken in the field Zp.)

For each z ∈ X and each k ∈ Zr = {0, 1, 2, . . . , r − 1}, let W kz = {t ∈ z : (F (gz))(t) = k}. Since z is the

disjoint union z =⋃k∈Zr

W kz there must be at least one k ∈ Zr such that

∑t∈Wk

z(F (gz))(t) �= 0. Let nz be the

least such element of Zr. Wnzz will be a finite, non-empty subset of z. Further r is not a divisor of |Wnz

z |, for if|Wnz

z | = m = r · i, where i is an integer, then∑t∈Wnz

z(F (gz))(t) =

∑t∈Wnz

znz = m · nz = (r · i) · nz = 0,

since the sum in being taken in Zr.Therefore, if we let h(z) = Wnz

z for each z ∈ Z , we have a choice function of the type required by SMC.

It is known that in the second Fraenkel model of ZFU (the model N2 in [4]) MC is true but SMC is false. Wetherefore obtain the following

Corollary 3.5 In the theory ZFU, MC does not imply VPM.

We note that the deductive strength of VPM∗ (which could be formulated as “Every vector space is a directsummand of a vector space with a basis”) is an open problem. We conjecture that VPM∗ is not provable in ZFU.

The following diagram summarizes what is known in the theory ZFU.

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MCVPM∗

P ∗ ⇒ P

AC VBT + SMC

VPM

SMC

VBT

Here P∗ ⇒ P is the assertion that every projective∗ module is projective, and a double arrowhead means that theimplication is not reversible.

References

[1] A. Blass, Existence of a basis imples the axiom of choice. In: Axiomatic Set Theory (Baumgartner, Martin and Shelah,eds.), Contemporary Mathematics Series, vol. 31, pp. 31 – 33 (American Mathematical Society, Providence 1984).

[2] D. Dummit and R. Foote, Abstract Algebra, Second Edition (Prentic Hall, Upper Saddle River (NJ) 1999).[3] U. Felgner and T. Jech, Variants of the axiom of choice in set theory with atoms. Fund. Math. 79, 79 – 85 (1973).[4] P. Howard and J. E. Rubin, Consequences of the Axiom of Choice. Math. Surveys and Monographs, vol. 59 (American

Mathematical Society, Providence 1996).[5] T. Jech, The Axiom of Choice (North-Holland Publ. Comp., Amsterdam 1973).[6] H. Lauchli, Auswahlaxiom in der Algebra. Comment. Math. Helv. 37, 1 – 18 (1963).[7] A. Levy, Axioms of multiple choice. Fund. Math. 50, 475 – 483 (1962).

c© 2005 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim