Commutative reduced filial rings
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.
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Інститут прикладної математики і механіки НАН України
2007
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irk-123456789-1523612019-06-11T01:25:12Z Commutative reduced filial rings Andruszkiewicz, R.R. Sobolewska, M. 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. 2007 Article Commutative reduced filial rings / R.R. Andruszkiewicz, M. Sobolewska// Algebra and Discrete Mathematics. — 2007. — Vol. 6, № 3. — С. 18–26. — Бібліогр.: 7 назв. — англ. 1726-3255 2000 Mathematics Subject Classification:16D25, 16D70, 13G05. http://dspace.nbuv.gov.ua/handle/123456789/152361 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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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. |
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Andruszkiewicz, R.R. Sobolewska, M. |
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Andruszkiewicz, R.R. Sobolewska, M. Commutative reduced filial rings Algebra and Discrete Mathematics |
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Andruszkiewicz, R.R. Sobolewska, M. |
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Andruszkiewicz, R.R. |
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Commutative reduced filial rings |
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Commutative reduced filial rings |
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Commutative reduced filial rings |
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Commutative reduced filial rings |
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Commutative reduced filial rings |
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commutative reduced filial rings |
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Інститут прикладної математики і механіки НАН України |
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2007 |
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http://dspace.nbuv.gov.ua/handle/123456789/152361 |
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Commutative reduced filial rings / R.R. Andruszkiewicz, M. Sobolewska// Algebra and Discrete Mathematics. — 2007. — Vol. 6, № 3. — С. 18–26. — Бібліогр.: 7 назв. — англ. |
| series |
Algebra and Discrete Mathematics |
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AT andruszkiewiczrr commutativereducedfilialrings AT sobolewskam commutativereducedfilialrings |
| first_indexed |
2025-07-13T02:54:45Z |
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2025-07-13T02:54:45Z |
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1837498671555411968 |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Number 3. (2007). pp. 18 – 26
c© Journal “Algebra and Discrete Mathematics”
Commutative reduced filial rings
Ryszard R. Andruszkiewicz and Magdalena Sobolewska
Communicated by V. V. Kirichenko
Abstract. 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.
Introduction
All rings in this paper are associative but we do not assume that each
ring has an identity element. By Z we denote the ring of integers and by
N the set of positive integers. Moreover, by P we denote the set of all
prime integers.
We say that a ring R is filial (left filial) when for every I, J , if I is an
ideal (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 there
mostly commutative rings) and were continued in [2], [6], [7], [1], [5].
Systematic studies of left filial rings were started in [4]. In particular
a structure theorem describing semiprime left filial was obtained (see
Theorem 1) there and it was shown that semiprime left filial rings are
filial. In [1] the complete classification and the method of construction of
commutative filial domains was given. The classification was proceed by
considering the set Π(R) = {p ∈ P : p is not a unit in R}. It was shown
that for an arbitrary subset Π of the set of prime numbers, a ring R is a
filial integral domain of characteristic 0 with Π(R) = Π if and only if R
2000 Mathematics Subject Classification: 16D25, 16D70, 13G05.
Key words and phrases: ideal, filial ring, reduced ring.
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.R. R. Andruszkiewicz, M. Sobolewska 19
is 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 every
a ∈ K, a = (ap)p∈Π we have ap ∈ Zp for almost all p ∈ Π and Qp is the
quotient field of the p-adic integers Zp.
In this paper we generalize methods and results obtained in [1] for
reduced commutative rings.
To denote that I is an ideal of a ring R, we write I ⊳R. Given a ring
R, we denote by R+ the additive group of R.
The class of filial rings is closed under taking homomorphic images
and 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 closed
under direct sums and extensions.
1. General properties of CRF -rings
Proposition 1. A commutative ring R is filial if and only if for every
a ∈ 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. Since
Ra2 + 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 Proposition
2.1 of [1].
A ring R containing no non-zero nilpotent is called reduced, i.e. for
every a ∈ R if a2 = 0 then a = 0. We say that R is a CRF -ring when R
is a commutative reduced filial ring. A ring R is called strongly regular if
for every a ∈ R, a ∈ Ra2. Every strongly regular ring is reduced and the
class S of all strongly regular rings is a radical class. It is easy to see that
if 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 R
are equivalent:
(i) R is reduced and left filial,
(ii) R contains an ideal I such that I is strongly regular and R/I is a
CRF -ring,
(iii) R/S(R) is a CRF -ring.
Lemma 1. The additive group of every S-semisimple CRF -ring R is
torsion-free.
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.20 Commutative reduced filial rings
Proof. 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 of
p-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 Proposition
1, Ra = Ra2 + Za2. Suppose that Ra2 ∩ Za2 6= 0. Thus there exists
n ∈ N such that na2 ∈ Ra2, so na2 = xa2 for some x ∈ R. Therefore
a(na − xa) = 0, thus (na − xa)2 = 0. Moreover R is reduced, hence
na − xa = 0 and na ∈ Ra. Since the group R+ is torsion-free and
a 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. But
Ra4 + Za4 ⊆ Ra3, hence Ra2 ⊆ Ra3 ⊆ Ra2, so Ra2 = Ra3. Therefore
a3 = ya3 for some y ∈ R. Hence a2(a − ya) = 0, so (a − ya)3 = 0 and
a = ya. Consequently 0 6= a ∈ Ra ∩ Za, a contradiction.
Theorem 2. Let R be a CRF -ring. Then for every a ∈ R there exists
n ∈ N such that na ∈ Ra2.
Proof. Suppose R+ is torsion-free. If a = 0 we can take n = 1. Let
a 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 one
obtains 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 R̄ = R/S(R). Then by
Lemma 2, R̄ is a torsion-free CRF -ring. Therefore for ā = a + S(R) by a
first part of the proof, there exists n ∈ N such that n · ā ∈ R̄ · ā2. Hence
na− ra2 = s ∈ S(R) for some r ∈ R, s ∈ S(R). But S(R) ∈ S, so s = bs2
for some b ∈ S(R). Consequently na − ra2 = b · (na − ra2)2 ∈ Ra2, so
na ∈ Ra2.
For every torsion-free ring R we denote by Π(R) the set
Π(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.
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.R. R. Andruszkiewicz, M. Sobolewska 21
Lemma 4. Let A be a non-zero ideal of a commutative filial domain R of
characteristic 0. Then Π(A) = Π(R) and there exists t ∈ S(Π(A)) such
that A = tR.
Proof. By Proposition 2.5 of [1] there exists a filial integral domain P of
characteristic 0 such that R ⊳ P . Hence A ⊳ R and R ⊳ P , so A ⊳ P by
filiality of P .
Now, by Theorem 3.3 of [1] there exist m, n ∈ S(Π(P )) such that
R = 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 · t
for 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) we
have Π(A) = Π(R).
2. General properties of the radical class Tp
Let p be a prime number. We denote by Tp the class of all rings R such
that pR+ = R+. Let us observe that Tp is a radical class. For every ring
R ∈ Tp and for every n ∈ N, pnR = R. Moreover, if R is torsion-free then
Tp(R) =
⋂
∞
n=1
pnR.
Remark 1. Let R be a torsion-free ring. For every prime p, p ∈ Π(R) if
and only if Tp(R) 6= R.
Theorem 3. Let p be a prime and let R be a torsion-free ring. Then the
ring R/Tp(R) is torsion-free.
Proof. Take any x ∈ R such that m · x ∈ Tp(R) for some m ∈ N. Then
m = 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. Hence
x ∈
⋂
∞
n=1
pnR, so x ∈ Tp(R).
Proposition 2. Let R be a torsion-free CRF -ring. Then for every prime
p the ring R/Tp(R) is reduced.
Proof. Take any a ∈ R such that a2 ∈ Tp(R). By Theorem 2 there exists
n ∈ 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 that
Tp(A) = 0 and Tp(B) = 0 for some prime p. Then A ⊕ B is not filial.
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.22 Commutative reduced filial rings
Proof. Suppose A⊕B is filial. By the assumption there exist a ∈ A \ pA
and 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 for
any 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+ are
torsion-free, so aα = px + ka and 0 = py + kb.
If p 6 |k, then there exist r, s ∈ Z such that kr + ps = 1. Hence
b = krb + psb = r · (−py) + psb = p(sb − ry) ∈ pB, a contradiction.
Therefore p | k and since aα = px + ka, we have aα ∈ pA. Consequently
aA ⊆ pA. (1)
Similarly, bB ⊆ pB.
Take any n ∈ N such that aA ⊆ pnA i bB ⊆ pnB. We prove that
aA ⊆ 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 by
filiality 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 some
x ∈ 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 and
by (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 = 0
and bB = 0. In particular a2 = 0 and b2 = 0, thus a = 0 and b = 0, a
contradiction.
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.
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.R. R. Andruszkiewicz, M. Sobolewska 23
Proof. Suppose R is not a domain. Then there exist non-zero elements
a, b ∈ R such that a · b = 0. Hence Ra and Rb are non-zero ideals
of R and (Ra ∩ Rb)2 ⊆ Ra · Rb = 0. So Ra ∩ Rb = 0, because R is
reduced. 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 every
prime p ∈ Π(R), Tp(R) is a prime ideal of R.
Proof. Denote R̄ = R/Tp(R). By Remark 1, R 6= Tp(R), so R̄ is a non-
zero commutative filial ring. Theorem 3 and Proposition 2 imply that R̄
is reduced and torsion-free. Moreover Tp(R̄) = 0. So by Theorem 5, R̄ is
a domain. Consequently, Tp(R) is a prime ideal of R.
3. Main results
Proposition 3. Let R be a torsion-free commutative reduced ring. Then
R is filial if and only if for every a ∈ R:
(i) Ra + Za = pRa + Za for every p ∈ P and (ii) ma ∈ Ra2 for some
m ∈ 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 is
torsion-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, if
k1, . . . , 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 + Za
for every n ∈ N. Since by (ii) there exist m ∈ N and b ∈ R such
that ma = ba2 and moreover Ra + Za = mRa + Za, so Ra + Za =
Rba2 + Za ⊆ Ra2 + Za ⊆ Ra2 + Za2 + Za ⊆ Ra + Za. Consequently
Ra + 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 an
identity. Then R is filial if and only if:
(i) R = pR + Z · 1 for every p ∈ P and (ii) for every a ∈ R there exists
m ∈ N such that ma ∈ Ra2.
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.24 Commutative reduced filial rings
Lemma 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 that
ax = kx for every x ∈ A. Moreover A = pA + Za and |A/pA| = p for
every 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 3.1
of [1], Π(P ) 6= ∅. Therefore by Lemma 4, Π(A) = Π(P ) and there exists
k ∈ S(Π(A)) such that A = kP . So a = k · 1 ∈ A, which means that
ax = kx for every x ∈ A. Take any p ∈ Π(A). Corollary 1 implies that
P = 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 ∈ N
such 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 that
there exists a minimal natural number n ∈ A. Suppose that n ∈ pA
for some p ∈ Π(A). Then there exists c ∈ A such that n = pc. If
n = pk for some k ∈ N, then k ∈ A, a contradiction. So, (p, n) = 1
and 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 every
p ∈ 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 that
Π(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 by
Theorems 3 and 6 we have that R̄ = R/Tp(R) is a non-zero commutative
filial domain of characteristic 0. Since Tp(R) ⊆ pR and pR̄ 6= R̄, thus
R/pR ∼= R̄/pR̄ and |R/pR| = p by Lemma 5.
Conversely, suppose |R/pR| = p for every p ∈ Π(R) and for every
a ∈ R there exists m ∈ N such that ma ∈ Ra2. By Proposition 3 it
suffices to prove that Ra + Za = pRa + Za for all a ∈ R and p ∈ Π(R).
Take any p ∈ Π(R). Let R̄ = R/Tp(R). Then by Theorem 3, R̄ is torsion-
free. Moreover pR̄ 6= R̄, so p ∈ Π(R̄). It is easy check that qR̄ = R̄ for
every q ∈ Π(R̄). Take any x ∈ R such that x2 ∈ Tp(R). Then mx ∈ Rx2
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.R. R. Andruszkiewicz, M. Sobolewska 25
for some m ∈ N. Thus mx ∈ Tp(R). Consequently, x ∈ Tp(R) and R̄ is
reduced. Take any a, b ∈ R \ Tp(R). Suppose that ab ∈ Tp(R). Let n, m
be 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. By
an 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 have
that y2 ∈ pkR. Consequently, y2 ∈ Tp(R), so y ∈ Tp(R), a contradiction.
Therefore R̄ is a domain. Lemma 6 implies that R̄ is filial. Now, by
Lemma 5 there exist c ∈ R and k ∈ N such that cx − kx ∈ Tp(R) for
every x ∈ R and R = pR+Zc. Take any a ∈ R. Then s = ac−ka ∈ Tp(R)
and ls = zs2 for some l ∈ N and z ∈ R. Thus ls ∈ Tp(R)s. Which gives
that there exists l0 ∈ N such that (p, l0) = 1 and l0s ∈ Tp(R)s. Hence
l0s ∈ pRs, so l0s ∈ pRa and l0ac ∈ pRa+Za. Moreover, pac ∈ pRa+Za.
Consequently, ac ∈ pRa + Za. Since R = pR + Zc, which gives that
Ra ⊆ pRa + Za. Therefore, Ra + Za = pRa + Za, and the proof is
completed.
Theorem 8. Let I be an ideal of a commutative ring R. Let I and R/I
be torsion-free CRF -rings. If Π(I)∩Π(R/I) = ∅ then R is a torsion-free
CRF -ring and Π(R) = Π(I) ∪ Π(R/I).
Proof. Obviously, R is a torsion-free commutative reduced ring and for
every 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 and
p ∈ Π(I). Therefore Π(R) ⊆ Π(I)∪Π(R/I). Take any p ∈ Π(I). Suppose
p 6∈ Π(R). Then R = pR and I ⊆ pR. Moreover R/I is torsion-free, so
I = 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), so
pR + I = R. Moreover pI 6= I and R/I is torsion-free. Which means
that 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 is
torsion-free then Π(I) ∩ Π(R/I) = ∅.
Proof. By assumptions, I and R/I are torsion-free CRF -rings. Suppose
there exists p ∈ Π(I)∩Π(R/I). Then pI 6= I and pR + I 6= R. Therefore
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.26 Commutative reduced filial rings
pR 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 ∈ T
there exists a torsion-free CRF -ring Rt such that Π(Rt) 6= ∅. If for every
distinct t, s ∈ T , Π(Rt)∩Π(Rs) = ∅, then R =
⊕
t∈T Rt is a torsion-free
CRF -ring and Π(R) =
⋃
t∈T Π(Rt).
References
[1] R. R. Andruszkiewicz, The classification of integral domains in which the relation
of 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 information
Ryszard R.
Andruszkiewicz
Institute of Mathematics,
University of Bia lystok,
ul. Akademicka 2, 15-267 Bia lystok, Poland
E-Mail: randrusz@math.uwb.edu.pl
URL: math.uwb.edu.pl/∼randrusz/
Magdalena
Sobolewska
Institute of Mathematics,
University of Bia lystok,
ul. Akademicka 2, 15-267 Bia lystok, Poland
E-Mail: magdas@math.uwb.edu.pl
URL: math.uwb.edu.pl/∼magdas/
Received by the editors: 16.07.2007
and in final form 27.01.2008.
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