Let $R$ be a commutativeunitary ring of prime characteristic $p$ and let $G$ be an Abelian group. We calculateonly in terms of $R$ and $G$ (and their sections)Warfield $p$-invariants of thequotient group $V(RG)/G$, that is, the group of all normalized units $V(RG)$ in the groupring $RG$ modulo $G$. This supplies recent results of ours in (Extr. Math., 2005),(Collect. Math., 2008) and (J. Algebra Appl., 2008)

Danchev, Peter V.

English

Università del Salento: ESE - Salento University Publishing

Note di Matematica 29, n. 2, 2009, 213–218.Warfield Invariants of V (RG)/GPeter V. Danchev i13, General Kutuzov Str., block 7, floor 2, flat 44003 Plovdiv, Bulgaria.pvdanchev@yahoo.comReceived: 07/07/2008; accepted: 27/11/2008.Abstract. Let R be a commutative unitary ring of prime characteristic p and let G be anAbelian group. We calculate only in terms of R and G (and their sections) Warfield p-invariantsof the quotient group V (RG)/G, that is, the group of all normalized units V (RG) in the groupring RG modulo G. This supplies recent results of ours in (Extr. Math., 2005), (Collect. Math.,2008) and (J. Algebra Appl., 2008).Keywords: Warfield p-invariants, nice subgroups, cardinalities, inequalities, idempotents.MSC 2000 classification: 16S34, 16U60, 20K21.1 IntroductionThroughout the present article, suppose that R is a commutative unitaryring of prime characteristic p, fixed for the duration, and G is an Abelian group,written multiplicatively as is customary when discussing group rings, with p-primary component Gp and torsion part Gt. As usual, RG denotes the groupring of G over R with group of normalized invertible elements V (RG) and itsp-component of torsion Vp(RG). Moreover, let we define inductively, Gp0 = G,Gpα= (Gpα−1)p when α is isolated and Gpα= ∩β<αGpβ when α is limit. Byanalogy Rp0= R, Rpα= (Rpα−1)p when α is isolated and Rpα= ∩β<αRpβ whenα is limit. We shall say that the ring R is perfect if R = Rp. For any set M ,we let |M | designate its cardinality, and ζd designate the primitive d-th root ofunity whenever d is a positive integer.All other unexplained explicitly notations and notions are standard andfollow essentially the classical ones stated in ([5], [6] and [8]).The goal of this paper, that we pursue, is to calculate only in terms of Rand GWarfield p-invariants of V (RG)/G, defined for an arbitrary multiplicativeAbelian group A in the following way (compare with [9]):Wα,p(A) = rank(Apα/(Apα+1Apαp )),iThe author would like to thank the referee for the helpful suggestions made as well asto thank the Editor, Professor Fr. de Giovanni, for his efforts and patience in processing thiswork.Note di MatematicaNote Mat. 29 (2009), n. 2, 213-218ISSN 1123-2536, e-ISSN 1590-0932DOI 10.1285/i15900932v29n2p213http://siba-ese.unisalento.it, © 2009 Università del Salento_________________________________________________________________214 P.V. Danchevwhere α is an ordinal.It easily follows thatWα,p(A) = |Apα/(Apα+1Apαp )| when |Apα/(Apα+1Apαp )| ≥ℵ0 or Wα,p(A) = logp|Apα/(Apα+1Apαp )| otherwise.Our calculations illustrated in the sequel naturally arise for applicable pur-poses and are helpful for the isomorphism description of the factor-groupV (RG)/G. In fact, Warfield p-invariants, together with Ulm-Kaplansky invari-ants, determine, up to isomorphism, p-mixed Warfield groups (e.g., [7]).It is worthwhile noticing that in [1]-[4] we have computed Warfield p-invari-ants of V (RG) under various restrictions on R and G. These computations willbe used here because as it will be proved below, we can restrict in some instancesWarfield p-invariants of V (RG)/G to the Warfield p-invariants of V (RG).2 PreliminariesBefore stating and proving our main result, we need some preparatory ma-chineries.1 Lemma. For every ordinal number α, the following two identities hold:(a) G ∩ V pα(RG) = Gpα;(b) (V (RG)/G)pα= V pα(RG)G/G.Proof. (a) Since it is straightforward that V pα(RG) = V (RpαGpα), theequality now follows without any difficulty.(b) It suffices to show that ∩β<α(V pβ (RG)G) = [∩β<α(V pβ (RG))]G =V pα(RG)G for each limit α. In fact, take x ∈ ∩β<α(V pβ (RG)G) = ∩β<α(V (RpβGpβ)G), hence x = (r1a1 + · · · + rsas)g = (f1b1 + · · · + fsbs)h = · · · , whereri ∈ Rpβ , ai ∈ Gpβ , fi ∈ Rpγ , bi ∈ Gpγ , i ∈ [1, s], β < γ < α; g, h ∈ G. Now,we obtain that ri = fi and gai = hbi, whence aia−1j = bib−1j ∈ Gpγ. Writingx = ga1(r1 + · · ·+ rsasa−11 ), we observe that x ∈ GV (RpγGpγ) = GV pγ(RG).Since the support is finite whereas the number of equalities is not becauseα is infinite being limit, we may assume that all relations are of the above type.That is why, x ∈ (∩γ<αV pγ (RG))G = V pα(RG)G as required. QEDThe next assertion appeared in ([1], Lemma 2). Nevertheless, for the reader’sconvenience and for the completeness of the exposition we shall provide a proof.2 Lemma. For each ordinal number α the following equality holds:Gpα ∩ (V pα+1(RG)V pαp (RG)) = Gpα+1Gpαp .Warfield Invariants of V (RG)/G 215Proof. Since it is routinely checked that V pα(RG) = V (RpαGpα), we maywrite g = uv, where g ∈ Gpα , u ∈ V pα+1(RG) = V (Rpα+1Gpα+1) and v ∈V pαp (RG) = Vp(RpαGpα). Therefore, g(r1a1 + · · · + rsas) = f1b1 + · · · + fsbsand ri = fi with gai = bi, for each i ∈ [1, s], where ri ∈ Rpα+1 , ai ∈ Gpα+1 andfi ∈ Rpα , bi ∈ Gpα . Since f1b1+· · ·+fsbs ∈ Vp(RpαGpα), there is an index, say j,such that bj ∈ Gpαp . Thus gaj = bj secures that g = bja−1j ∈ Gpαp Gpα+1. QEDThe next statement may be found in ([5], p. 157, Exercise 14) as well.3 Lemma. [Dlab] Let A be an Abelian multiplicative group with finite rankand B ≤ A. Then B is neat in A (i.e., B ∩ pA = pB) if and only if r(A) =r(B) + r(A/B).The following corresponding claim is also useful.4 Corollary. [[5], p. 105, Exercise 4] If A is a multiplicative Abelian groupand B ≤ A is a direct factor of A, then r(A) = r(B) + r(A/B).3 Main ResultsWe are now in a position to prove the following5 Theorem. Suppose G is an Abelian group and R is a commutative unitaryring of prime characteristic p without zero divisors. Then, for each ordinal α,the following holds:(1) Wα,p(V (RG)/G) = Wα,p(V (RG))−Wα,p(G)when Wα,p(V (RG)/G) < ℵ0. Thus(1’) Wα,p(V (RG)/G) =∑d/|Gt/Gp|a(d).Wα,p(G/∐l =pGl)−Wα,p(G)where a(d) = |{g ∈ Gt/Gp : order(g) = d}|/(R(ζd) : R) provided that R is aperfect field.(2) Wα,p(V (RG)/G) = Wα,p(V (RG))when Wα,p(V (RG)/G) ≥ ℵ0. Thus(2’) Wα,p(V (RG)/G) = |Gt/Gp|Wα,p(G)provided that R is perfect.Proof. By definition we writeWα,p = rank((V (RG)/G)pα/((V (RG)/G)pα+1(V (RG)/G)pαp )).216 P.V. DanchevBut according to Lemma 1 we may write(V (RG)/G)pα= (V pα(RG)G)/G,(V (RG)/G)pα+1= (V pα+1(RG)G)/Gand (V (RG)/G)pαp = (Vpαp (RG)G)/G. Therefore, using the modular law from[5], we obtain(V (RG)/G)pα/((V (RG)/G)pα+1(V (RG)/G)pαp )= (V pα(RG)G)/G/(V pα+1(RG)V pαp (RG)G)/G∼= (V pα(RG)G)/(V pα+1(RG)V pαp (RG)G)∼= V pα(RG)/(V pα(RG) ∩ [GV pα+1(RG)V pαp (RG)])= V pα(RG)/[V pα+1(RG)V pαp (RG)(G ∩ V pα(RG))]= V pα(RG)/(V pα+1(RG)V pαp (RG)Gpα)∼= V pα(RG)/(V pα+1(RG)V pαp (RG))/(V pα+1(RG)V pαp (RG)Gpα)/(V pα+1(RG)V pαp (RG)).But(V pα+1(RG)V pαp (RG)Gpα)/(V pα+1(RG)V pαp (RG))∼= Gpα/[Gpα ∩ (V pα+1(RG)V pαp (RG))] = Gpα/(Gpα+1Gpαp )by using Lemma 2.Furthermore, since V pα(RG)/(V pα+1(RG)V pαp (RG)Gpα) is an epimorphicimage of the quotient group V pα(RG)/(V pα+1(RG)V pαp (RG)), we observe thatWα,p(V (RG)/G) ≤ Wα,p(V (RG)).Next, we shall show that Wα,p(V (RG)/G) ≥ Wα,p(G) whenever Gt = Gp.In fact, we consider the element e = (1/|C|)∑c∈C rcc ∈ RC ≤ RGq ⊆ RGpα+t ,for any t ∈ IN, where |C| < ℵ0; clearly |C| inverts in R since char(R) = p. Itis not hard to verify that e is an idempotent, i.e., e2 = e. Let g, h ∈ Gpα withgGpα+1Gpαp = hGpα+1Gpαp . Construct the elements xg = eg + (1 − e) and xh =eh+ (1− e). Apparently, xg, xh ∈ V (RG). We claim that xgGpαV pα+1(RG)V pαp(RG) = xhGpαV pα+1(RG)V pαp (RG). If not, xgx−1h = xgxh−1 = (eg + (1 −e))(eh−1+(1− e)) = egh−1+(1− e) = ea+(1− e) ∈ GpαV pα+1(RG)V pαp (RG),where we denote a = gh−1 ∈ Gpα+1Gpαp . By our assumption there exists anatural k such that (ea + (1 − e))pk = eapk + (1 − e) ∈ Gpα+kV pα+k+1(RG) =Gpα+kV (Rpα+k+1Gpα+k+1). Writing e =∑c∈C fcc, we obtain that∑c∈C fccapk+Warfield Invariants of V (RG)/G 2171−∑c∈C fcc ∈ Gpα+kV (Rpα+k+1Gpα+k+1); fc ∈ R. Furthermore, ∑c∈C fccapk +1 −∑c∈C fcc = dpk ∑v∈Gpα fvvpk+1 = ∑v∈Gpα fvdpkvpk+1 , where fv ∈ R andd ∈ Gpα . Thus, dpkvpk+1 ∈ C ⊆ Gpα+k+1 for some v ∈ Gpα , and hence dpk ∈Gpα+k+1. Therefore, capk ∈ Gpα+k+1 and so apk ∈ Gpα+k+1 because c ∈ Gpα+k+1 .Now, apk= bpk+1with b ∈ Gpα , i.e., (ab−p)pk = 1 and ab−p ∈ Gpαp . Consequently,a ∈ Gpα+1Gpαp which is the desired contradiction.Since V pα(RG)/(V pα+1(RG)V pαp (RG)) is a group bounded by p, all its sub-groups are pure and so they are direct factors (see, for example, [5], Theo-rem 27.5). That is why, by what we have just shown above, we may writeV pα(RG)/(V pα+1(RG)V pαp (RG)) ∼= (Gpα/(Gpα+1Gpαp )) × ((V (RG)/G)pα//((V (RG)/G)pα+1(V (RG)/G)pαp )). Consequently, employing Lemma 3 andCorollary 4 (see also [5], p. 157, Exercise 14 and p. 105, Exercise 4), we de-duce thatrank(V pα(RG)/(V pα+1(RG)V pαp (RG))) = rank(Gpα/Gpα+1Gpαp )+rank((V (RG)/G)pα/((V (RG)/G)pα+1(V (RG)/G)pαp )),i.e., Wα,p(V (RG)) = Wα,p(G) + Wα,p(V (RG)/G). By what we have alreadyshown above when Gt = Gp, if Wα,p(V (RG)/G) is finite, then Wα,p(G) is finite,whence Wα,p(V (RG)) is finite and thus Wα,p(V (RG)/G) = Wα,p(V (RG)) −Wα,p(G) whenever Gt = Gp. Note that when Gt = Gp we know via [1] thatWα,p(V (RG)) = Wα,p(G) and that Wα,p(V (RG)/G) = 0. So, the same formulais true even in this case. Further, we apply [3] and [4] to complete point (1’).Let us now Wα,p(V (RG)/G) be infinite; thus Gt = Gp. By virtue of theinequality Wα,p(V (RG)/G) ≥ Wα,p(G) established above we obtain thatWα,p(V (RG)/G) = Wα,p(V (RG)). Finally, we can apply [2] and [3] to concludethat point (2’) is valid. QEDReferences[1] P. V. Danchev: Warfield invariants in abelian group rings, Extr. Math., 20 (2005),233–241.[2] P. V. Danchev: Warfield invariants in abelian group algebras, Collect. Math., 59 (2008),255–262.[3] P. V. Danchev: Warfield invariants in commutative group algebras, J. Algebra Appl., 7(2008), 337–346.[4] P. V. Danchev: Warfield invariants in commutative group rings, J. Algebra Appl., 8(2009), 829–836.[5] L. Fuchs: Infinite Abelian Groups, v. I, Mir, Moskva (1974) (translated from English inRussian).218 P.V. Danchev[6] L. Fuchs: Infinite Abelian Groups, v. II, Mir, Moskva (1977) (translated from Englishin Russian).[7] R. Hunter, F. Richman: Global Warfield groups, Trans. Amer. Math. Soc., 266 (1981),555–572.[8] S. Lang: Algebra, Mir, Moskva, (1968) (translated from English in Russian).[9] R. B. Warfield Jr.: Classification theorems for p-groups and modules over a discretevaluation ring, Bull. Amer. Math. Soc., 78 (1972), 88–92.

Warfield Invariants of $V(RG)/G$

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Warfield Invariants of $V(RG)/G$

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Warfield Invariants of V(RG)/GV(RG)/GV(RG)/G

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Warfield Invariants of $V(RG)/G$