PII. S0161171201010274 http://ijmms.hindawi.com © Hindawi Publishing Corp.
SUBGROUPS OF FINITE INDEX IN AN ADDITIVE
GROUP OF A RING
DOOSTALI MOJDEH and S. HASSAN HASHEMI
(Received 15 March 2000 and in revised form 10 June 2000)
Abstract.IfKis an infinite field andG⊆Kis a subgroup of finite index in an additive group, thenK∗=G∗G∗−1whereG∗denotes the set of all invertible elements inGand G∗−1denotes all inverses of elements ofG∗. Similar results hold for various fields, division rings and rings.
2000 Mathematics Subject Classification. 16Kxx.
1. Introduction. LetRbe a ring (not necessarily commutative) with a unit element 1 andR∗denotes the multiplicative group of invertible elements ofR. In [9] Leep and Shapiro proved that ifGis a subgroup of index 3 in the multiplicative groupF∗, then G+G=F. In [2] Berrizbetia proved that ifFis a field andG⊆F∗is a subgroup of finite indexn, then there is a positive integerm, that depends onn, so that if charF=0 or charF≥m, thenG−G=F. In [1] Bergelson and Shapiro proved that, for various ring R, ifGis a subgroup of finite index ofR∗, thenG−G=R. In [14] Turnwald proved that ifGis a subgroup of finite indexnin the multiplicative group of a division ringFthen G−G=F or|F|< (n+1)4+4n, and if|F|> (n−1)2and−1 is a sum of elements ofG then every element ofF has this property; the bound(n−1)2is optimal for infinitely manyn. The theories which have important role in studying of the above were Ramsey theory, measure theory and number theory, (cf. [4,7,15]). Furthermore in [1] the roles of multiplication and addition were switched, and it was shown that
Proposition1.1(see [1, Proposition 2.14]). LetKbe an infinite field andG⊆Ka
subgroup of finite index in additive group. ThenG∗G∗−1=K∗whereG∗=G\{0}; that is, for everyc∈K∗there existg1, g2∈Gsuch thatc=g1/g2.
Corollary1.2. IfDis an infinite division ring then the above result is satisfied.
In this paper, the roles of multiplication and addition are switched and it is shown thatProposition 1.1andCorollary 1.2hold for various fields, division rings and rings. Now letG⊆Rbe a subgroup of finite index in an additive group,G∗be the set of all invertible elements inG,G∗−1= {g−1:g∈G∗}andG∗G∗−1= {g1g2−1:g1, g2∈G∗}.
2. G∗G∗−1-ring. LetKbe a ring or field andG⊆Kbe a subgroup of finite index in an additive group, then it is not necessary thatG∗G∗−1=R∗ or evenG∗,G∗−1, and G∗G∗−1have group structure. Note the following statements.
(i) LetF=Fp2, andG=Fp, thenG∗=Fp∗andG∗G∗− 1=
Fp∗≠F∗.
whereF8has 8 elements. PutG= {0, α, α2, α+α2}soG∗= {α, α2, α+α2}. It is clear thatG∗is not a group, becauseG∗does not contain the unit element 1. ButG∗G∗−1 is a subgroup of the multiplicative groupF8∗. Furthermore,G∗G∗−1=F8∗.
(iii) Letβbe a root of the polynomialx4−x+1 over the splitting fieldZ2(β)=F16. PutG= {0,1, β,1+β}, soG∗= {1, β,1+β},G∗−1= {1, β3+1, β3+β2+β}and therefore G∗G∗−1= {1, β3+1, β3+β2+β, β,1+β+β2+β3,1+β, β3}. It is clear thatG∗ is not a group butG∗G∗−1is a proper subgroup ofF16∗.
(iv) LetR=Z/nZwherenis not a prime number. IfG⊂Ris a proper subgroup in an additive group then it is clear thatG∗= ∅andG∗G∗−1= ∅≠R∗.
(v) LetS=Z/nZwherenis a natural number,R=S[x]andHis a proper subgroup ofS. IfG= {f (x)=a0+a1x+ ··· +akxk:ai∈S, a0∈H}, soG⊆Ris a subgroup of finite index in an additive group andG∗= ∅. If the square of every prime number does not divide nand a0∈S butat ∈H, for finitely manyt >0, thenG⊆R is a subgroup of finite index in an additive group,G∗=R∗andG∗G∗−1=R∗.
(vi) LetR=Z[x]andG= {f (x)=a0+a1x+ ··· +anxn:a0∈mZ(m >1), ai∈ Z(i≥1)}. SoG⊆Ris a subgroup of indexmin an additive group. It is clear thatG∗= ∅ =G∗G∗−1≠R∗= {1,−1}. If for finitely many nonzero indicesi’s,ai∈miZ(mi> 1), thenG∗=R∗andG∗G∗−1=R∗.
(vii) LetQbe the set of rational numbers andv2the 2-adic valuation onQ. Then R= {x∈Q:v2(x)≥0} = {m/n∈Q:(n,2)=1}is a valuation ring (cf. [3,10,11,12] or [13]). IfG= {2m/n∈R:(n,2)=1}, thenGis a subgroup of finite index 2 in an additive group where 0+Gand 1+Gare two distinct cosetsGinR. It is easy to see thatG∗= ∅,R∗= {m/n:m=2k+1, n=2l+1}, andG∗G∗−1≠R∗.
The above statement can be shown for anyp-adic valuation ring inQ.
ByProposition 1.1,Corollary 1.2, and the previous statements, the following ques-tion may be raised.
Question2.1. IfF is a finite field or a ring andGis a subgroup of finite index in
an additive group, mustG∗G∗−1=F∗?
We will answer the question for all finite fields and some rings.
IfF is a finite field and|F|is sufficiently large to the index ofG, in other words,G is sufficiently large, thenG∗G∗−1=F∗, see the following result.
Theorem2.2. (i)LetDbe a division ring withcharD=pandG⊆Dbe a subgroup
of indexpkin an additive group. If|D| ≥p2k+1, thenG∗G∗−1=D∗.
(ii) IfGis a subgroup of finite indexn≥pkin a division ringDand|D| =p2k, then G∗G∗−1≠D∗.
Proof. (i) Fixc∈D∗. Let thegi’s be distinct elements inG∗(|G|> pk+1). We form
the cosets(cg1+G), . . . , (cgpk+1+G). By the pigeonhole principle at least two cosets are equal. Socgi+G=cgj+G⇒0≠c(gi−gj)∈G⇒cg=g⇒c=g/g∈G∗G∗−1.
(ii) Since|D| =p2khence|D∗| =p2k−1. By hypothesis[D:G]= |D|/|G| ≥pk, so
|G| ≤pk and therefore,|G∗−1| = |G∗| ≥pk−1, so we have,|G∗G∗−1| ≤(pk−1)2= p2k−2pk+1< p2k−1= |D∗|soG∗G∗−1≠D∗.
We now give the result which generalizes Proposition 1.1, Corollary 1.2, and Theorem 2.2(i).
Lemma2.4. LetRbe a ring and let S be a subset ofR with invertible differences,
that is,a−b∈R∗for any distinct elementsa, b∈S.
(i) SupposeG⊆R is a subgroup of indexnin an additive group. If|S|> n2then G∗G∗−1=R∗.
(ii) If|S| = ∞thenG∗G∗−1=R∗.
Proof. (i) Letr∈R∗be any element. By the pigeonhole principle there exists, t∈S
such thats−t=aandr s−r tboth lie inG∗. Sor=ba−1∈G∗G∗−1
, as claimed. (ii) This part is an immediate consequence of part (i).
Apply the lemma withS=Kfor the proof ofProposition 1.1, withS =Dfor the proof ofProposition 1.1andTheorem 2.2(i).
We now state the following definition which is a key concept in the paper. This is the analog of [1, Definition 0.1].
Definition2.5. A ringR is aG∗G∗−1-ring, ifG∗G∗−1=R∗ for every subgroup G⊆Rof finite index in an additive group.
IfRis a ring which is a divisible group, thenRhas no additive subgroups of finite index (cf. [6]). Combining this statement,Lemma 2.4, andDefinition 2.5we obtain the following result.
Proposition2.6. IfDis an infinite division ring, then every ringRwhich contains
a copy of D is aG∗G∗−1-ring. In particularD[x], D[[x]], Mn(D), Mn(D[x]), and Mn(D[[x]])areG∗G∗−1-rings.
Proof. If char(D)is zero then every ring that contains a copy ofDis a divisible
group and hence R∗ =G∗G∗−1. If the char(D)≠0, Lemma 2.4 implies that R∗ = G∗G∗−1.
Remark 2.7. The converse of Definition 2.5 does not necessarily hold. Let R= Q[x],G=Q, thenG∗G∗−1=R∗. ButGis not of finite index in an additive group.
3. Properties ofG∗G∗−1-ring. In this section, some properties of theG∗G∗−1-ring is verified.
Proposition3.1. LetR be a commutative ring andI an ideal ofRsuch thatR/I
is aG∗G∗−1-ring. IfIdoes not contain any additive subgroup of finite index and every element of1+Iis invertible, thenRis aG∗G∗−1-ring.
Proof. Let G ⊆ R be a subgroup of finite index in an additive group. Since (G+I)/GI/(G∩I) so |I/(G∩I)|<∞and hence I∩G=I. Choosex ∈R∗, then x+I=(g1+I)(g2+I)−1. It is easily seen thatg1, g2∈G∗. Sox=g1g−1
2 +afor some a∈I. Butx=(g1+ag2)g2−1whereg1+ag2∈G∗, that is,R∗=G∗G∗−
1 .
Theorem 3.2. Let R be a commutative ring, R[x] the polynomial ring andI = {g(x)=a1x+a2x2+ ··· +a
nxn|ai (i≥1)is nilpotent}. IfI does not contain any
additive subgroup of finite index and R is a G∗G∗−1-ring, thenR[x] also has that property.
Proof. We have(R[x]/I)∗=((R+I)/I)∗= {r+I|r∈R∗}. SupposeG/I⊆R[x]/I
is a subgroup of finite index in an additive group, thenGis an additive subgroup of finite index inR[x]and|R/(G∩R)| = |(R+G)/G|<∞. By hypothesis forr∈R∗there existg1, g2∈G∗∩R∗such thatr=g1g2−1. Sor+I=g1g2−1+I=(g1+I)(g−21+I)∈ (G/I)∗(G/I)∗−1, that is,(R[x]/I)∗=(G/I)∗(G/I)∗−1. NowProposition 3.1completes the proof.
As an immediate consequence we obtain the following result.
Corollary3.3. LetRbe a commutative ring without any nonzero nilpotent
ele-ments. IfRis aG∗G∗−1-ring then so isR[x].
The converse ofTheorem 3.2holds in general, see the following result.
Theorem3.4. LetRbe a commutative ring. IfR[x]is aG∗G∗−1-ring thenRalso
has that property.
Proof. LetG⊆Rbe a subgroup of finite index in an additive group. PutH= {a0+ r1x1+ ··· +r
kxk:kis a nonnegative integer, a0∈G ri∈R i >0}. It is easily seen that,H⊆R[x]is a subgroup of finite index in an additive group andH∗∩R∗=G∗. Since(R[x])∗=H∗H∗−1thereforeR∗=(R[x])∗∩R∗=(H∗H∗−1)∩R∗=G∗G∗−1. Thus the proof is complete.
Here, we give a necessary condition for infiniteR;G∗G∗−1=R∗, this condition is not sufficient. We also verify the behavior ofG∗G∗−1-ring under homomorphisms.
Theorem3.5. IfR is aG∗G∗−1-ring. Assume thatS is a nontrivial homomorphic
image ofRwith homomorphismϕ:R→Sthen
(i) S is infinite.
(ii) Assumeϕ−1{1S} = {1R}. IfR∗is aG∗G∗−1
-ring , thenS∗also aG∗G∗−1-ring.
Proof. (i) SupposeS is a finite ring. LetG=kerϕ·G⊆R is a subgroup of finite
index in an additive group, because R/GS. Then G∗G∗−1= R∗ therefore 1R∈ G∗G∗−1and 1S=ϕ(1R)∈ϕ(G∗G∗−1)=ϕ(G∗)ϕ(G∗−1)⊆ϕ(G)ϕ(G∗−1)=0 there-foreS=0 which is a contradiction, and the proof is complete.
(ii) LetG⊆S be a subgroup of finite index in an additive group. PutH=ϕ−1(G), thenHis a subgroup ofR. Now define the following homomorphism
α:R → S
G, α(x)=ϕ(x)+G, (3.1) soαis also surjective and by the first isomorphism theoremR/HS/G. SinceS/Gis finite then so isR/Hand thusHis of finite index. Then by hypothesisH∗H∗−1=R∗ now we haveϕ(H∗)ϕ(H∗−1)=ϕ(H∗H∗−1)=ϕ(R∗)=S∗soG∗G∗−1=S∗, and this implies thatS∗is aG∗G∗−1-ring.
We now verify the behavior ofG∗G∗−1-rings under products.
Theorem3.6. SupposeR=R1×R2, ifR1∗andR∗2 isG∗G∗−1-ring then so isR∗.
Proof. SupposeG⊆R=R1×R2is a subgroup of finite index ofR. PutA1= {a∈ R1:(a,0)∈G}. Now define
α:R →R1×R2
G , α(a)=(a,0)+G, (3.2)
soA1=kerα. It implies thatA1⊆R1is a subgroup of finite index in an additive group. ThereforeA∗1A∗1−
1
=R∗1.Similarly, we defineA2inR2, soA∗2A∗2− 1
=R2∗. Now we have A1×A2= {(a, b)|(a,0), (0, b)∈G} = {(a,0)+(0, b)|(a,0), (0, b)∈G} ⊆G+G⊆G and also(A1×A∗2)(A1×A∗2)−
1
=(A∗1×A∗2)(A∗1− 1
×A∗2− 1
)=A∗1A∗1− 1
×A∗2A∗2− 1
=R1∗× R2∗=R∗. SinceA1×A2⊆GthenG∗G∗−1=R∗and thusR∗is aG∗G∗−1-group.
Theorem3.7. LetRbe a ring,Iits ideal and every element of1+Iis invertible. If RisG∗G∗−1-ring thenR/Iis alsoG∗G∗−1-ring.
Proof. LetG/I ⊆R/I be a subgroup of finite index in an additive group, then G⊆Ris a subgroup of finite index in an additive group. Chooser+I∈(R/I)∗where r∈R∗ andr=g1g2−1wheregi∈G∗, i=1,2. Therefore,r+I=(g1+I)(g2+I)−1∈ (G/I)∗(G/I)∗−1.
Theorems3.5,3.6, the properties of isomorphism,Proposition 2.6, Artin-Wederburn theorem (cf. [8]), andTheorem 3.7imply the following result.
Corollary3.8. (i)IfRR1×R2thenR1∗andR∗2 areG∗G∗−1-rings if and only if R∗is aG∗G∗−1-ring.
(ii) Every semisimple ring which has no finite homomorphic image is aG∗G∗−1-ring.
(iii) LetRbe aG∗G∗−1-ring andJthe Jacobson radical ofR. ThenSis aG∗G∗−1-ring.
Remark3.9. IfSis aG∗G∗−1-ring andRis a subring ofSthenRis not necessarily
aG∗G∗−1-ring. So Ifϕ:R→S is a monomorphism andS is aG∗G∗−1-ring thenRis not necessarily aG∗G∗−1-ring.
We end this section by verifying whetherD∗=G∗G∗−1Dis an infinite division ring andG=F+[D, D]whereF denotes the center ofDand[D, D]denotes the additive commutator subgroup ofD, (cf. [5]). As an example see the following example.
Example3.10. Suppose thatD=Q(i, j, k)is the rational quaternion, by a simple
investigation one can see that[D, D] =ai+bj+ck fora, b, c ∈Q, therefore G= F+[D, D]=Dand soG∗G∗−1=D∗.
This also holds for real quaternions. But in general we have the following result. This question is answered for a finite-dimensional division algebra (or more gener-ally central algebra).
Lemma3.11. LetDbe a finite-dimensional division (or, more generally, central
Proof. Letd1, d2, . . . , dn2 be a basis ofDofF vector space; heren=deg(D). Let
T0be then2−1-dimensional ofF-subspace ofDconsisting of trace-zero elements. Clearly[D, D]⊆T0. Thus it is enough to show that dimF[D, D]≥n2−1. LetK be a splitting field ofD. ThenD⊗FK=Mn(K). It is easy to see that[Mn(K), Mn(K)]is precisely the set ofn×n-matrices of trace zero. On the other hand, this set is spanned by elements[di, dj], asi, j=1,2, . . . , n2.Thusn2−1 of these elements are linearly independent overKand, hence, overF. This proves the inequality dimF[D, D]≥n2−1, as desired.
We therefore conclude that dimF[D, D]=n2−1, while dimF(D)=n2. Thus
F+[D, D]=
D, if char(F )does not dividen,
[D, D], if char(F )dividesn. (3.3)
IfDis noncommutative (i.e., of degreen≥2) then the following lemma shows that D∗=[D, D]∗([D, D])∗−1in any characteristic.
Lemma3.12. LetD be a finite-dimensional division algebra of degreen≥2with
centerF and letGbe ad-dimensionalF-vector subspace ofD.
(a)Assume2d > n2. ThenD∗=G∗G∗−1
.
(b)AssumeG=[D, D]. ThenD∗=G∗G∗−1.
Proof. (a) Leta∈D∗. Since 2d > n2, thed-dimensional F-vector spacesG and aGhave a nontrivial intersection inD, that is,g1=ag2for someg1, g2∈G∗. Then a=g1g2−1, as desired.
(b) ByLemma 3.12,d=dimF[D, D]=n2−1. SinceDis noncommutative,n≥2 and thus 2d=2n2−2> n2. Now apply part (a).
Question3.13. (1) IfRis not a finite homomorphic image, mustR∗ be infinite?
MustRcontain an infinite subset with invertible differences?
(2) Is there a ring with no finite homomorphic image, but with some finite index subgroupGavoiding all units:G∗= ∅?
(3) IfR is aG∗G∗−1-ring then mustR∗ be infinite? IfR is aG∗G∗−1-ring mustR contain an infinite sets with invertible differences?
(4) IfRis aG∗G∗−1-ring, then must the matrix ringMn(R)also have that property? Conversely, ifMn(R)is aG∗G∗−1-ring, then mustRbe aG∗G∗−1-ring?
(5) If R is a G∗G∗−1-ring and R is a subring of a ring S then must S also be a G∗G∗−1-ring?
(6) IfR/Iis aG∗G∗−1-ring and 1+Iis invertible elements then mustR also have that property?
(7) LetDbe an infinite (algebraic) division algebra over its centerF. IfG=F+[D, D]. IsD∗=G∗G∗−1?
Acknowledgement. The authors would like to thank the referees for their
help-ful comments.
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Doostali Mojdeh: Department of Mathematics, Faculty of Basic Sciences, University of Mazandaran, Babolsar, Iran
E-mail address:[email protected]
