Uniformly 2-absorbing primary ideals of commutative rings
In this study, we introduce the concept of “uniformly 2-absorbing primary ideals” of commutative rings, which imposes a certain boundedness condition on the usual notion of 2-absorbing primary ideals of commutative rings. Then we investigate some properties of uniformly 2-absorbing primary ideals of...
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irk-123456789-1885172023-03-04T01:27:13Z Uniformly 2-absorbing primary ideals of commutative rings Mostafanasab, H. Tekir, Ü. Ulucak, G. In this study, we introduce the concept of “uniformly 2-absorbing primary ideals” of commutative rings, which imposes a certain boundedness condition on the usual notion of 2-absorbing primary ideals of commutative rings. Then we investigate some properties of uniformly 2-absorbing primary ideals of commutative rings with examples. Also, we investigate a specific kind of uniformly 2-absorbing primary ideals by the name of “special 2-absorbing primary ideals” 2020 Article Uniformly 2-absorbing primary ideals of commutative rings / H. Mostafanasab, Ü. Tekir, G. Ulucak // Algebra and Discrete Mathematics. — 2020. — Vol. 29, № 2. — С. 221–240. — Бібліогр.: 16 назв. — англ. 1726-3255 DOI:10.12958/adm476 2010 MSC: Primary 13A15; Secondary 13E05, 13F05 http://dspace.nbuv.gov.ua/handle/123456789/188517 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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In this study, we introduce the concept of “uniformly 2-absorbing primary ideals” of commutative rings, which imposes a certain boundedness condition on the usual notion of 2-absorbing primary ideals of commutative rings. Then we investigate some properties of uniformly 2-absorbing primary ideals of commutative rings with examples. Also, we investigate a specific kind of uniformly 2-absorbing primary ideals by the name of “special 2-absorbing primary ideals” |
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Article |
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Mostafanasab, H. Tekir, Ü. Ulucak, G. |
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Mostafanasab, H. Tekir, Ü. Ulucak, G. Uniformly 2-absorbing primary ideals of commutative rings Algebra and Discrete Mathematics |
| author_facet |
Mostafanasab, H. Tekir, Ü. Ulucak, G. |
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Mostafanasab, H. |
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Uniformly 2-absorbing primary ideals of commutative rings |
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Uniformly 2-absorbing primary ideals of commutative rings |
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Uniformly 2-absorbing primary ideals of commutative rings |
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Uniformly 2-absorbing primary ideals of commutative rings |
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Uniformly 2-absorbing primary ideals of commutative rings |
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uniformly 2-absorbing primary ideals of commutative rings |
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Інститут прикладної математики і механіки НАН України |
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2020 |
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http://dspace.nbuv.gov.ua/handle/123456789/188517 |
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Uniformly 2-absorbing primary ideals of commutative rings / H. Mostafanasab, Ü. Tekir, G. Ulucak // Algebra and Discrete Mathematics. — 2020. — Vol. 29, № 2. — С. 221–240. — Бібліогр.: 16 назв. — англ. |
| series |
Algebra and Discrete Mathematics |
| work_keys_str_mv |
AT mostafanasabh uniformly2absorbingprimaryidealsofcommutativerings AT tekiru uniformly2absorbingprimaryidealsofcommutativerings AT ulucakg uniformly2absorbingprimaryidealsofcommutativerings |
| first_indexed |
2025-07-16T10:36:40Z |
| last_indexed |
2025-07-16T10:36:40Z |
| _version_ |
1837799520668221440 |
| fulltext |
“adm-n2” — 2020/7/8 — 8:15 — page 221 — #81
© Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 29 (2020). Number 2, pp. 221–240
DOI:10.12958/adm476
Uniformly 2-absorbing primary ideals of
commutative rings
H. Mostafanasab, Ü. Tekir, and G. Ulucak
Communicated by V. A. Artamonov
Abstract. In this study, we introduce the concept of “uni-
formly 2-absorbing primary ideals” of commutative rings, which
imposes a certain boundedness condition on the usual notion of
2-absorbing primary ideals of commutative rings. Then we inves-
tigate some properties of uniformly 2-absorbing primary ideals of
commutative rings with examples. Also, we investigate a specific
kind of uniformly 2-absorbing primary ideals by the name of “special
2-absorbing primary ideals”.
Introduction
Throughout this paper, we assume that all rings are commutative with
1 6= 0. Let R be a commutative ring. An ideal I of R is a proper ideal if
I 6= R. Then ZI(R) = {r ∈ R | rs ∈ I for some s ∈ R\I} for a proper
ideal I of R. Additively, if I is an ideal of R, then the radical of I is given
by
√
I = {r ∈ R | rn ∈ I for some positive integer n}. Let I, J be two
ideals of R. We will denote by (I :R J), the set of all r ∈ R such that
rJ ⊆ I.
Cox and Hetzel have introduced uniformly primary ideals of a commu-
tative ring with nonzero identity in [6]. They said that a proper ideal Q
of a commutative ring R is uniformly primary if there exists a positive
integer n such that whenever r, s ∈ R satisfy rs ∈ Q and r /∈ Q, then
2010 MSC: Primary 13A15; Secondary 13E05, 13F05.
Key words and phrases: uniformly 2-absorbing primary ideal, Noether strongly
2-absorbing primary ideal, 2-absorbing primary ideal.
https://doi.org/10.12958/adm476
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222 Uniformly 2-absorbing Primary Ideals
sn ∈ Q. A uniformly primary ideal Q has order N and write ordR(Q) = N ,
or simply ord(Q) = N if the ring R is understood, if N is the smallest
positive integer for which the aforementioned property holds.
Badawi [3] said that a proper ideal I of R is a 2-absorbing ideal of R
if whenever a, b, c ∈ R and abc ∈ I, then ab ∈ I or ac ∈ I or bc ∈ I. He
proved that I is a 2-absorbing ideal of R if and only if whenever I1, I2, I3
are ideals of R with I1I2I3 ⊆ I, then I1I2 ⊆ I or I1I3 ⊆ I or I2I3 ⊆ I.
Anderson and Badawi [1] generalized the notion of 2-absorbing ideals to
n-absorbing ideals. A proper ideal I of R is called an n-absorbing (resp. a
strongly n-absorbing) ideal if whenever x1 · · ·xn+1 ∈ I for x1, . . . , xn+1 ∈ R
(resp. I1 · · · In+1 ⊆ I for ideals I1, . . . , In+1 of R), then there are n of the
xi’s (resp. n of the Ii’s) whose product is in I. Badawi et. al. [4] defined
a proper ideal I of R to be a 2-absorbing primary ideal of R if whenever
a, b, c ∈ R and abc ∈ I, then either ab ∈ I or ac ∈
√
I or bc ∈
√
I. Let
I be a 2-absorbing primary ideal of R. Then P =
√
I is a 2-absorbing
ideal of R by [4, Theorem 2.2]. We say that I is a P -2-absorbing primary
ideal of R. For more studies concerning 2-absorbing (submodules) ideals
we refer to [5, 9, 10, 15, 16]. These concepts motivate us to introduce a
generalization of uniformly primary ideals. A proper ideal Q of R is said
to be a uniformly 2-absorbing primary ideal of R if there exists a positive
integer n such that whenever a, b, c ∈ R satisfy abc ∈ Q, ab /∈ Q and
ac /∈ √
Q, then (bc)n ∈ Q. In particular, if for n = 1 the above property
holds, then we say that Q is a special 2-absorbing primary ideal of R.
In section 2, we introduce the concepts of uniformly 2-absorbing pri-
mary ideals and Noether strongly 2-absorbing primary ideals. Then we
investigate the relationship between uniformly 2-absorbing primary ideals,
Noether strongly 2-absorbing primary ideals and 2-absorbing primary
ideals. After that, in Theorem 2 we characterize uniformly 2-absorbing pri-
mary ideals. We show that if Q1, Q2 are uniformly primary ideals of a ring
R, then Q1 ∩Q2 and Q1Q2 are uniformly 2-absorbing primary ideals of R,
Theorem 4. Let R = R1 ×R2, where R1 and R2 are rings with 1 6= 0. It is
shown (Theorem 5) that a proper ideal Q of R is a uniformly 2-absorbing
primary ideal of R if and only if either Q = Q1 ×R2 for some uniformly
2-absorbing primary ideal Q1 of R1 or Q = R1 ×Q2 for some uniformly
2-absorbing primary ideal Q2 of R2 or Q = Q1 ×Q2 for some uniformly
primary ideal Q1 of R1 and some uniformly primary ideal Q2 of R2.
In section 3, we give some properties of special 2-absorbing primary
ideals. For example, in Theorem 7 we show that Q is a special 2-absorbing
primary ideal of R if and only if for every ideals I, J,K of R, IJK ⊆ Q
implies that either IJ ⊆ √
Q or IK ⊆ Q or JK ⊆ Q. We prove that
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H. Mostafanasab, Ü. Tekir, G. Ulucak 223
if Q is a special 2-absorbing primary ideal of R and x ∈ R\√Q, then
(Q :R x) is a special 2-absorbing primary ideal of R, Theorem 8. It is
proved (Theorem 9) that an irreducible ideal Q of R is special 2-absorbing
primary if and only if (Q :R x) = (Q :R x2) for every x ∈ R\√Q. Let
R be a Prüfer domain and I be an ideal of R. In Corollary 10 we show
that Q is a special 2-absorbing primary ideal of R if and only if Q[X] is a
special 2-absorbing primary ideal of R[X].
1. Uniformly 2-absorbing primary ideals
Let Q be a P -primary ideal of R. We recall from [6] that Q is a Noether
strongly primary ideal of R if Pn ⊆ Q for some positive integer n. We say
that N is the exponent of Q if N is the smallest positive integer for which
the above property holds and it is denoted by e(Q) = N .
Definition 1. Let Q be a proper ideal of a ring R.
1) Q is a uniformly 2-absorbing primary ideal of R if there exists a
positive integer n such that whenever a, b, c ∈ R satisfy abc ∈ Q,
ab /∈ Q and ac /∈ √
Q, then (bc)n ∈ Q. We call that N is order of Q
if N is the smallest positive integer for which the above property
holds and it is denoted by 2- ordR(Q) = N or 2- ord(Q) = N .
2) P -2-absorbing primary ideal Q is a Noether strongly 2-absorbing
primary ideal of R if Pn ⊆ Q for some positive integer n. We say
that N is the exponent of Q if N is the smallest positive integer for
which the above property holds and it is denoted by 2-e(Q) = N .
A valuation ring is an integral domain V such that for every element
x of its field of fractions K, at least one of x or x−1 belongs to K.
Proposition 1. Let V be a valuation ring with the quotient field K and
let Q be a proper ideal of V . The following conditions are equivalent:
1) Q is a uniformly 2-absorbing primary ideal of V ;
2) There exists a positive integer n such that for every x, y, z ∈ K
whenever xyz ∈ Q and xy /∈ Q, then xz ∈ √
Q or (yz)n ∈ Q.
Proof. (1)⇒(2) Assume that Q is a uniformly 2-absorbing primary ideal
of V . Let xyz ∈ Q for some x, y, z ∈ K such that xy /∈ Q. If z /∈ V , then
z−1 ∈ V , since V is valuation. So xyzz−1 = xy ∈ Q, a contradiction.
Hence z ∈ V . If x, y ∈ V , then there is nothing to prove. If y /∈ V , then
xz ∈ Q ⊆ √
Q, and if x /∈ V , then yz ∈ Q. Consequently we have the
claim.
(2)⇒(1) It is clear.
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224 Uniformly 2-absorbing Primary Ideals
Proposition 2. Let Q1, Q2 be two Noether strongly primary ideals of
a ring R. Then Q1 ∩ Q2 and Q1Q2 are Noether strongly 2-absorbing
primary ideals of R such that 2-e(Q1 ∩ Q2) 6 max{e(Q1), e(Q2)} and
2-e(Q1Q2) 6 e(Q1) + e(Q2).
Proof. Since Q1, Q2 are primary ideals of R, then Q1 ∩Q2 and Q1Q2 are
2-absorbing primary ideals of R, by [4, Theorem 2.4].
Proposition 3. If Q is a uniformly 2-absorbing primary ideal of R, then
Q is a 2-absorbing primary ideal of R.
Proof. Straightforward.
Proposition 4. Let R be a ring and Q be a proper ideal of R.
1) If Q is a 2-absorbing ideal of R, then
(a) Q is a Noether strongly 2-absorbing primary ideal with 2-e(Q) 6 2.
(b) Q is a uniformly 2-absorbing primary ideal with 2-ord(Q) = 1.
2) If Q is a uniformly primary ideal of R, then it is a uniformly 2-
absorbing primary ideal with 2-ord(Q) = 1.
Proof. (1) (a) If Q is a 2-absorbing ideal, then it is a 2-absorbing primary
ideal and (
√
Q)2 ⊆ Q, by [3, Theorem 2.4].
(b) It is evident.
(2) Let Q be a uniformly primary ideal of R and let abc ∈ Q for some
a, b, c ∈ R such that ac /∈ √
Q. Since Q is uniformly primary, abc ∈ Q and
ac /∈ √
Q, then b ∈ Q. Therefore ab ∈ Q or bc ∈ Q. Consequently Q is a
uniformly 2-absorbing primary ideal with 2-ord(Q) = 1.
Example 1. Let R = K[X,Y ] where K is a field. Then Q =
(X2, XY, Y 2)R is a Noether strongly (X,Y )R-primary ideal of R and so
it is a Noether strongly 2-absorbing primary ideal of R.
Proposition 5. If Q is a Noether strongly 2-absorbing primary ideal of R,
then Q is a uniformly 2-absorbing primary ideal of R and 2- ord(Q) 62-
e(Q).
Proof. Let Q be a Noether strongly 2-absorbing primary ideal of R. Now,
let a, b, c ∈ R such that abc ∈ Q, ab /∈ Q, ac /∈ √
Q. Then bc ∈ √
Q since
Q is a 2-absorbing primary ideal of R. Thus (bc)2-e(Q) ∈ (
√
Q)2-e(Q) ⊆ Q.
Therefore, Q is a uniformly 2-absorbing primary ideal and also 2- ord(Q) 6
2-e(Q).
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H. Mostafanasab, Ü. Tekir, G. Ulucak 225
In the following example, we show that the converse of Proposition 5
is not true. We make use of [6, Example 6 and Example 7]
Example 2. Let R be a ring of characteristic 2 and T = R[X] where X =
{X1, X2, X3, . . . } is a set of indeterminates over R. Let Q = ({X2
i }∞i=1)T .
By [6, Example 7] Q is a uniformly P -primary ideal of T with ordT (Q) = 1
where P = (X)T . Then Q is a uniformly 2-absorbing primary ideal of T
with 2-ordT (Q) = 1, by Proposition 4(2). But Q is not a Noether strongly
2-absorbing primary ideal since for every positive integer n, Pn * Q.
Remark 1. Every 2-absorbing ideal of a ring R is a uniformly 2-absorbing
primary ideal, but the converse does not necessarily hold. For example, let
p, q be two distinct prime numbers. Then p2qZ is a 2-absorbing primary
ideal of Z, [4, Corollary 2.12]. On the other hand (
√
p2qZ)2 = p2q2Z ⊆
p2qZ, and so p2qZ is a Noether strongly 2-absorbing primary ideal of Z.
Hence Proposition 5 implies that p2qZ is a uniformly 2-absorbing primary
ideal. But, notice that p2q ∈ p2qZ and neither p2 ∈ p2qZ nor pq ∈ p2qZ
which shows that p2qZ is not a 2-absorbing ideal of Z. Also, it is easy to
see that p2qZ is not primary and so it is not a uniformly primary ideal
of Z. Consequently the two concepts of uniformly primary ideals and of
uniformly 2-absorbing primary ideals are different in general.
Proposition 6. Let R be a ring and Q be a proper ideal of R. If Q is
a uniformly 2-absorbing primary ideal of R, then one of the following
conditions must hold:
1)
√
Q = p is a prime ideal.
2)
√
Q = p ∩ q, where p and q are the only distinct prime ideals of R
that are minimal over Q.
Proof. Use [4, Theorem 2.3].
Let R be a ring and I be an ideal of R. We denote by I [n] the ideal of
R generated by the n-th powers of all elements of I. If n! is a unit in R,
then I [n] = In, see [2].
Theorem 1. Let Q be a proper ideal of R. Then the following conditions
are equivalent:
1) Q is uniformly primary;
2) There exists a positive integer n such that for every ideals I, J of R,
IJ ⊆ Q implies that either I ⊆ Q or J [n] ⊆ Q;
3) There exists a positive integer n such that for every a ∈ R either
a ∈ Q or (Q :R a)[n] ⊆ Q;
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226 Uniformly 2-absorbing Primary Ideals
4) There exists a positive integer n such that for every a ∈ R either
an ∈ Q or (Q :R a) = Q.
Proof. (1)⇒(2) Suppose that Q is uniformly primary with ord(Q) = n.
Let IJ ⊆ Q for some ideals I, J of R. Assume that neither I ⊆ Q nor
J [n] ⊆ Q. Then there exist elements a ∈ I\Q and bn ∈ J [n]\Q, where
b ∈ J . Since ab ∈ IJ ⊆ Q, then either a ∈ Q or bn ∈ Q, which is a
contradiction. Therefore either I ⊆ Q or J [n] ⊆ Q.
(2)⇒(3) Note that a(Q :R a) ⊆ Q for every a ∈ R.
(3)⇒(1) and (1)⇔(4) have easy verifications.
Corollary 1. Let R be a ring. Suppose that n! is a unit in R for every
positive integer n, and Q is a proper ideal of R. The following conditions
are equivalent:
1) Q is uniformly primary;
2) There exists a positive integer n such that for every ideals I, J of R,
IJ ⊆ Q implies that either I ⊆ Q or Jn ⊆ Q;
3) There exists a positive integer n such that for every a ∈ R either
a ∈ Q or (Q :R a)n ⊆ Q;
4) There exists a positive integer n such that for every a ∈ R either
an ∈ Q or (Q :R a) = Q.
In the following theorem we characterize uniformly 2-absorbing primary
ideals.
Theorem 2. Let Q be a proper ideal of R. Then the following conditions
are equivalent:
1) Q is uniformly 2-absorbing primary;
2) There exists a positive integer n such that for every a, b ∈ R either
(ab)n ∈ Q or (Q :R ab) ⊆ (Q :R a) ∪ (
√
Q :R b);
3) There exists a positive integer n such that for every a, b ∈ R either
(ab)n ∈ Q or (Q :R ab) = (Q :R a) or (Q :R ab) ⊆ (
√
Q :R b);
4) There exists a positive integer n such that for every a, b ∈ R and
every ideal I of R, abI ⊆ Q implies that either aI ⊆ Q or bI ⊆ √
Q
or (ab)n ∈ Q;
5) There exists a positive integer n such that for every a, b ∈ R either
ab ∈ Q or (Q :R ab)[n] ⊆ (
√
Q :R a) ∪ (Q :R bn);
6) There exists a positive integer n such that for every a, b ∈ R either
ab ∈ Q or (Q :R ab)[n] ⊆ (
√
Q :R a) or (Q :R ab)[n] ⊆ (Q :R bn).
Proof. (1)⇒(2) Suppose that Q is uniformly 2-absorbing primary with
2- ord(Q) = n. Assume that a, b ∈ R such that (ab)n /∈ Q. Let x ∈ (Q:Rab).
“adm-n2” — 2020/7/8 — 8:15 — page 227 — #87
H. Mostafanasab, Ü. Tekir, G. Ulucak 227
Thus xab ∈ Q, and so either xa ∈ Q or xb ∈ √
Q. Hence x ∈ (Q :R a) or
x ∈ (
√
Q :R b) which shows that (Q :R ab) ⊆ (Q :R a) ∪ (
√
Q :R b).
(2)⇒(3) By the fact that if an ideal is a subset of the union of two
ideals, then it is a subset of one of them.
(3)⇒(4) Suppose that n is a positive number which exists by part (3).
Let a, b ∈ R and I be an ideal of R such that abI ⊆ Q and (ab)n /∈ Q.
Then I ⊆ (Q :R ab), and so I ⊆ (Q :R a) or I ⊆ (
√
Q :R b), by (3).
Consequently aI ⊆ Q or bI ⊆ √
Q.
(4)⇒(1) Is easy.
(1)⇒(5) Suppose that Q is uniformly 2-absorbing primary with
2- ord(Q) = n. Assume that a, b ∈ R such that ab /∈ Q. Let x ∈ (Q :R ab).
Then abx ∈ Q. So ax ∈ √
Q or (bx)n ∈ Q. Hence xn ∈ (
√
Q :R a) or
xn ∈ (Q :R bn). Consequently (Q :R ab)[n] ⊆ (
√
Q :R a) ∪ (Q :R bn).
(5)⇒(6) Is similar to the proof of (2)⇒(3).
(6)⇒(1) Assume (6). Let abc ∈ Q for some a, b, c ∈ R such that ab /∈ Q.
Then c ∈ (Q :R ab) and thus cn ∈ (Q :R ab)[n]. So, by part (6) we have
that cn ∈ (
√
Q :R a) or cn ∈ (Q :R bn). Therefore ac ∈ √
Q or (bc)n ∈ Q,
and so Q is uniformly 2-absorbing primary.
Corollary 2. Let R be a ring. Suppose that n! is a unit in R for every
positive integer n, and Q is a proper ideal of R. The following conditions
are equivalent:
1) Q is uniformly 2-absorbing primary;
2) There exists a positive integer n such that for every a, b ∈ R either
ab ∈ Q or (Q :R ab)n ⊆ (
√
Q :R a) ∪ (Q :R bn);
3) There exists a positive integer n such that for every a, b ∈ R either
ab ∈ Q or (Q :R ab)n ⊆ (
√
Q :R a) or (Q :R ab)n ⊆ (Q :R bn).
Proposition 7. Let Q be a uniformly 2-absorbing primary ideal of R and
x ∈ R\Q be idempotent. The following conditions hold:
1) (
√
Q :R x) =
√
(Q :R x).
2) (Q :R x) is a uniformly 2-absorbing primary ideal of R with 2-
ord((Q :R x)) 6 2-ord(Q).
Proof. (1) Is easy.
(2) Suppose that 2-ord(Q) = n. Let abc ∈ (Q :R x) for some a, b, c ∈ R.
Then a(bc)x ∈ Q and so either abc ∈ Q or ax ∈ √
Q or (bc)nx ∈ Q. If
abc ∈ Q, then either ab ∈ Q ⊆ (Q :R x) or ac ∈ √
Q ⊆
√
(Q :R x) or
(bc)n ∈ Q ⊆ (Q :R x). If ax ∈ √
Q, then ac ∈ (
√
Q :R x) =
√
(Q :R x) by
part (1). In the third case we have (bc)n ∈ (Q :R x). Hence (Q :R x) is a
uniformly 2-absorbing primary ideal of R with 2-ord((Q :R x)) 6 n.
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228 Uniformly 2-absorbing Primary Ideals
Proposition 8. Let I be a proper ideal of a ring R.
1)
√
I is a 2-absorbing ideal of R.
2) For every a, b, c ∈ R, abc ∈ I implies that ab ∈
√
I or ac ∈
√
I or
bc ∈
√
I;
3)
√
I is a 2-absorbing primary ideal of R;
4)
√
I is a Noether 2-absorbing primary ideal of R (2-e(
√
I) = 1);
5)
√
I is a uniformly 2-absorbing primary ideal of R.
Proof. (1)⇒(2) It is trivial.
(2)⇒(1) Let xyz ∈
√
I for some x, y, z ∈ R. Then there exists a
positive integer m such that xmymzm ∈ I. So, the hypothesis in (2)
implies that xmym ∈
√
I or xmzm ∈
√
I or ymzm ∈
√
I. Hence xy ∈
√
I
or xz ∈
√
I or yz ∈
√
I which shows that
√
I is a 2-absorbing ideal.
(1)⇔(3) and (3)⇒(4) are clear.
(4)⇒(5) By Proposition 5.
(5)⇒(3) It is easy.
Proposition 9. If Q1 is a uniformly P -primary ideal of R and Q2 is
a uniformly P -2-absorbing primary ideal of R such that Q1 ⊆ Q2, then
2- ord(Q2) 6 ord(Q1).
Proof. Let ord(Q1) = m and 2- ord(Q2) = n. Then there are a, b, c ∈ R
such that abc ∈ Q2, ab /∈ Q2, ac /∈ √
Q2 and (bc)n ∈ Q2 but (bc)n−1 /∈ Q2.
Thus bc ∈ √
Q2 =
√
Q1. Hence (bc)m ∈ Q1 ⊆ Q2 by [6, Proposition 8].
Therefore, n > m− 1 and so n > m.
Theorem 3. Let R be a ring and {Qi}i∈I be a chain of uniformly P -2-
absorbing primary ideals such that maxi∈I{2- ord(Qi)} = n, where n is a
positive integer. Then Q =
⋂
i∈I
Qi is a uniformly P -2-absorbing primary
ideal of R with 2-ord(Q) 6 n.
Proof. It is clear that
√
Q =
⋂
i∈I
√
Qi = P . Let a, b, c ∈ R such that
abc ∈ Q, ab /∈ Q and (bc)n /∈ Q. Since {Qi}i∈I is a chain, there exists some
k ∈ I such that ab /∈ Qk and (bc)n /∈ Qk. On the other handQk is uniformly
2-absorbing primary with 2-ord(Qk) 6 n, thus ac ∈ √
Qi =
√
Q, and so
Q is a uniformly 2-absorbing primary ideal of R with 2-ord(Q) 6 n.
In the following remark, we show that if Q1 and Q2 are uniformly
2-absorbing primary ideals of R, then Q1 ∩Q2 need not be a uniformly
2-absorbing primary ideal of R.
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H. Mostafanasab, Ü. Tekir, G. Ulucak 229
Remark 2. Let p, q, r be distinct prime numbers. Then p2qZ and rZ are
uniformly 2-absorbing primary ideals of Z. Notice that p2qr ∈ p2qZ ∩ rZ
and neither p2q ∈ p2qZ ∩ rZ nor p2r ∈
√
p2qZ ∩ rZ = pZ ∩ qZ ∩ rZ nor
qr ∈
√
p2qZ ∩ rZ = pZ ∩ qZ ∩ rZ. Hence p2qZ ∩ rZ is not a 2-absorbing
primary ideal of Z which shows that it is not a uniformly 2-absorbing
primary ideal of Z.
Theorem 4. Let Q1, Q2 be uniformly primary ideals of a ring R.
1) Q1 ∩ Q2 is a uniformly 2-absorbing primary ideal of R with 2-
ord(Q1 ∩Q2) 6 max{ord(Q1), ord(Q2)}.
2) Q1Q2 is a uniformly 2-absorbing primary ideal of R with
2-ord(Q1Q2) 6 ord(Q1) + ord(Q2).
Proof. (1) Let Q1, Q2 be uniformly primary. Set n = max{ord(Q1),
ord(Q2)}. Assume that for some a, b, c ∈ R, abc ∈ Q1 ∩Q2, ab /∈ Q1 ∩Q2
and ac /∈ √
Q1 ∩Q2. Since Q1 and Q2 are primary ideals of R, then Q1∩Q2
is 2-absorbing primary by [4, Theorem 2.4]. Therefore bc ∈ √
Q1 ∩Q2 =√
Q1 ∩
√
Q2. By [6, Proposition 8] we have that (bc)ord(Q1) ∈ Q1 and
(bc)ord(Q2) ∈ Q2. Hence (bc)n ∈ Q1 ∩ Q2 which shows that Q1 ∩ Q2 is
uniformly 2-absorbing primary and 2-ord(Q1 ∩Q2) 6 n.
(2) Similar to the proof in (1).
We recall from [7], if R is an integral domain and P is a prime ideal
of R that can be generated by a regular sequence of R, then, for each
positive integer n, the ideal Pn is a P -primary ideal of R.
Lemma 1. ([6, Corollary 4]) Let R be a ring and P be a prime ideal of
R. If Pn is a P -primary ideal of R for some positive integer n, then Pn
is a uniformly primary ideal of R with ord(Pn) 6 n.
Corollary 3. Let R be a ring and P1, P2 be prime ideals of R. If Pn
1 is a
P1-primary ideal of R for some positive integer n and Pm
2 is a P2-primary
ideal of R for some positive integer m, then Pn
1 P
m
2 and Pn
1 ∩ Pm
2 are
uniformly 2-absorbing primary ideals of R with 2-ord(Pn
1 P
m
2 ) 6 n +m
and 2-ord(Pn
1 ∩ Pm
2 ) 6 max{n,m}.
Proof. By Theorem 4 and Lemma 1.
Proposition 10. Let f : R −→ R′ be a homomorphism of commutative
rings. Then the following statements hold:
1) If Q′ is a uniformly 2-absorbing primary ideal of R′, then f−1(Q′) is
a uniformly 2-absorbing primary ideal of R with 2- ordR(f
−1(Q′)) 6
2- ordR′(Q′).
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230 Uniformly 2-absorbing Primary Ideals
2) If f is an epimorphism and Q is a uniformly 2-absorbing primary
ideal of R containing ker(f), then f(Q) is a uniformly 2-absorbing
primary ideal of R′ with 2- ordR′(f(Q)) 6 2- ordR(Q).
Proof. (1) Set N = 2- ordR′(Q′). Let a, b, c ∈ R such that abc ∈ f−1(Q′),
ab /∈ f−1(Q′) and ac /∈
√
f−1(Q′) = f−1(
√
Q′). Then f(abc) =
f(a)f(b)f(c) ∈ Q′, f(ab) = f(a)f(b) /∈ Q′ and f(ac) = f(a)f(c) /∈
√
Q′.
Since Q′ is a uniformly 2-absorbing primary ideal of R′, then fN (bc) ∈ Q′.
Then f((bc)N ) ∈ Q′ and so (bc)N ∈ f−1(Q′). Thus f−1(Q′) is a uniformly
2-absorbing primary ideal of R with 2- ordR(f
−1(Q′)) 6 N = 2- ordR′(Q′).
(2) Set N = 2- ordR(Q). Let a, b, c ∈ R′ such that abc ∈ f(Q), ab /∈
f(Q) and ac /∈
√
f(Q). Since f is an epimorphism, then there exist
x, y, z ∈ R such that f(x) = a, f(y) = b and f(z) = c. Then f(xyz) = abc
∈ f(Q), f(xy) = ab /∈ f(Q) and f(xz) = ac /∈
√
f(Q). Since ker(f) ⊆ Q,
then xyz ∈ Q. Also xy /∈ Q, and xz /∈ √
Q, since f(
√
Q) ⊆
√
f(Q).
Then (yz)N ∈ Q since Q is a uniformly 2-absorbing primary ideal of
R. Thus f((yz)N ) = (f(y)f(z))N = (bc)N ∈ f(Q). Therefore, f(Q) is a
uniformly 2-absorbing primary ideal of R′. Moreover 2- ordR′(f(Q)) 6
N = 2- ordR(Q).
As an immediate consequence of Proposition 10 we have the following
result:
Corollary 4. Let R be a ring and Q be an ideal of R.
1) If R′ is a subring of R and Q is a uniformly 2-absorbing primary
ideal of R, then Q ∩R′ is a uniformly 2-absorbing primary ideal of
R′ with 2- ordR′(Q ∩R′) 6 2- ordR(Q).
2) Let I be an ideal of R with I ⊆ Q. Then Q is a uniformly 2-absorbing
primary ideal of R if and only if Q/I is a uniformly 2-absorbing
primary ideal of R/I.
Corollary 5. Let Q be an ideal of a ring R. Then 〈Q,X〉 is a uniformly 2-
absorbing primary ideal of R[X] if and only if Q is a uniformly 2-absorbing
primary ideal of R.
Proof. By Corollary 4(2) and regarding the isomorphism 〈Q,X〉/〈X〉 ≃ Q
in R[X]/〈X〉 ≃ R we have the result.
Corollary 6. Let R be a ring, Q a proper ideal of R and X = {Xi}i∈I a
collection of indeterminates over R. If QR[X] is a uniformly 2-absorbing
primary ideal of R[X], then Q is a uniformly 2-absorbing primary ideal
of R with 2- ordR(Q) 6 2- ordR[X](QR[X]).
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H. Mostafanasab, Ü. Tekir, G. Ulucak 231
Proof. It is clear from Corollary 4(1).
Proposition 11. Let S be a multiplicatively closed subset of R and Q be
a proper ideal of R. Then the following conditions hold:
1) If Q is a uniformly 2-absorbing primary ideal of R such that Q∩S =
∅, then S−1Q is a uniformly 2-absorbing primary ideal of S−1R
with 2- ord(S−1Q) 6 2- ord(Q).
2) If S−1Q is a uniformly 2-absorbing primary ideal of S−1R and
S ∩ZQ(R) = ∅, then Q is a uniformly 2-absorbing primary ideal of
R with 2- ord(Q) 6 2- ord(S−1Q).
Proof. (1) Set N := 2- ord(Q). Let a, b, c ∈ R and s, t, k ∈ S such that
a
s
b
t
c
k
∈ S−1Q, a
s
b
t
/∈ S−1Q, a
s
c
k
/∈
√
S−1Q = S−1
√
Q. Thus there is u ∈ S
such that uabc ∈ Q. By assumptions we have that uab /∈ Q and uac /∈ √
Q.
Since Q is a uniformly 2-absorbing primary ideal of R, then (bc)N ∈ Q.
Hence ( b
t
c
k
)N ∈ S−1Q. Consequently, S−1Q is a uniformly 2-absorbing
primary ideal of S−1R and 2- ord(S−1Q) 6 N = 2- ord(Q).
(2) Set N := 2- ord(S−1Q). Let a, b, c ∈ R such that abc ∈ Q, ab /∈
Q and ac /∈ √
Q. Then abc
1 = a
1
b
1
c
1 ∈ S−1Q, ab
1 = a
1
b
1 /∈ S−1Q and
ac
1 = a
1
c
1 /∈
√
S−1Q = S−1
√
Q, because S∩ZQ(R) = ∅ and S∩Z√
Q(R) =
∅. Since S−1Q is a uniformly 2-absorbing primary ideal of S−1R, then
( b1
c
1)
N = (bc)N
1 ∈ S−1Q. Then there exists u ∈ S such that u(bc)N ∈ Q.
Hence (bc)N ∈ Q because S ∩ ZQ(R) = ∅. Thus Q is a uniformly 2-
absorbing primary ideal of R and 2- ord(Q) 6 N = 2- ord(S−1Q).
Proposition 12. Let Q be a 2-absorbing primary ideal of a ring R and
P =
√
Q be a finitely generated ideal of R. Then Q is a Noether strongly
2-absorbing primary ideal of R. Thus Q is a uniformly 2-absorbing primary
ideal of R.
Proof. It is clear from [14, Lemma 8.21] and Proposition 5.
Corollary 7. Let R be a Noetherian ring and Q a proper ideal of R. Then
the following conditions are equivalent:
1) Q is a uniformly 2-absorbing primary ideal of R;
2) Q is a Noether strongly 2-absorbing primary ideal of R;
3) Q is a 2-absorbing primary ideal of R.
Proof. Apply Proposition 5 and Proposition 12.
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232 Uniformly 2-absorbing Primary Ideals
We recall from [8] the construction of idealization of a module. Let R
be a ring and M be an R-module. Then R(+)M = R×M is a ring with
identity (1, 0) under addition defined by (r,m) + (s, n) = (r + s,m+ n)
and multiplication defined by (r,m)(s, n) = (rs, rn + sm). Note that√
I(+)M =
√
I(+)M .
Proposition 13. Let R be a ring, Q be a proper ideal of R and M be an
R-module. The following conditions are equivalent:
1) Q(+)M is a uniformly 2-absorbing primary ideal of R(+)M ;
2) Q is a uniformly 2-absorbing primary ideal of R.
Proof. The proof is routine.
Theorem 5. Let R = R1 × R2, where R1 and R2 are rings with 1 6= 0.
Let Q be a proper ideal of R. Then the following conditions are equivalent:
1) Q is a uniformly 2-absorbing primary ideal of R;
2) Either Q = Q1 ×R2 for some uniformly 2-absorbing primary ideal
Q1 of R1 or Q = R1 ×Q2 for some uniformly 2-absorbing primary
ideal Q2 of R2 or Q = Q1 ×Q2 for some uniformly primary ideal
Q1 of R1 and some uniformly primary ideal Q2 of R2.
Proof. (1)⇒(2) Assume that Q is a uniformly 2-absorbing primary ideal of
R with 2-ordR(Q) = n. We know that Q is in the form of Q1×Q2 for some
ideal Q1 of R1 and some ideal Q2 of R2. Suppose that Q2 = R2. Since Q
is a proper ideal of R, Q1 6= R1. Let R′ = R
{0}×R2
. Then Q′ = Q
{0}×R2
is a
uniformly 2-absorbing primary ideal of R′ by Corollary 4(2). Since R′ is
ring-isomorphic to R1 and Q1 ≃ Q′, Q1 is a uniformly 2-absorbing primary
ideal of R1. Suppose that Q1 = R1. Since Q is a proper ideal of R,Q2 6= R2.
By a similar argument as in the previous case,Q2 is a uniformly 2-absorbing
primary ideal of R2. Hence assume that Q1 6= R1 and Q2 6= R2. We claim
that Q1 is a uniformly primary ideal of R1. Assume that x, y ∈ R1 such
that xy ∈ Q1 but x /∈ Q1. Notice that (x, 1)(1, 0)(y, 1) = (xy, 0) ∈ Q,
but neither (x, 1)(1, 0) = (x, 0) ∈ Q nor (x, 1)(y, 1) = (xy, 1) ∈ √
Q. So
[(1, 0)(y, 1)]n = (yn, 0) ∈ Q. Therefore yn ∈ Q1. Thus Q1 is a uniformly
primary ideal of R1 with ordR1(Q1) 6 n. Now, we claim that Q2 is a
uniformly primary ideal of R2. Suppose that for some z, w ∈ R2, zw ∈ Q2
but z /∈ Q2. Notice that (1, z)(0, 1)(1, w) = (0, zw) ∈ Q, but neither
(1, z)(0, 1) = (0, z) ∈ Q nor (1, z)(1, w) = (1, zw) ∈ √
Q. Therefore
[(0, 1)(1, w)]n = (0, wn) ∈ Q, and so wn ∈ Q2 which shows that Q2
is a uniformly primary ideal of R2 with ordR2(Q2) 6 n. Consequently
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H. Mostafanasab, Ü. Tekir, G. Ulucak 233
when Q1 6= R1 and Q2 6= R2 we have that max{ordR1(Q1), ordR2(Q2)} 6
2- ordR(Q).
(2)⇒(1) If Q = Q1 × R2 for some uniformly 2-absorbing primary
ideal Q1 of R1, or Q = R1 ×Q2 for some uniformly 2-absorbing primary
ideal Q2 of R2, then it is clear that Q is a uniformly 2-absorbing primary
ideal of R. Hence assume that Q = Q1 ×Q2 for some uniformly primary
ideal Q1 of R1 and some uniformly primary ideal Q2 of R2. Then Q′
1 =
Q1 × R2 and Q′
2 = R1 × Q2 are uniformly primary ideals of R with
ordR(Q
′
1) 6 ordR1(Q1) and ordR(Q
′
2) 6 ordR2(Q2). Hence Q′
1 ∩ Q′
2 =
Q1 × Q2 = Q is a uniformly 2-absorbing primary ideal of R with 2-
ordR(Q) 6 max{ordR1(Q1), ordR2(Q2)} by Theorem 4.
Lemma 2. Let R = R1 ×R2 × · · · ×Rn, where R1, R2, . . . , Rn are rings
with 1 6= 0. A proper ideal Q of R is a uniformly primary ideal of R if and
only if Q = ×n
i=1Qi such that for some k ∈ {1, 2, . . . , n}, Qk is a uniformly
primary ideal of Rk, and Qi = Ri for every i ∈ {1, 2, . . . , n}\{k}.
Proof. (⇒) Let Q be a uniformly primary ideal of R with ordR(Q) = m.
We know Q = ×n
i=1Qi where for every 1 6 i 6 n, Qi is an ideal of Ri,
respectively. Assume that Qr is a proper ideal of Rr and Qs is a proper
ideal of Rs for some 1 6 r < s 6 n. Since
(0, . . . , 0,
r-th
︷︸︸︷
1Rr
, 0, . . . , 0)(0, . . . , 0,
s-th
︷︸︸︷
1Rs
, 0, . . . , 0) = (0, . . . , 0) ∈ Q,
then either (0, . . . , 0,
r-th
︷︸︸︷
1Rr
, 0, . . . , 0) ∈ Q or (0, . . . , 0,
s-th
︷︸︸︷
1Rs
, 0, . . . , 0)m ∈ Q,
which is a contradiction. Hence exactly one of the Qi’s is proper, say Qk.
Now, we show that Qk is a uniformly primary ideal of Rk. Let ab ∈ Qk
for some a, b ∈ Rk such that a /∈ Qk. Therefore
(0, . . . , 0,
k-th
︷︸︸︷
a , 0, . . . , 0)(0, . . . , 0,
k-th
︷︸︸︷
b , 0, . . . , 0)
= (0, . . . , 0,
k-th
︷︸︸︷
ab , 0, . . . , 0) ∈ Q,
but (0, . . . , 0,
k-th
︷︸︸︷
a , 0, . . . , 0) /∈ Q, and so (0, . . . , 0,
k-th
︷︸︸︷
b , 0, . . . , 0)m ∈ Q.
Thus bm ∈ Qk which implies that Qk is a uniformly primary ideals of Rk
with ordRk
(Qk) 6 m.
(⇐) Is easy.
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234 Uniformly 2-absorbing Primary Ideals
Theorem 6. Let R = R1 × R2 × · · · × Rn, where 2 6 n < ∞, and
R1, R2, . . . , Rn are rings with 1 6= 0. For a proper ideal Q of R the following
conditions are equivalent:
1) Q is a uniformly 2-absorbing primary ideal of R.
2) Either Q = ×n
t=1Qt such that for some k ∈ {1, 2, . . . , n}, Qk is
a uniformly 2-absorbing primary ideal of Rk, and Qt = Rt for
every t ∈ {1, 2, . . . , n}\{k} or Q = ×n
t=1Qt such that for some
k,m ∈ {1, 2, . . . , n}, Qk is a uniformly primary ideal of Rk, Qm
is a uniformly primary ideal of Rm, and Qt = Rt for every t ∈
{1, 2, . . . , n}\{k,m}.
Proof. We use induction on n. For n = 2 the result holds by Theorem
5 . Then let 3 6 n < ∞ and suppose that the result is valid when
K = R1 × · · · ×Rn−1. We show that the result holds when R = K ×Rn.
By Theorem 5, Q is a uniformly 2-absorbing primary ideal of R if and
only if either Q = L×Rn for some uniformly 2-absorbing primary ideal
L of K or Q = K × Ln for some uniformly 2-absorbing primary ideal Ln
of Rn or Q = L× Ln for some uniformly primary ideal L of K and some
uniformly primary ideal Ln of Rn. Notice that by Lemma 2, a proper ideal
L of K is a uniformly primary ideal of K if and only if L = ×n−1
t=1 Qt such
that for some k ∈ {1, 2, . . . , n− 1}, Qk is a uniformly primary ideal of Rk,
and Qt = Rt for every t ∈ {1, 2, . . . , n− 1}\{k}. Consequently we reach
the claim.
2. Special 2-absorbing primary ideals
Definition 2. We say that a proper ideal Q of a ring R is special 2-
absorbing primary if it is uniformly 2-absorbing primary with 2-ord(Q) = 1.
Remark 3. By Proposition 4(2), every primary ideal is a special 2-
absorbing primary ideal. But the converse is not true in general. For exam-
ple, let p, q be two distinct prime numbers. Then pqZ is a 2-absorbing ideal
of Z and so it is a special 2-absorbing primary ideal of Z, by Proposition
4(1). Clearly pqZ is not primary.
Recall that a prime ideal p of R is called divided prime if p ⊂ xR for
every x ∈ R\p.
Proposition 14. Let Q be a special 2-absorbing primary ideal of R such
that
√
Q = p is a divided prime ideal of R. Then Q is a p-primary ideal
of R.
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H. Mostafanasab, Ü. Tekir, G. Ulucak 235
Proof. Let xy ∈ Q for some x, y ∈ R such that y /∈ p. Then x ∈ p. Since
p is a divided prime ideal, p ⊂ yR and so there exists r ∈ R such that
x = ry. Hence xy = ry2 ∈ Q. Since Q is special 2-absorbing primary and
y /∈ p, then x = ry ∈ Q. Consequently Q is a p-primary ideal of R.
Remark 4. Let p, q be distinct prime numbers. Then by [4, Theorem
2.4] we can deduce that pZ∩q2Z is a 2-absorbing primary ideal of Z. Since
pq2 ∈ pZ ∩ q2Z, pq /∈ pZ ∩ q2Z and q2 /∈ pZ ∩ qZ, then pZ ∩ q2Z is not a
special 2-absorbing primary ideal of Z.
Notice that for n = 1 we have that I [n] = I.
Theorem 7. Let Q be a proper ideal of R. Then the following conditions
are equivalent:
1) Q is special 2-absorbing primary;
2) For every a, b ∈ R either ab ∈ Q or (Q :R ab) = (Q :R a) or
(Q :R ab) ⊆ (
√
Q :R b);
3) For every a, b ∈ R and every ideal I of R, abI ⊆ Q implies that
either ab ∈ Q or aI ⊆ Q or bI ⊆ √
Q;
4) For every a ∈ R and every ideal I of R either aI ⊆ Q or (Q :R
aI) ⊆ (Q :R a) ∪ (
√
Q :R I);
5) For every a ∈ R and every ideal I of R either aI ⊆ Q or (Q :R
aI) = (Q :R a) or (Q :R aI) ⊆ (
√
Q :R I);
6) For every a ∈ R and every ideals I, J of R, aIJ ⊆ Q implies that
either aI ⊆ Q or IJ ⊆ √
Q or aJ ⊆ Q;
7) For every ideals I, J of R either IJ ⊆ √
Q or (Q :R IJ) ⊆ (Q :R
I) ∪ (Q :R J);
8) For every ideals I, J of R either IJ ⊆ √
Q or (Q :R IJ) = (Q :R I)
or (Q :R IJ) = (Q :R J);
9) For every ideals I, J,K of R, IJK ⊆ Q implies that either IJ ⊆ √
Q
or IK ⊆ Q or JK ⊆ Q.
Proof. (1)⇔(2)⇔(3) By Theorem 2.
(3)⇒(4) Let a ∈ R and I be an ideal of R such that aI * Q. Suppose
that x ∈ (Q :R aI). Then axI ⊆ Q, and so by part (3) we have that x ∈
(Q :R a) or x ∈ (
√
Q :R I). Therefore (Q :R aI) ⊆ (Q :R a) ∪ (
√
Q :R I).
(4)⇒(5)⇒(6)⇒(7)⇒(8)⇒(9)⇒(1) Have straightforward proofs.
Theorem 8. Let Q be a special 2-absorbing primary ideal of R and
x ∈ R\√Q. The following conditions hold:
1) (Q :R x) = (Q :R xn) for every n > 2.
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236 Uniformly 2-absorbing Primary Ideals
2) (
√
Q :R x) =
√
(Q :R x).
3) (Q :R x) is a special 2-absorbing primary ideal of R.
Proof. (1) Clearly (Q :R x) ⊆ (Q :R xn) for every n > 2. For the converse
inclusion we use induction on n. First we get n = 2. Let r ∈ (Q :R x2).
Then rx2 ∈ Q, and so either rx ∈ Q or x2 ∈ √
Q. Notice that x2 ∈ √
Q
implies that x ∈ √
Q which is a contradiction. Therefore rx ∈ Q and so
r ∈ (Q :R x). Therefore (Q :R x) = (Q :R x2). Now, assume n > 2 and
suppose that the claim holds for n− 1, i.e. (Q :R x) = (Q :R xn−1). Let
r ∈ (Q :R xn). Then rxn ∈ Q. Since x /∈ √
Q, then we have either rxn−1 ∈
Q or rx ∈ Q. Both two cases implies that r ∈ (Q :R x). Consequently
(Q :R x) = (Q :R xn).
(2) It is easy to investigate that
√
(Q :R x) ⊆ (
√
Q :R x). Let r ∈
(
√
Q :R x). Then there exists a positive integer m such that (rx)m ∈ Q.
So, by part (1) we have that rm ∈ (Q :R x). Hence r ∈
√
(Q :R x). Thus
(
√
Q :R x) =
√
(Q :R x).
(3) Let abc ∈ (Q :R x) for some a, b, c ∈ R. Then ax(bc) ∈ Q and so
ax ∈ Q or abc ∈ Q or bcx ∈ √
Q. In the first case, we have ab ∈ (Q :R x).
If abc ∈ Q, then either ab ∈ Q ⊆ (Q :R x) or ac ∈ Q ⊆ (Q :R x) or bc ∈√
Q ⊆
√
(Q :R x). In the third case we have bc ∈ (
√
Q :R x) =
√
(Q :R x)
by part (2). Therefore (Q :R x) is a special 2-absorbing primary ideal
of R.
Theorem 9. Let Q be an irreducible ideal of R. Then Q is special 2-
absorbing primary if and only if (Q :R x) = (Q :R x2) for every x ∈ R\√Q.
Proof. (⇒) By Theorem 8.
(⇐) Let abc ∈ Q for some a, b, c ∈ R such that neither ab ∈ Q nor
ac ∈ Q nor bc ∈ √
Q. We search for a contradiction. Since bc /∈ √
Q,
then b /∈ √
Q. So, by our hypothesis we have (Q :R b) = (Q :R b2). Let
r ∈ (Q+Rab)∩ (Q+Rac). Then there are q1, q2 ∈ Q and r1, r2 ∈ R such
that r = q1+ r1ab = q2+ r2ac. Hence q1b+ r1ab
2 = q2b+ r2abc ∈ Q. Thus
r1ab
2 ∈ Q, i.e., r1a ∈ (Q :R b2) = (Q :R b). Therefore r1ab ∈ Q and so
r = q1 + r1ab ∈ Q. Then Q = (Q+Rab) ∩ (Q+Rac), which contradicts
the assumption that Q is irreducible.
A ring R is said to be a Boolean ring if x = x2 for all x ∈ R. It is
famous that every prime ideal in a Boolean ring R is maximal. Notice
that every ideal of a Boolean ring R is radical. So, every (uniformly)
2-absorbing primary ideal of R is a 2-absorbing ideal of R.
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H. Mostafanasab, Ü. Tekir, G. Ulucak 237
Corollary 8. Let R be a Boolean ring. Then every irreducible ideal of R
is a maximal ideal.
Proof. Let I be an irreducible ideal of R. Thus, Theorem 9 implies that I
is special 2-absorbing primary. Therefore by Proposition 6, either I =
√
I
is a maximal ideal or is the intersection of two distinct maximal ideals.
Since I is irreducible, then I cannot be in the second form. Hence I is a
maximal ideal.
Proposition 15. Let Q be a special 2-absorbing primary ideal of R and
p, q be distinct prime ideals of R.
1) If
√
Q = p, then {(Q :R x) | x ∈ R\p} is a totally ordered set.
2) If
√
Q = p ∩ q, then {(Q :R x) | x ∈ R\p ∪ q} is a totally ordered
set.
Proof. (1) Let x, y ∈ R\p. Then xy ∈ R\p. It is clear that (Q :R x)∪(Q :R
y) ⊆ (Q :R xy). Assume that r ∈ (Q :R xy). Therefore rxy ∈ Q, whence
rx ∈ Q or ry ∈ Q, because xy /∈ √
Q. Consequently (Q :R xy) = (Q :R
x) ∪ (Q :R y). Thus, either (Q :R xy) = (Q :R x) or (Q :R xy) = (Q :R y),
and so either (Q :R y) ⊆ (Q :R x) or (Q :R x) ⊆ (Q :R y).
(2) Is similar to the proof of (1).
Corollary 9. Let f : R −→ R′ be a homomorphism of commutative rings.
Then the following statements hold:
1) If Q′ is a special 2-absorbing primary ideal of R′, then f−1(Q′) is a
special 2-absorbing primary ideal of R.
2) If f is an epimorphism and Q is a special 2-absorbing primary ideal
of R containing ker(f), then f(Q) is a special 2-absorbing primary
ideal of R′.
Proof. By Proposition 10.
Let R be a ring with identity. We recall that if f = a0+a1X+· · ·+atX
t
is a polynomial on the ring R, then content of f is defined as the ideal
of R, generated by the coefficients of f , i.e. c(f) = (a0, a1, . . . , at). Let T
be an R-algebra and c the function from T to the ideals of R defined by
c(f) = ∩{I | I is an ideal of R and f ∈ IT} known as the content of f .
Note that the content function c is nothing but the generalization of the
content of a polynomial f ∈ R[X]. The R-algebra T is called a content
R-algebra if the following conditions hold:
1) For all f ∈ T , f ∈ c(f)T .
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238 Uniformly 2-absorbing Primary Ideals
2) (Faithful flatness) For any r ∈ R and f ∈ T , the equation c(rf) =
rc(f) holds and c(1T ) = R.
3) (Dedekind-Mertens content formula) For each f, g ∈ T , there exists
a natural number n such that c(f)nc(g) = c(f)n−1c(fg).
For more information on content algebras and their examples we refer to
[11], [12] and [13]. In [10] Nasehpour gave the definition of a Gaussian
R-algebra as follows: Let T be an R-algebra such that f ∈ c(f)T for all
f ∈ T . T is said to be a Gaussian R-algebra if c(fg) = c(f)c(g), for all
f, g ∈ T .
Example 3. ([10]) Let T be a content R-algebra such that R is a Prüfer
domain. Since every nonzero finitely generated ideal of R is a cancella-
tion ideal of R, the Dedekind-Mertens content formula causes T to be a
Gaussian R-algebra.
Theorem 10. Let R be a Prüfer domain, T a content R-algebra and Q
an ideal of R. Then Q is a special 2-absorbing primary ideal of R if and
only if QT is a special 2-absorbing primary ideal of T .
Proof. (⇒) Assume that Q is a special 2-absorbing primary ideal of R.
Let fgh ∈ QT for some f, g, h ∈ T . Then c(fgh) ⊆ Q. Since R is a Prüfer
domain and T is a content R-algebra, then T is a Gaussian R-algebra.
Therefore c(fgh) = c(f)c(g)c(h) ⊆ Q. Since Q is a special 2-absorbing
primary ideal of R, Theorem 7 implies that either c(f)c(g) = c(fg) ⊆ Q or
c(f)c(h) = c(fh) ⊆ Q or c(g)c(h) = c(gh) ⊆ √
Q. So fg ∈ c(fg)T ⊆ QT
or fh ∈ c(fh)T ⊆ QT or gh ∈ c(gh)T ⊆ √
QT ⊆ √
QT . Consequently
QT is a special 2-absorbing primary ideal of T .
(⇐) Note that since T is a content R-algebra, QT ∩R = Q for every
ideal Q of R. Now, apply Corollary 4(1).
The algebra of all polynomials over an arbitrary ring with an arbitrary
number of indeterminates is an example of content algebras.
Corollary 10. Let R be a Prüfer domain and Q be an ideal of R. Then
Q is a special 2-absorbing primary ideal of R if and only if Q[X] is a
special 2-absorbing primary ideal of R[X].
Corollary 11. Let S be a multiplicatively closed subset of R and Q be a
proper ideal of R. Then the following conditions hold:
1) If Q is a special 2-absorbing primary ideal of R such that Q∩S = ∅,
then S−1Q is a special 2-absorbing primary ideal of S−1R with
2- ord(S−1Q) 6 2- ord(Q).
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H. Mostafanasab, Ü. Tekir, G. Ulucak 239
2) If S−1Q is a special 2-absorbing primary ideal of S−1R and S ∩
ZQ(R) = ∅, then Q is a special 2-absorbing primary ideal of R with
2- ord(Q) 6 2- ord(S−1Q).
Proof. By Proposition 11.
In view of Theorem 5 and its proof, we have the following result.
Corollary 12. Let R = R1 ×R2, where R1 and R2 are rings with 1 6= 0.
Let Q be a proper ideal of R. Then the following conditions are equivalent:
1) Q is a special 2-absorbing primary ideal of R;
2) Either Q = Q1 ×R2 for some special 2-absorbing primary ideal Q1
of R1 or Q = R1 ×Q2 for some special 2-absorbing primary ideal
Q2 of R2 or Q = Q1 ×Q2 for some prime ideal Q1 of R1 and some
prime ideal Q2 of R2.
Corollary 13. Let R = R1 ×R2, where R1 and R2 are rings with 1 6= 0.
Suppose that Q1 is a proper ideal of R1 and Q2 is a proper ideal of R2.
Then Q1 ×Q2 is a special 2-absorbing primary ideal of R if and only if it
is a 2-absorbing ideal of R.
Proof. See Corollary 12 and apply [1, Theorem 4.7].
Corollary 14. Let R = R1 × R2 × · · · × Rn, where 2 6 n < ∞, and
R1, R2, . . . , Rn are rings with 1 6= 0. For a proper ideal Q of R the following
conditions are equivalent:
1) Q is a special 2-absorbing primary ideal of R.
2) Either Q = ×n
t=1Qt such that for some k ∈ {1, 2, . . . , n}, Qk is
a special 2-absorbing primary ideal of Rk, and Qt = Rt for every
t ∈ {1, 2, . . . , n}\{k} or Q = ×n
t=1Qt such that for some k,m ∈
{1, 2, . . . , n}, Qk is a prime ideal of Rk, Qm is a prime ideal of Rm,
and Qt = Rt for every t ∈ {1, 2, . . . , n}\{k,m}.
Proof. By Theorem 6.
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Contact information
Hojjat
Mostafanasab
Department of Mathematics and Applications,
University of Mohaghegh Ardabili, P.O. Box
179, Ardabil, Iran
E-Mail(s): h.mostafanasab@gmail.com
Ünsal Tekir Department of Mathematics, Faculty of Science
and Arts, Marmara University 34722, Istanbul,
Turkey
E-Mail(s): utekir@marmara.edu.tr
Gülşen Ulucak Department of Mathematics, Gebze Technical
University, P. K. 14141400 Gebze, Kocaeli,
Turkey
E-Mail(s): gulsenulucak@gtu.edu.tr
Received by the editors: 10.06.2017.
H. Mostafanasab, Ü. Tekir, G. Ulucak
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