A NEW PROOF OF A THEOREM OF BALCERZYK, BIA LYNICKI-BIRULA AND LO´ S

C O L L O Q U I U M M A T H E M A T I C U M

VOL. LXX 1996 FASC. 2

A NEW PROOF OF A THEOREM OF BALCERZYK, BIA LYNICKI-BIRULA AND LO´ S

BY

JOHN D. O’N E I L L (DETROIT, MICHIGAN)

Let Z ^κ be the direct product of κ copies of the integers Z, where κ is less than the least measurable cardinal number. In [1] the authors cited in the title showed, among other things, that, if A is a direct summand of the abelian group Z ^κ , then A ∼ = Z ^α for some cardinal number α. A key ingredient in their proof was showing that Z ^κ as well as Z ^(κ) , the direct sum of κ copies of Z, were Z-dual. An abelian group G is Z-dual (in modern terminology Z-reflexive) if G is isomorphic to Hom(Hom(G, Z), Z). About the same time R. L. Nunke in [3] obtained the same result using similar arguments. We here present a proof of their result which avoids the use of duality and Hom and seems to us to be more natural.

Unless otherwise indicated our notation and terminology is the same as in [2]. We rely heavily on the following version of a famous Theorem of Lo´ s (see Theorem 94.4 in [2]): if P and S are the direct product and direct sum respectively of torsion-free abelian groups G i , i ∈ κ, where κ is less than the least measurable cardinal number, and η : P → Z is a homomorphism, then

(a) η(G i ) = 0 for almost all i, and (b) if η(S) = 0, then η(P ) = 0.

Theorem 1 (see the Corollary to Theorem 1 in [1]; also, Theorem 5 in [3]). Let Z ^κ = Q

i∈κ he _i i = A ⊕ B, where κ is a set of cardinality less than the least measurable cardinal number. Then A (similarly B) is a direct product of copies of Z.

We begin with two lemmas whereby in Proposition 4 we reduce Theo- rem 1 to the countable case. This case is treated in Proposition 5. A formal proof of Theorem 1 follows. We conclude with some generalizations of The- orem 1.

1991 Mathematics Subject Classification: Primary 20K25.

Key words and phrases: abelian group, direct product, torsion-free, slender group, measurable cardinal number.

[191]

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Lemma 2. The set κ can be ordered as an ordinal such that , if P s equals Q

i≥s he _i i for each s ∈ κ, then, for s = 0 or a limit ordinal , P _s = A s ⊕ B _s with A s ⊆ A and B s ⊆ B.

P r o o f. Let π : Z ^κ → A and π _i : Z ^κ → A → he _i i be the natural projections. By Lo´ s’ Theorem for each i ∈ κ, π i (e j ) = 0 for almost all j ∈ κ. We totally order κ as follows. Assume for ordinal j that we have chosen a subset I = {i : i < j} from κ and that κ − I is nonempty. Choose j from κ − I such that π i (e j ) 6= 0 for the least possible i ∈ I; if no such i exists let j be arbitrary. Now let s be 0 or a limit ordinal in κ. By our ordering of κ we have π i (P s ) = 0 for all i < s by Lo´ s’ Theorem. Let A s = π(P s ) and let B s be the projection of P s to B. Since A s is contained in P s , so is B s . Therefore P s = A s ⊕ B _s .

Let S be the set consisting of 0 and all limit ordinals in κ. For each s in S define P (s) = Q

s≤i<s+ω he _i i where ω = {0, 1, 2, . . .}. In what follows, P s , A s and B s are as in Lemma 2. However, if s happens to be maximal in κ, then s + ω is not in S, so we set P s+ω , A s+ω and B s+ω all equal to 0. We also note that, if s is maximal in κ, then P (s) could have finite rank.

Lemma 3. For each s in S,

(a) A s = A(s) ⊕ A s+ω , where A(s) = A s ∩ (P (s) ⊕ B s+ω ), (b) B s = B(s) ⊕ B s+ω , where B(s) = B s ∩ (P (s) ⊕ A _s+ω ), and (c) P (s) ∼ = A(s) ⊕ B(s) ∼ = Z ^m for some m ≤ ω.

P r o o f. By Lemma 2, P s = A s ⊕ B _s and P s+ω = A s+ω ⊕ B _s+ω . Also, P s

equals P (s) ⊕ A s+ω ⊕ B s+ω . Thus, since A s ⊇ A s+ω , (a) is true; similarly, (b) is true. For (c) note that we now have A(s) ⊕ B(s) ∼ = P s //P s+ω ∼ = Z ^m for some m ≤ ω.

Proposition 4. A ∼ = Q

s∈S A(s), where A(s) is as in Lemma 3.

P r o o f. Let a s ∈ A(s) for each s ∈ S. Each a _s equals π(p s ) for some p s ∈ P (s). Since Q

S P (s) is a product, for each i ∈ κ we have π i (a s ) = 0 for almost all s by Lo´ s’ Theorem. Thus, if a s = (a si ), i ∈ κ, in Z ^κ for each s, we can write P

s∈S a s for the element ( P

s∈S a si ), i ∈ κ, in Z ^κ . Let C = { P

s∈S a s : a s ∈ A(s)}. Suppose that P

s∈S a s = 0 but that a t 6= 0 for minimal t = s. Let φ _t : A → P (t) in Q

s∈S P (s) be the natural projection. Then 0 = φ t ( P

s∈S a s ) = φ t (a t ) = p t 6= 0 (since π(p _t ) = a t 6= 0), a contradiction. Hence C ∼ = Q

s∈S A(s), the external direct product. We now show that A = C. Since A s ⊆ A for all s, C ⊆ A by Lo´s’ Theorem. Let a ∈ A. We complete the proof by showing a ∈ C. Suppose for each s < t in S we have found a s ∈ A(s) such that a − P

r≤s a r is in A s+ω for each s. Then the element a − P

r<t a r is in A s+ω for each s < t since it equals (a − P

r≤s a r ) − P

s<r<t a r for each such s. So it is in A t . By Lemma 3(a),

NEW PROOF OF A THEOREM

193

for some a t in A(t), the element a − P

r≤t a r is in A t+ω . Inductively then we can find a s ∈ A(s) for each s ∈ S such that a − P

r≤s a r is in A s+ω for each s. Since the element a − P

s∈S a s equals (a − P

r≤s a r ) − P

r>s a r for each s, it is in T

s∈S A s+ω , which is 0. Therefore a = P

s∈S a s , an element in C.

Our proof of the next result is well known but, since it does not seem to be in standard references, we include it for completeness.

Proposition 5. If Z ^ω = A ⊕ B, then A ∼ = Z ^m for some m ≤ ω.

P r o o f. Write Z ^ω = {(x n ) : n ∈ ω, x n ∈ Z}. For each s ∈ ω, let A _s equal {(x n ) ∈ A : x n = 0 for n < s}. Choose a s ∈ A s for each s such that its s-component is the least possible positive integer; if no such element exists, let a s = 0. Observe that, if x ∈ A s , then x − ma s is in A s+1 for some integer m. Write a s = (a sn ), n ∈ ω. Since a sn is 0 whenever s > n, we can identify P

s∈ω m s a s (m s ∈ Z) with the element ( P

s∈ω m s a sn ), n ∈ ω, in Z ^ω . Let C be the set of all such elements. By our choice of each a s it follows that C ∼ = Q

s∈ω ha s i ∼ = Z ^m for some m ≤ ω. Also, C ⊆ A. Let a ∈ A. We complete the proof by showing a ∈ C. By induction we can find m s ∈ Z for each s such that a − P

r≤s m r a r is in A s+1 for each s. It follows easily then that a = P

s∈S m s a s , an element in C.

P r o o f o f T h e o r e m 1. By Proposition 4 and Lemma 3, A is isomor- phic to Q

s∈S A(s), where each A(s) is isomorphic to a direct summand of Z ^ω . Proposition 5 completes the proof.

Generalizations. (a) Suppose P = Q

i∈κ G i , where each G i is isomor- phic to a proper subgroup of the rational numbers and where κ is less than the least measurable cardinal number. In [4] we showed that any direct summand of P is a direct product of rank-one abelian groups. This gener- alized all the results in [1]. The proof in [4] was much more difficult than the proof of Theorem 1 above but still avoided the use of duality and Hom.

(b) In [5] by other means we showed that Theorem 1 can be extended to the case where κ is the least measurable cardinal number.

REFERENCES

[1] S. B a l c e r z y k, A. B i a l y n i c k i-B i r u l a and J. L o´ s, On direct decompositions of complete direct sums of groups of rank 1, Bull. Acad. Polon. Sci. S´ er. Sci. Math.

Astronom. Phys. 9 (1961), 451–454.

[2] L. F u c h s, Infinite Abelian Groups, Vol. II, Academic Press, New York, 1973.

[3] R. J. N u n k e, On direct products of infinite cyclic groups, Proc. Amer. Math. Soc.

13 (1962), 66–71.

[4] J. D. O’N e i l l, A theorem on direct products of slender modules, Rend. Sem. Mat.

Univ. Padova 78 (1987), 261–266.

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[5] J. D. O’N e i l l, Direct summands of Z

for large κ, in: Abelian Groups and Related Topics, R. G¨ obel, P. Hill and W. Liebert (eds.), Contemp. Math. 171, Amer. Math.

Soc., 1994, 313–323.

Mathematics Department University of Detroit Mercy Detroit, Michigan 48219 U.S.A.

E-mail: oneilljd@udmercy.edu

Re¸ cu par la R´ edaction le 22.8.1994;

en version modifi´ ee le 25.5.1995