A ring R is filial when for every I, J, if I is an ideal of J and J is an ideal of R then I is an ideal of R. Several characterizations and results on structure of commutative reduced filial rings are obtained

Andruszkiewicz, R.R.

Sobolewska, M.

Algebra and Discrete Mathematics

Наукова електронна бібліотека періодичних видань НАН України (Vernadsky National Library of Ukraine)

Journal Algebra Discrete Math.Algebra and Discrete Mathematics RESEARCH ARTICLENumber 3. (2007). pp. 18 – 26c© Journal “Algebra and Discrete Mathematics”Commutative reduced filial ringsRyszard R. Andruszkiewicz and Magdalena SobolewskaCommunicated by V. V. KirichenkoAbstract. A ring R is filial when for every I, J , if I is anideal of J and J is an ideal of R then I is an ideal of R. Severalcharacterizations and results on structure of commutative reducedfilial rings are obtained.IntroductionAll rings in this paper are associative but we do not assume that eachring has an identity element. By Z we denote the ring of integers and byN the set of positive integers. Moreover, by P we denote the set of allprime integers.We say that a ring R is filial (left filial) when for every I, J , if I is anideal (left ideal) of J and J is an ideal (left ideal) of R then I is an ideal(left ideal) of R.Filial rings appeared independently in some several papers. System-atic investigations of them were begun by Ehrlich [3] (she studied theremostly commutative rings) and were continued in [2], [6], [7], [1], [5].Systematic studies of left filial rings were started in [4]. In particulara structure theorem describing semiprime left filial was obtained (seeTheorem 1) there and it was shown that semiprime left filial rings arefilial. In [1] the complete classification and the method of construction ofcommutative filial domains was given. The classification was proceed byconsidering the set Π(R) = {p ∈ P : p is not a unit in R}. It was shownthat for an arbitrary subset Π of the set of prime numbers, a ring R is afilial integral domain of characteristic 0 with Π(R) = Π if and only if R2000 Mathematics Subject Classification: 16D25, 16D70, 13G05.Key words and phrases: ideal, filial ring, reduced ring.Journal Algebra Discrete Math.R. R. Andruszkiewicz, M. Sobolewska 19is isomorphic to a subring of QΠ =∏{Qp : p ∈ Π} of the form K ∩ ZΠ,where ZΠ =∏{Zp : p ∈ Π}, K is a subfield of QΠ such that for everya ∈ K, a = (ap)p∈Π we have ap ∈ Zp for almost all p ∈ Π and Qp is thequotient field of the p-adic integers Zp.In this paper we generalize methods and results obtained in [1] forreduced commutative rings.To denote that I is an ideal of a ring R, we write I ⊳R. Given a ringR, we denote by R+ the additive group of R.The class of filial rings is closed under taking homomorphic imagesand ideals. Obviously, Z is a filial ring. However, as it was noted in [2],the ring Z ⊕ Z is not filial. Hence the class of filial rings is not closedunder direct sums and extensions.1. General properties of CRF -ringsProposition 1. A commutative ring R is filial if and only if for everya ∈ R, Ra = Ra2 + Za2 +Ra ∩ Za.Proof. Suppose R is filial and let a ∈ R. Then by Proposition 2.1 of[1], Za + Ra = Ra2 + Za2 + Za. Thus Ra ⊆ Ra2 + Za2 + Za. SinceRa2+Za2 ⊆ Ra, so by the modularity of the lattice of subgroups of R+,Ra = Ra2 + Za2 + (Ra ∩ Za).Conversely, let a ∈ R. Then Ra = Ra2+Za2+Ra∩Za, so Za+Ra =Ra2+Za2+Ra∩Za = Ra2+Za2+Za. Therefore R is filial by Proposition2.1 of [1].A ring R containing no non-zero nilpotent is called reduced, i.e. forevery a ∈ R if a2 = 0 then a = 0. We say that R is a CRF -ring when Ris a commutative reduced filial ring. A ring R is called strongly regular iffor every a ∈ R, a ∈ Ra2. Every strongly regular ring is reduced and theclass S of all strongly regular rings is a radical class. It is easy to see thatif R 6= 0 is a commutative domain and R is not a field then S(R) = 0.Every commutative strongly regular ring is a CRF -ring by Proposi-tion 1.Theorem 1 ([4], Theorem 3.4). The following conditions on a ring Rare equivalent:(i) R is reduced and left filial,(ii) R contains an ideal I such that I is strongly regular and R/I is aCRF -ring,(iii) R/S(R) is a CRF -ring.Lemma 1. The additive group of every S-semisimple CRF -ring R istorsion-free.Journal Algebra Discrete Math.20 Commutative reduced filial ringsProof. Suppose R+ is not torsion-free. Then there exist p ∈ P and 0 6=a ∈ R such that pa = 0. Thus Ra is a non-zero algebra over a field ofp-elements and Ra⊳R, so Ra is filial. Therefore by Theorem 4.1 of [5],Ra ∈ S, hence Ra ⊆ S(R) = 0, and Ra = 0, a contradiction.Applying Theorem 1 and Lemma 1, one obtains the following:Lemma 2. Let R be a CRF -ring. Then R/S(R) is a torsion-free CRF -ring.Lemma 3. Let R be a torsion-free CRF -ring. Then for every 0 6= a ∈ R,Ra ∩ Za 6= 0.Proof. Suppose for some 0 6= a ∈ R, Ra ∩ Za = 0. Then by Proposition1, Ra = Ra2 + Za2. Suppose that Ra2 ∩ Za2 6= 0. Thus there existsn ∈ N such that na2 ∈ Ra2, so na2 = xa2 for some x ∈ R. Thereforea(na − xa) = 0, thus (na − xa)2 = 0. Moreover R is reduced, hencena − xa = 0 and na ∈ Ra. Since the group R+ is torsion-free anda 6= 0, we have na 6= 0. Thus na ∈ Ra ∩ Za = 0, a contradiction.Therefore Ra2 ∩ Za2 = 0 and by Proposition 1, Ra2 = Ra4 + Za4. ButRa4 + Za4 ⊆ Ra3, hence Ra2 ⊆ Ra3 ⊆ Ra2, so Ra2 = Ra3. Thereforea3 = ya3 for some y ∈ R. Hence a2(a − ya) = 0, so (a − ya)3 = 0 anda = ya. Consequently 0 6= a ∈ Ra ∩ Za, a contradiction.Theorem 2. Let R be a CRF -ring. Then for every a ∈ R there existsn ∈ N such that na ∈ Ra2.Proof. Suppose R+ is torsion-free. If a = 0 we can take n = 1. Leta 6= 0. Then, by Lemma 3, there exists m ∈ N such that ma ∈ Ra.Moreover Ra⊳R, so Ra is filial. Applying Lemma 3 to the ring Ra oneobtains that n · (ma) ∈ Ra(ma) for some n ∈ N. Hence (nm)a ∈ mRa2.Consequently na ∈ Ra2.Suppose now that R+ is not torsion-free. Let R¯ = R/S(R). Then byLemma 2, R¯ is a torsion-free CRF -ring. Therefore for a¯ = a+S(R) by afirst part of the proof, there exists n ∈ N such that n · a¯ ∈ R¯ · a¯2. Hencena− ra2 = s ∈ S(R) for some r ∈ R, s ∈ S(R). But S(R) ∈ S, so s = bs2for some b ∈ S(R). Consequently na − ra2 = b · (na − ra2)2 ∈ Ra2, sona ∈ Ra2.For every torsion-free ring R we denote by Π(R) the setΠ(R) = {p ∈ P : pR 6= R} .If R has an identity then Π(R) = {p ∈ P : p is not a unit in R}. More-over, let S(X) be the least multiplicative subset of N containing X ⊆ N.Journal Algebra Discrete Math.R. R. Andruszkiewicz, M. Sobolewska 21Lemma 4. Let A be a non-zero ideal of a commutative filial domain R ofcharacteristic 0. Then Π(A) = Π(R) and there exists t ∈ S(Π(A)) suchthat A = tR.Proof. By Proposition 2.5 of [1] there exists a filial integral domain P ofcharacteristic 0 such that R⊳ P . Hence A⊳R and R⊳ P , so A⊳ P byfiliality of P .Now, by Theorem 3.3 of [1] there exist m,n ∈ S(Π(P )) such thatR = mP and A = nP . But A ⊆ R, then n ∈ mP and m | n in P .Therefore by Proposition 3.4 of [1], m | n in Z. Consequently n = m · tfor some t ∈ N, thus t ∈ S(Π(P )) by definition of S(Π(P )) and A =nP = mtP = tR. Since R+ is torsion-free, then by definition of Π(R) wehave Π(A) = Π(R).2. General properties of the radical class TpLet p be a prime number. We denote by Tp the class of all rings R suchthat pR+ = R+. Let us observe that Tp is a radical class. For every ringR ∈ Tp and for every n ∈ N, pnR = R. Moreover, if R is torsion-free thenTp(R) =⋂∞n=1 pnR.Remark 1. Let R be a torsion-free ring. For every prime p, p ∈ Π(R) ifand only if Tp(R) 6= R.Theorem 3. Let p be a prime and let R be a torsion-free ring. Then thering R/Tp(R) is torsion-free.Proof. Take any x ∈ R such that m · x ∈ Tp(R) for some m ∈ N. Thenm = pαk for some α ∈ N ∪ {0}, k ∈ N such that p 6 |k. Since Tp(R) =pαTp(R), so pα(k ·x) ∈ pαTp(R), thus k ·x ∈ Tp(R), because R+ is torsion-free. Let n ∈ N. Then there exist integers ln, kn such that pnln+k·kn = 1.Consequently x = pn(lnx) + kn · (k · x) ∈ pnR + pnTp(R) ⊆ pnR. Hencex ∈⋂∞n=1 pnR, so x ∈ Tp(R).Proposition 2. Let R be a torsion-free CRF -ring. Then for every primep the ring R/Tp(R) is reduced.Proof. Take any a ∈ R such that a2 ∈ Tp(R). By Theorem 2 there existsn ∈ N such that na ∈ Ra2 ⊆ Tp(R). Hence by Theorem 3, a ∈ Tp(R).Therefore the ring R/Tp(R) is reduced.Theorem 4. Let A and B are non-zero torsion-free CRF -rings such thatTp(A) = 0 and Tp(B) = 0 for some prime p. Then A⊕B is not filial.Journal Algebra Discrete Math.22 Commutative reduced filial ringsProof. Suppose A⊕B is filial. By the assumption there exist a ∈ A \ pAand b ∈ B \ pB. Observe that p2A⊕ p2B+Z(pa, pb)⊳ pA⊕ pB⊳A⊕B.So by filiality of A⊕B, p2A⊕ p2B+Z(pa, pb)⊳A⊕B. In particular forany α ∈ A(paα, 0) = (α, 0) · (pa, pb) ∈ p2A⊕ p2B + Z(pa, pb).Hence there exist x ∈ A, y ∈ B, k ∈ Z such that (paα, 0) = (p2x, p2y) +k · (pa, pb), so paα = p2x+ kpa and 0 = p2y + kpb. But A+ and B+ aretorsion-free, so aα = px+ ka and 0 = py + kb.If p 6 |k, then there exist r, s ∈ Z such that kr + ps = 1. Henceb = krb+ psb = r · (−py) + psb = p(sb− ry) ∈ pB, a cont adiction.Therefore p | k and since aα = px+ ka, we have aα ∈ pA. ConsequentlyaA ⊆ pA. (1)Similarly, bB ⊆ pB.Take any n ∈ N such that aA ⊆ pnA i bB ⊆ pnB. We prove thataA ⊆ pn+1A and bB ⊆ pn+1B.By (1) we have that pn+2A⊕pn+2B+Z(pa, pb)⊳pn+1A⊕pn+1B+Z(pa, pb)and pn+1A⊕ pn+1B+Z(pa, pb)⊳ pnA⊕ pnB+Z(pa, pb)⊳A⊕B. So byfiliality of A⊕B:pn+2A⊕ pn+2B + Z(pa, pb)⊳A⊕B. (2)In particular for α ∈ A, (paα, 0) = (pn+2x, pn+2y) + k · (pa, pb) for somex ∈ A, y ∈ B, k ∈ Z. But A+ and B+ are torsion-free, so:aα = pn+1x+ ka and 0 = pn+1y + kb. (3)If pn+1 6 |k, then k = pβ · l for some β ∈ N0, l ∈ Z, p 6 |l and β < n + 1.Thus by (3), l ·b ∈ pB, so b ∈ pB, a contradiction. Therefore pn+1 | k andby (3), aα ∈ pn+1A. Consequently aA ⊆ pn+1A. Similarly, bB ⊆ pn+1B.Therefore aA ⊆ pmA and bB ⊆ pmB for every m ∈ N. Hence aA ⊆⋂∞m=1 pmA = Tp(A) = 0 and bB ⊆⋂∞m=1 pmB = Tp(B) = 0, so aA = 0and bB = 0. In particular a2 = 0 and b2 = 0, thus a = 0 and b = 0, acontradiction.Theorem 5. Let R be a non-zero torsion-free CRF -ring such that Tp(R)= 0 for some prime p. Then R is a domain.Journal Algebra Discrete Math.R. R. Andruszkiewicz, M. Sobolewska 23Proof. Suppose R is not a domain. Then there exist non-zero elementsa, b ∈ R such that a · b = 0. Hence Ra and Rb are non-zero idealsof R and (Ra ∩Rb)2 ⊆ Ra · Rb = 0. So Ra ∩ Rb = 0, because R isreduced. Moreover Tp(Ra) = 0 and Tp(Rb) = 0, thus by Theorem 4,Ra+ Rb = Ra⊕ Rb is not filial. But Ra⊕ Rb⊳ R, so Ra⊕ Rb is filial,a contradiction.Theorem 6. Let R be a non-zero torsion-free CRF -ring. Then for everyprime p ∈ Π(R), Tp(R) is a prime ideal of R.Proof. Denote R¯ = R/Tp(R). By Remark 1, R 6= Tp(R), so R¯ is a non-zero commutative filial ring. Theorem 3 and Proposition 2 imply that R¯is reduced and torsion-free. Moreover Tp(R¯) = 0. So by Theorem 5, R¯ isa domain. Consequently, Tp(R) is a prime ideal of R.3. Main resultsProposition 3. Let R be a torsion-free commutative reduced ring. ThenR is filial if and only if for every a ∈ R:(i) Ra + Za = pRa + Za for every p ∈ P and (ii) ma ∈ Ra2 for somem ∈ N.Proof. Suppose R is filial. Then (ii) holds by Theorem 2.5. Moreover,Rp2a2 + Zp2a2 + Zpa ⊳ Rpa + Zpa ⊳ R, so Rp2a2 + Zp2a2 + Zpa ⊳ R,by filiality of R. Hence Rp2a2 + Zp2a2 + Zpa = Rpa + Zpa. But R istorsion-free, so Ra + Za = Rpa2 + Zpa2 + Za. Therefore Ra + Za ⊆pRa+ Za ⊆ Ra+ Za. Thus Ra+ Za = pRa+ Za.Conversely, let (i) and (ii) holds. If k, l ∈ N are such that Ra +Za = kRa + Za = lRa + Za, then Ra + Za = (kl)Ra + Za. Hence, ifk1, . . . , ks ∈ N and Ra+Za = kiRa+Za for i = 1, . . . , s, then Ra+Za =(k1 · . . . · ki)Ra+Za. Therefore by (i) we have that Ra+Za = nRa+Zafor every n ∈ N. Since by (ii) there exist m ∈ N and b ∈ R suchthat ma = ba2 and moreover Ra + Za = mRa + Za, so Ra + Za =Rba2 + Za ⊆ Ra2 + Za ⊆ Ra2 + Za2 + Za ⊆ Ra + Za. ConsequentlyRa+ Za = Ra2 + Za2 + Za and R is filial by Proposition 2.1 of [1].Applying Proposition 3, one immediately obtain the following.Corollary 1. Let R be a torsion-free commutative reduced ring with anidentity. Then R is filial if and only if:(i) R = pR + Z · 1 for every p ∈ P and (ii) for every a ∈ R there existsm ∈ N such that ma ∈ Ra2.Journal Algebra Discrete Math.24 Commutative reduced filial ringsLemma 5. Let A be a non-zero commutative filial domain of character-istic 0 which is not a field. Then there exist a ∈ A and k ∈ N such thatax = kx for every x ∈ A. Moreover A = pA + Za and |A/pA| = p forevery p ∈ Π(A).Proof. From Proposition 2.5 of [1] there exists a filial domain P of char-acteristic 0 such that A ⊳ P . Thus P is not a field and by Theorem 3.1of [1], Π(P ) 6= ∅. Therefore by Lemma 4, Π(A) = Π(P ) and there existsk ∈ S(Π(A)) such that A = kP . So a = k · 1 ∈ A, which means thatax = kx for every x ∈ A. Take any p ∈ Π(A). Corollary 1 implies thatP = pP + Z · 1. Hence A = pA + Z · (k · 1) = pA + Za. Thus a 6∈ pA.Consequently, |A/pA| = p.Lemma 6. Let A be a non-zero commutative domain of characteristic 0.If |A/pA| = p for every p ∈ Π(A) and for every x ∈ A there exists m ∈ Nsuch that mx ∈ Ax2, then A is filial.Proof. If Π(A) = ∅ then A is a Q-algebra, so x ∈ Ax2 for every x ∈ A.Hence A is a strongly regular ring and A is filial. Now, let Π(A) 6= ∅.Take any 0 6= a ∈ A. Then there exists m ∈ N such that ma ∈ Aa2.Thus ma = ba2 for some b ∈ A. Hence m = ba, which means thatthere exists a minimal natural number n ∈ A. Suppose that n ∈ pAfor some p ∈ Π(A). Then there exists c ∈ A such that n = pc. Ifn = pk for some k ∈ N, then k ∈ A, a contradiction. So, (p, n) = 1and there exist u, v ∈ Z such that nu + pv = 1. Therefore for x ∈ A,x = (nu)x+ p(vx) = p(ucx) + p(vx) ∈ pA and A = pA, a contradiction.Consequently, n 6∈ pA for every p ∈ Π(A). Hence A = pA+Zm for everyp ∈ Π(A). Thus Aa+Za = pAa+Za and A is filial by Proposition 3.Theorem 7. Let R be a torsion-free commutative reduced ring such thatΠ(R) 6= ∅. Then R is filial if and only if |R/pR| = p for every p ∈ Π(R)and for every a ∈ R there exists m ∈ N such that ma ∈ Ra2.Proof. Suppose R is filial. By Proposition 3 it suffices to prove that|R/pR| = p for every p ∈ Π(R). So, take any p ∈ Π(R). Then byTheorems 3 and 6 we have that R¯ = R/Tp(R) is a non-zero commutativefilial domain of characteristic 0. Since Tp(R) ⊆ pR and pR¯ 6= R¯, thusR/pR ∼= R¯/pR¯ and |R/pR| = p by Lemma 5.Conversely, suppose |R/pR| = p for every p ∈ Π(R) and for everya ∈ R there exists m ∈ N such that ma ∈ Ra2. By Proposition 3 itsuffices to prove that Ra + Za = pRa + Za for all a ∈ R and p ∈ Π(R).Take any p ∈ Π(R). Let R¯ = R/Tp(R). Then by Theorem 3, R¯ is torsion-free. Moreover pR¯ 6= R¯, so p ∈ Π(R¯). It is easy check that qR¯ = R¯ forevery q ∈ Π(R¯). Take any x ∈ R such that x2 ∈ Tp(R). Then mx ∈ Rx2Journal Algebra Discrete Math.R. R. Andruszkiewicz, M. Sobolewska 25for some m ∈ N. Thus mx ∈ Tp(R). Consequently, x ∈ Tp(R) and R¯ isreduced. Take any a, b ∈ R \ Tp(R). Suppose that ab ∈ Tp(R). Let n,mbe the minimal non-negative integers such that a ∈ pnR and b ∈ pmR,respectively. Therefore a = pnx and b = pmy for some x, y ∈ R \ pR.Obviously, xy ∈ Tp(R). Moreover, |R/pR| = p, hence R = pR + Zx. Byan easy induction, R = pkR + Zx for every k ∈ N. Take any k ∈ N.Then y = pkr + lx for some r ∈ R and l ∈ Z. Since xy ∈ Tp(R), we havethat y2 ∈ pkR. Consequently, y2 ∈ Tp(R), so y ∈ Tp(R), a contradiction.Therefore R¯ is a domain. Lemma 6 implies that R¯ is filial. Now, byLemma 5 there exist c ∈ R and k ∈ N such that cx − kx ∈ Tp(R) forevery x ∈ R and R = pR+Zc. Take any a ∈ R. Th n s = ac−ka ∈ Tp(R)and ls = zs2 for some l ∈ N and z ∈ R. Thus ls ∈ Tp(R)s. Which givesthat there exists l0 ∈ N such that (p, l0) = 1 and l0s ∈ Tp(R)s. Hencel0s ∈ pRs, so l0s ∈ pRa and l0ac ∈ pRa+Za. Moreover, pac ∈ pRa+Za.Consequently, ac ∈ pRa + Za. Since R = pR + Zc, which gives thatRa ⊆ pRa + Za. Therefore, Ra + Za = pRa + Za, and the proof iscompleted.Theorem 8. Let I be an ideal of a commutative ring R. Let I and R/Ibe torsion-free CRF -rings. If Π(I)∩Π(R/I) = ∅ then R is a torsion-freeCRF -ring and Π(R) = Π(I) ∪Π(R/I).Proof. Obviously, R is a torsion-free commutative reduced ring and forevery a ∈ R there exists m ∈ N such that ma ∈ Ra2. Take any p ∈ Π(R).Suppose p 6∈ Π(R/I). Then R = pR+I. Since R 6= pR, hence I 6= pI andp ∈ Π(I). Therefore Π(R) ⊆ Π(I)∪Π(R/I). Take any p ∈ Π(I). Supposep 6∈ Π(R). Then R = pR and I ⊆ pR. Moreover R/I is torsion-free, soI = pI, a contradiction. Hence Π(I) ⊆ Π(R). Take any p ∈ Π(R/I).Then p(R/I) 6= R/I, so pR 6= R. Which implies that Π(R/I) ⊆ Π(R).Consequently Π(R) = Π(I) ∪Π(R/I).Now, take any p ∈ Π(I). Then by assumptions, p 6∈ Π(R/I), sopR + I = R. Moreover pI 6= I and R/I is torsion-free. Which meansthat I ∩ pR = pI. Consequently, R/pR ∼= I/pI and |R/pR| = p.Finally, take any p ∈ Π(R/I). Then pR+I 6= R and, by assumptions,pI = I. Hence I ⊆ pR and R/pR ∼= (R/I)/p(R/I). Consequently,|R/pR| = p.Therefore R is filial, by Theorem 7, and the proof is completed.Corollary 2. Let I be an ideal of a torsion-free CRF -ring R. If R/I istorsion-free then Π(I) ∩Π(R/I) = ∅.Proof. By assumptions, I and R/I are torsion-free CRF -rings. Supposethere exists p ∈ Π(I)∩Π(R/I). Then pI 6= I and pR+ I 6= R. ThereforeJournal Algebra Discrete Math.26 Commutative reduced filial ringspR 6= R and p ∈ Π(R). So, |R/pR| = p, by Theorem 7. Consequently,pR is a maximal ideal of R. Moreover, R/I is torsion-free, so I 6⊆ pR.Therefore R = pR+ I, a contradiction.As an immediate consequence of Theorem 7 one obtains the following.Corollary 3. Let T be a non-empty subset of N such that for every t ∈ Tthere exists a torsion-free CRF -ring Rt such that Π(Rt) 6= ∅. If for everydistinct t, s ∈ T , Π(Rt)∩Π(Rs) = ∅, then R =⊕t∈T Rt is a torsion-freeCRF -ring and Π(R) =⋃t∈T Π(Rt).References[1] R. R. Andruszkiewicz, The classification of integral domains in which the relationof being an ideal is transitive, Comm. Algebra, N. 31, 2003, No. 5, pp. 2067-2093.[2] R. R. Andruszkiewicz, E. R. Puczy lowski, On filial rings, Portugal. Math., N. 45,1988, No. 2, pp. 139-149.[3] G. Ehrlich, Filial rings, Portugal. Math., N. 42, 1983/1984, pp. 185-194.[4] M. Filipowicz, E. R. Puczy lowski, Left filial rings, Algebra Colloq., N. 11, 2004,No.3, pp. 335-344.[5] M. Filipowicz, E. R. Puczy lowski, On filial and left filial rings, Publ. Math. De-brecen, N. 66, 2005, No. 3-4, pp. 257-267.[6] A. D. Sands, On ideals in over-rings, Publ. Math. Debrecen, N. 35, 1988, pp.273-279.[7] S. Veldsman, Extensions and ideals of rings, Publ. Math. Debrecen, N. 38, 1991,pp. 297-309.Contact informationRyszard R.AndruszkiewiczInstitute of Mathematics,University of Bia lystok,ul. Akademicka 2, 15-267 Bia lystok, PolandE-Mail: randrusz@math.uwb.edu.plURL: math.uwb.edu.pl/∼randrusz/MagdalenaSobolewskaInstitute of Mathematics,University of Bia lystok,ul. Akademicka 2, 15-267 Bia lystok, PolandE-Mail: magdas@math.uwb.edu.plURL: math.uwb.edu.pl/∼magdas/Received by the editors: 16.07.2007and in final form 27.01.2008.

Commutative reduced filial rings

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Commutative reduced filial rings

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Наукова електронна бібліотека періодичних видань НАН України (Vernadsky National Library of Ukraine)