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Tomus 48 (2012), 107–112

ON THE SUBGROUPS OF COMPLETELY DECOMPOSABLE TORSION-FREE GROUPS THAT ARE IDEALS

IN EVERY RING

A. M. Aghdam, F. Karimi, and A. Najafizadeh

Abstract. In this paper we consider completely decomposable torsion-free groups and we determine the subgroups which are ideals in every ring over such groups.

1. Introduction

All groups considered here are abelian, with addition as the group operation.

Given an abelian groupA, we callRa ring overA if the groupAis isomorphic to the additive group ofR. In this situation we writeR= (A,∗), where∗denotes the ring multiplication. This multiplication is not assumed to be associative. Every group may be turned into a ring in a trivial way, by setting all products equal to zero; such a ring is called a zero-ring. If this is the only multiplication overA, thenAis said to be a nil group. Fried [2] gives a criterion for the subgroups of an abelian group that are ideals in every ring. In [4], the authors use the type set of a rank one or an indecomposable rank two abelian groupA, to give necessary and sufficient conditions for the subgroups ofAwhich are ideals in every ring overA. In this paper, we discuss about the subgroups of completely decomposable torsion-free groups which are ideals in every ring.

2. Notations and preliminaries

LetA be a torsion-free abelian group. Given a primep, thep-height ofx∈A, denoted byh^A_p(x), is the largest integerksuch thatp^k dividesxinA; if no such maximal integer exists, we set h^A_p(x) = ∞. Now let p₁, p₂, . . . be an increasing sequence of all primes. Then the sequence

χA(x) = (h^A_p1(x), h^A_p2(x), . . . , h^A_pn(x), . . .),

is said to be the height-sequence ofx. We omit the subscriptAif no ambiguity arises.

For any two height-sequencesχ= (k1, k2, . . . , kn, . . .) andµ= (l1, l2, . . . , ln, . . .) we setχ≥µifkn≥lnfor alln. Moreover,χandµwill be considered equivalent if

2010Mathematics Subject Classification: primary 20K15.

Key words and phrases: completely decomposable group, type, ring.

Received October 10, 2011. Editor J. Trlifaj.

DOI: 10.5817/AM2012-2-107

P

n|kn−l_n|is finite [we set∞ − ∞= 0]. An equivalence class of height-sequences is called a type. Ifχ(x) belongs to the typet, then we say thatxis of type t. For two types t₁,t₂ we havet₁≤t₂ if there existsχ∈t₁ andµ∈t₂ such thatχ≤µ.

The type set ofA is the partially ordered set of types, i.e., T(A) ={t(x)|x∈A\0}.

A torsion-free groupAin which all non-zero elements are of the same typet is called homogeneous of type t, or t-homogeneous. For example every rank one groupA is homogeneous. We use the symbolt(A) for the type set of a rank one group A, which is indeed the type of any non-zero element of A. For two types t₁= (l₁, l₂, . . .) andt₂= (k₁, k₂, . . .) we set

t₁∩t₂= (min{l1,k₁},min{l2,k₂}, . . .) and

t₁t₂= (l₁+k₁, l₂+k₂, . . .). IfC is a subgroup ofA andS a subset ofC, then

hSi^C_∗ ={a∈C|na∈ hSi; for some 06=n∈Z}

is the pure subgroup of C generated by S. Moreover, we set hSi∗ = hSi^A_∗. A torsion-free groupAis completely decomposable ifA is the direct sum of rank one groups. A proper subgroup CofA is called strongly nil if for any ring (A,·) over Awe have a·c=c·a= 0 for alla∈Aand for allc∈C. If Cis not strongly nil then it said to be a strongly non-nil subgroup ofA. IfA=⊕i∈IA_i is a completely decomposable group and S={xi|x_i∈A_i}i∈I is a maximal independent set ofA, then for any subgroupCofAwe defineU_i^C:={βi∈Q|β_ix_i∈C}andU_i:=U_i^Ai.

We end this section with the following proposition which is used later.

Proposition 2.1. Let A = ⊕i∈IA_i be a completely decomposable group which supports a non-zero ring R = (A,·). Let S = {x_i | x_i ∈ A_i}_i∈I be a maximal independent set of A andU_i={u_i∈Q|u_ix_i∈A_i}for all i∈I. Then

(1) If there existr,s∈I such thatxr·xs=P

i∈Iαr,s,ixi thenαr,s,iUrUs⊆Ui

for all i∈I.

(2) In (1)if αr,s,t 6= 0for somet∈I, then the multiplication∗ defined as xi∗xj=

(αr,s,txt (i, j) = (r, s) ;

0 otherwise,

yields a non-zero ring over A.

Proof. Straightforward.

3. Completely decomposable homogeneous groups

Theorem 3.1. Let A=⊕ⁿ_i=iAi be a completely decomposable and homogeneous group of rankn. If A is non-nil, thenA contains no non-trivial subgroup of rank less than nwhich is an ideal in every ring onA.

Proof. Let t(A) =t. Then t² =t sinceA is non-nil. Now suppose that x_i ∈A_i and{x1, x₂, . . . , x_n}be a maximal independent set ofA. Hencet(U_i) =t=t(U_iU_j) for alli, j∈ {1,2, . . . , n}.

By the way of contradiction, suppose that C is a non-trivial subgroup of A with r(C) = k ≤ n−1 such that C is an ideal in every ring on A. Let 0 6=

c =Pn

i=1αixi be an element of C. Then there existsi∈ {1,2, . . . , n} such that αi 6= 0. Without loss of generality suppose that α1 6= 0. On the other hand t(U₁²) = t(U1) = t(U2) = · · · = t(Un), hence there exist some non-zero integer numbers m1, m2, . . . , mn, k1, k2, . . . , kn such that:

m₁U₁²=k₁U₁, m₂U₁²=k₂U₂, . . . , m_nU₁²=k_nU_n. Now we define ∗1 as follows

x_i∗1x_j=

(m₁x₁ if i=j= 1, 0 otherwise.

Clearly∗1yields a ring onAsuch thatc∗1x1=m1α1x1∈C. In fact ifu1=β1x1+

· · ·+β_nx_n, u₂=γ₁x₁+· · ·+γ_nx_n are two arbitrary elements ofA, thenu₁∗1u₂= m₁β₁γ₁x₁. But m₁β₁γ₁ ∈m₁U₁² =k₁U₁ ⊆U₁, hence∗1 is actually a ring on A.

Similarly we may define multiplications∗2,∗3, . . . ,∗n such that (A,∗2), . . . ,(A,∗n) are rings on Aand for alll= 2,3, . . . , nwe have

06=c∗lx1=mlα1xl∈C .

This implies thatr(C) =n, a contradiction.

4. Completely decomposable groups of rank nwhose typesets containsn maximal elements

Theorem 4.1. LetA=⊕ⁿ_i=1Ai be a completely decomposable group of rank n. Let S ={xi |xi ∈Ai, i= 1,2, . . . , n} be a maximal independent set of A such that t(xi) =ti are maximal elements in T(A)for all i= 1,2, . . . , n. Then

(1) Any rank one subgroup C which is an ideal in every ring on A is of the form C = U_i^C(mxi) with t²_i = ti, m ∈ Z\ {0} or C is generated by a rational combination of some elements in S with non-idempotent types.

Moreover,C in the first case is strongly non-nil and in the second case is strongly nil.

(2) Any subgroup C of rank k < n which is an ideal in every ring on A is generated by l(≤ k) rational multiples of some elements in S with idempotent types and k−l combinations with rational coefficients of some elements in S with non-idempotent types. Moreover, if l 6= 0 then C is strongly non-nil.

Proof. 1) LetC be any rank one subgroup ofAwhich is an ideal in every ring on Aand let c=Pn

i=1α_ix_i be a non-zero element ofC. We consider two cases. First suppose thatα_i6= 0, for somei∈ {1,2, . . . , n}witht²(x_i) =t(x_i). For example let

α₁6= 0 andt²(x₁) =t(x₁). This impliest(U₁²) =t(U₁). Hence, as in the proof of Theorem 3.1, there exists a non-zero integermsuch that

xi∗xj=

(mx1 if i=j= 1, 0 otherwise,

is a ring on A. Clearly, 0 6= c∗ x1 = α1mx1 ∈ C. Now if αj 6= 0 for some j 6= 1, then r(C) ≥ 2, which is a contradiction. Consequently, C = U₁^C(mx₁) for some m ∈ Z\ {0} and clearly C is strongly non-nil. In the second case let c=Pn

i=1β_ix_i, t²(x_i)6=t(x_i), β_i∈Q. Nowt(x_i) is maximal and non-idempotent, hence any ring on Asatisfies:x_ix_i=x_ix_j= 0 which yieldsC is strongly nil.

2) LetC be a rankksubgroup ofAwhich is an ideal in every ring onA and let {c1=α11x1+· · ·+α1nxn, c2=α21x1+· · ·+α2nxn, . . . , ck =αk1x1+· · ·+αknxn}be a maximal independent set ofC. If there existi∈ {1,2, . . . , k}andj∈ {1,2, . . . , n}

such that αij 6= 0 andt²(xj) = t(xj), then as in case (i) there exist a non-zero integer m and a ring on A with 06=cixj =αijmxj ∈C. Let αijm=βj, hence there existc⁰₂, . . . , c⁰_k ∈C such that {βjxj, c⁰₂, . . . , c⁰_k}is an independent set inC and for alli= 1,2, . . . , k,

c⁰_i=α⁰_i1x₁+· · ·+α⁰_ij−1x_j−1+α⁰_ij+1x_j+1+· · ·+α⁰_inx_n. Repeating this procedure we get a maximal independent set inC

{βj₁xj₁, . . . , βj_lxj_l, c⁰⁰₁, . . . , c⁰⁰_k−l},

such thatt²(x_j1) =t(x_j1), . . . , t²(x_jl) =t(x_jl) andc⁰⁰₁, . . . , c⁰⁰_k−l are rational combi- nations of some elements in S with non-idempotent types. Now the Final claim is

obvious.

5. Completely decomposable group of rankn whose typesets contains less than n maximal elements

Theorem 5.1. Let A =⊕ⁿ_i=1Ai be a completely decomposable group of rank n such that T(A) contains k < n maximal elements. Suppose that S ={xi | xi ∈ A, i= 1,2, . . . , n} be a maximal independent set ofA. Then

(1) For any rank one subgroupC which is an ideal in every ring onA we have one of the following three cases:

(a) C=U_i^C(mxi)for some non-zero integer mand t(xi)is idempotent.

Moreover, such a subgroup is strongly non-nil.

(b) C =hαxii^C_∗ with t²(xi) 6=t(xi). Moreover, if C is strongly non-nil then there exists(xi6=)xk∈S such thatt(xi)t(xk) =t(xi).

(c) C = hP

j∈J⊆{1,2,...,n}αjxji^C_∗,|J| ≥ 2, such that every t(xj) is of non-idempotent type. Moreover, such a subgroup is strongly nil.

(2) Any rankm(< n)subgroupC ofAwhich is an ideal in every ring on Ais C=hH₁⁰, H₂⁰, H₃⁰ |H_i⁰⊆H_i, i= 1,2,3i^C_∗, in which:

H1={ci=αixi|αi∈Q, t(xi)is maximal and idempotent}, H₂={c⁰_i= X

j∈J⊆{1,2,...,n}

α_ijx_j |α_i ∈Q, t(x_i)is not idempotent},

H₃={c⁰⁰_k =α_kx_k|α_i∈Q, t²(x_k) =t(x_k), t(x_k)is not maximal}.

Moreover, in this case ifc=P

j∈J⊆{1,2,...,n}αjxj∈H2∩C andαjxj6= 0 witht(xj)maximal andxjxk6= 0for somexk ∈S, thent(xk)t(xj) =t(xj).

Proof. 1-a) Let Cbe any rank one subgroup of Awhich is an ideal in every ring over A and c = Pn

i=1αixi ∈ C. If αi 6= 0, t²(xi) = t(xi), then by the proof of Theorem 3.1, there exists a non-zero integermsuch that

xr∗xs=

(mxi if r=s=i , 0 otherwise.

yields a ring onAsuch thatc·xi=mαixi∈C. Moreover, byr(C) = 1 we obtain αj= 0 for all j6=i. Consequently,C=U_i^C(mxi) which clearly is strongly non-nil.

b, c) Suppose thatC is strongly non-nil and any arbitrary element ofC is of the formαcwithα∈Qandc=P

j∈J⊆{1,2,...,n}αjxj such thatt(xj) is not idempotent.

If there exists exactly one index i ∈ J such that αi 6= 0, then αixixk ∈ C for any xk ∈ S such that t(xk) > t(xi). But t(αixixk) > t(xi) hence xixk = 0. If t(xk) =t(xi) then xixk= 0, becauset(xi) is not idempotent. On the other hand C is strongly non-nil and therefore 0 6= αixixk ∈ C for some xk ∈ S, hence t(αixixk) =t(xi). Butt(αixixk)≥t(xi)t(xk)≥t(xi) which yields the result. For the last case if at least two coefficients α_j are non-zero and c·x_k 6= 0, for some x_k∈S then there exists j∈J such thatx_jx_k6= 0. Now by Proposition 2.1 there exists a ring R= (A,∗) onAsuch thatc∗x_k is a non-zero rational multiple of an element inS which meansr(C)≥2, a desired contradiction.

2)LetC be any rankmsubgroup which is an ideal in every ring onA. Similarly as previous part, ifc=Pn

i=1αixi∈Candαi6= 0 for somei, witht²(xi) =t(xi) then a non-zero multiple ofxi is inC, i.e., there isβi∈Qwithβixi∈C. Hence such a generator ofC must be inH1 orH3. Consequently, as Theorem 4.1, any generator ofCis inH1,H2orH3. Moreover, if there existc=P

j∈J⊆{1,2,···,n}αjxj ∈H2∩C andαjxj 6= 0 witht(xj) maximal andxjxk 6= 0 for somexk ∈Switht(xk)< t(xj), then t(xkxj) ≥ t(xk)t(xj) ≥ t(xj). But t(xj) is maximal, hence we must have t(xkxj) =t(xj) and this completes the proof.

References

[1] Aghdam, A. M., Najafizadeh, A.,Square subgroups of rank two abelian groups, Colloq. Math.

117(1) (2009), 19–28.

[2] Fried, E.,On the subgroups of an abelian group that are ideals in every ring, Proc. Colloq.

Abelian groups, Budapest, 1964, pp. 51–55.

[3] Fuchs, L.,Infinite abelian groups, Academic Press, New York–London, 1973.

[4] Najafizadeh, A., Aghdam, A. M., Karimi, F.,On the ideals of torsion–free rings of rank one and two, Math. Notes (to appear).

Department of Mathematics, University of Tabriz, Tabriz, Iran E-mail:[email protected]

Department of Mathematics, University of Tabriz, Tabriz, Iran E-mail:[email protected]

Mathematics Department,

Payame Noor University, 19395-4697, Tehran, I.R. of IRAN E-mail:[email protected](corresponding author)