Abstract
Let R be a commutative ring with identity. Badawi (Bull Aust Math Soc 75(3), 417–429, 2007) introduced a generalization of prime ideals called 2-absorbing ideals, and this idea is further generalized in a paper by Anderson and Badawi (Commun Algebra 39(5), 1646–1672, 2011) to a concept called n-absorbing ideals. A proper ideal I of R is said to be an n-absorbing ideal if whenever \(x_1\ldots x_{n+1}\in I\) for \(x_1,\ldots ,\) \(x_{n+1}\in R\) then there are n of the \(x_i\)’s whose product is in I. It was conjectured by Anderson and Badawi (Commun Algebra 39(5), 1646–1672, 2011) that if I is an n-absorbing ideal of R then I is strongly n-absorbing (Conjecture 1) and \(Rad(I)^n\subseteq I\) (Conjecture 2). In Cahen et al. (in: Fontana et al., Commutative rings. Integer-valued polynomials, and polynomial function, Springer, New York, 2014, Problem 30c), it was conjectured also that I[X] is an n-absorbing ideal of the polynomial ring R[X] for each n-absorbing ideal of the ring R (Conjecture 3). In this paper we give an answer to (Conjecture 2) for \(n=3\), \(n=4\) and \(n=5\) and we prove that (Conjecture 1) and (Conjecture 3) hold in various classes of rings.
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1 Introduction
We assume throughout that all rings are commutative with \(1\ne 0\). In this paper, we study Anderson–Badawi conjectures. The concept of 2-absorbing ideals was introduced and investigated in Badawi (2007). Recall that a proper ideal I of R is called a 2-absorbing ideal of R if whenever a, b, \(c\in R\) and \(abc\in I\), then \(ab\in I\) or \(ac\in I\) or \(bc\in I\). More generally, let n be a positive integer, a proper ideal I of R is said to be an n-absorbing ideal if whenever \(x_1\ldots x_{n+1}\in I\) for \(x_1,\ldots ,\) \(x_{n+1}\in R\) then there are n of the \(x_i\)’s whose product is in I. And I is said to be a strongly n-absorbing ideal if whenever \(I_1\ldots I_{n+1}\subseteq I\) for ideals \(I_1,\ldots ,\) \(I_{n+1}\) of R, then the product of some n of the \(I_i\)’s is in I. Anderson and Badawi (2011) conjectured that every n-absorbing ideal of R is strongly n-absorbing (Conjecture 1) and \(Rad(I)^n\subseteq I\), where Rad(I) denotes the radical ideal of I (Conjecture 2).
In Sect. 2, we give an answer to (Conjecture 2) in the case where \(n=3\), \(n=4\) and \(n=5\). After that, we give some equivalent characterizations of n-absorbing ideals and we prove that (Conjecture 1) is true in the class of U-rings. Recall that a commutative ring R is said to be a U-ring provided R has the property that an ideal contained in a finite union of ideals must be contained in one of those ideals.
An ideal I of a ring R is an SFT (strong finite type) ideal if there exists an ideal F of finite type with \(F\subseteq I\) and an integer n such that for any \(a\in I\), \(a^n\in F\). A ring R is an SFT-ring if every ideal of R is SFT, which is equivalent to each prime ideal of R is SFT (Arnold 1973). We prove that if every nonzero proper ideal of a ring R is a 2-absorbing ideal of R then R is an SFT ring.
Finally, we prove that if n is an integer with \(n\ge 3\), then I is an n-absorbing ideal of R if and only if I[X] (respectively I[[X]]) is an n-absorbing ideal of R[X] (Conjecture 3) (respectively R[[X]]), if the ring R is a Gaussian ring (respectively Noetherian Gaussian ring) or the ring R is a pseudo-valuation domain (PVD).
We start by recalling some background material.
An integral domain R is said to be a valuation domain if x|y (in R) or y|x (in R) for every nonzero x, \(y\in R\). An integral domain R is called a Pr\(\ddot{u}\)fer domain if \(R_P\) is a valuation domain for each prime ideal P of R.
The content of a polynomial (respectively a power series) f over a commutative ring R is the ideal C(f) of R generated by all the coefficients of f. A commutative ring R is said to be a Gaussian (respectively P-Gaussian) ring if \(C(fg)=C(f)C(g)\) for every f and g in R[X] (respectively f and g in R[[X]]).
Let R be an integral domain with quotient field K. A prime ideal P of R is called strongly prime if whenever x, \(y\in K\) and \(xy\in P\) then \(x\in P\) or \(y\in P\). A domain R is called a pseudo-valuation domain if P is a strongly prime ideal for each prime ideal P of R.
A prime ideal P of a ring R is said to be a divided prime ideal if \(P\subset xR\) for every \(x\in R{\setminus } P\); thus a divided prime ideal is comparable to every ideal of R. An integral domain R is said to be a divided domain if every prime ideal of R is a divided prime ideal.
Let R be a ring, Spec(R) denotes the set of prime ideals of R and Nil(R) denotes the ideal of nilpotent elements of R. If I is a proper ideal of R, then \(Min_R(I)\) denotes the set of prime ideals of R minimal over I.
2 On the Anderson–Badawi conjectures
Let R be a commutative ring. Anderson and Badawi (2011) conjectured that every n-absorbing ideal of R is strongly n-absorbing (Conjecture 1) and \(Rad(I)^n\subseteq I\) (Conjecture 2). As observed in Anderson and Badawi (2011), it is easy to see that Conjecture 1 implies Conjecture 2. Conjecture 1 was proved for \(n=2\), see Anderson and Badawi (2011, Theorem 2.13). It was also verified for arbitrary n when R is a Pr\(\ddot{u}\)fer domain (Anderson and Badawi 2011, Corollary 6.9). Darani (2013, Theorem 4.2) proved that Conjecture 1 is true for all commutative rings with torsion-free additive group. Donadze (2016) gives answers for the two conjectures in special cases.
Moreover, Conjecture 2 is true in the case where I is an n-absorbing ideal with exactly n minimal prime ideals \(\{ P_1,\ldots , P_n\}\). In fact, by Anderson and Badawi (2011, Theorem 2.14) we have \(P_1\ldots P_n\subseteq I\). Since \(Rad(I)={\cap }_{P_i\in Min_R(I)}P_i\subseteq P_j\) for each \(1\le j\le n\), we have \(Rad(I)^n\subseteq I\). If in addition, the \(P_i\)’s are comaximal, then \(I=P_1\cap \cdots \cap P_n\), (Anderson and Badawi 2011, Corollary 2.15) so \(I[X]=P_1[X]\cap \cdots \cap P_n[X]\) (respectively \(I[[X]]=P_1[[X]]\cap \cdots \cap P_n[[X]]\)), which implies, by Anderson and Badawi (2011, Theorem 2.1), that I[X] (respectively I[[X]]) is an n-absorbing ideal of R[X] (respectively R[[X]]).
Theorem 2.1
Let I be a 3-absorbing ideal of R. Then \(Rad(I)^3\subseteq I\).
Proof
Let x, y, \(z\in Rad(I)\). First observe that \(x^2y^2\in I\). In fact, we have \(x^3\in I\) for all \(x\in Rad(I)\), by Anderson and Badawi (2011, Theorem 2.1). Since \(x^2y^2(x+y)=xxy^2(x+y)\in I\) and I is a 3-absorbing ideal, we conclude that either \(xy^2(x+y)\in I\) or \(x^2(x+y)\in I\) or \(x^2y^2\in I\), thus \(x^2y^2\in I\). Now, we prove that \(x^2y\in I\). Since \(x^2y(x^2+y)=xxy(x^2+y)\in I\) we have that \(xy(x^2+y)\in I\) or \(x^2(x^2+y)\in I\) or \(x^2y\in I\). So \(x^2y\in I\) or \(xy^2\in I\). If \(xy^2\in I\), since \(x^2y(x+y)=xxy(x+y)\in I\), we conclude that \(x^2y\in I\). Finally, since \(xyz(x+y+z)\in I\) we have \(xyz\in I\). \(\square \)
Theorem 2.2
Let I be a 4-absorbing ideal of R. Then \(Rad(I)^4\subseteq I\).
Proof
By Anderson and Badawi (2011, Theorem 2.1), \(x^4\in I\) for each \(x\in Rad(I)\). Now following these steps we get the result:
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Let \(x_1\), \(x_2\in Rad(I)\) then \(x_1^3x_2^3\in I\). In fact, we have \(x_1^3(x_1+x_2)x_2^3\in I\) and I is a 4-absorbing ideal.
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Let \(x_1\), \(x_2\in Rad(I)\) then \(x_1^3x_2^2\in I\). In fact, by the last step, as \(x_1^3x_2^3\in I\), then either \(x_1^3x_2^2\in I\) or \(x_1^2x_2^3\). If \(x_1^2x_2^3\in I\) and since \(x_1^3x_2^2(x_1+x_2)\in I\), we have the result.
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Let \(x_1\), \(x_2\in Rad(I)\) then \(x_1^2x_2^2\in I\) and \(x_1^3x_2\in I\). In fact, we have \(x_1^3x_2^2\in I\) (and \(x_1^2x_2^3\in I\)), then either \(x_1^2x_2^2\in I\) or \(x_1^3x_2\in I\). If \(x_1^2x_2^2\in I\), since \(x_1^3x_2(x_1+x_2)\in I\), we conclude that \(x_1^3x_2\in I\). If \(x_1^3x_2\in I\), since \(x_1^2(x_1+x_2)x_2^2\in I\), we conclude that \(x_1^2(x_1+x_2)x_2\in I\) or \(x_1^2x_2^2\in I\) or \(x_1(x_1+x_2)x_2^2\in I\). In the first and second cases, we get \(x_1^2x_2^2\in I\). In the last case, since \(x_1^2x_2^3\in I\) we have the result.
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Let \(x_1\), \(x_2\), \(x_3\in Rad(I)\) then \(x_1^2x_2^2x_3^2\in I\). In fact, it suffices to remark that \(x_1^2x_2^2x_3^2(x_1+x_2+x_3)\in I\).
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Let \(x_1\), \(x_2\), \(x_3\in Rad(I)\) then \(x_1^2x_2x_3\in I\), since \(x_1^2x_2x_3(x_2+x_3)\in I\).
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Let \(x_1\), \(x_2\), \(x_3\), \(x_4\in Rad(I)\) then \(x_1x_2x_3x_4\in I\). In fact, we have \(x_1x_2x_3x_4(x_1+x_2+x_3+x_4)\in I\) and since I is a 4-absorbing ideal, the result is clear.
\(\square \)
Theorem 2.3
Let I be a 5-absorbing ideal of R. Then \(Rad(I)^5\subseteq I\).
Proof
By Anderson and Badawi (2011, Theorem 2.1), \(x^5\in I\) for each \(x\in Rad(I)\).
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Let \(x_1\), \(x_2\in Rad(I)\) then \(x_1^4x_2^4\in I\), since \(x_1^4(x_1+x_2)x_2^4\in I\).
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Let \(x_1\), \(x_2\in Rad(I)\) then \(x_1^4x_2^3\in I\). In fact, we have \(x_1^4xx_2^4\in I\). Hence, either \(x_1^4x_2\in I\) or \(x_1^4x_2^3\in I\) or \(x_1^3x_2^4\in I\). If \(x_1^3x_2^4\in I\), we have either \(x_1^3x_2^3\in I\) or \(x_1^2x_2^4\in I\). Suppose that \(x_1^2x_2^4\in I\), then either \(x_1x_2^4\in I\) or \(x_1x_2^3\in I\). If \(x_1x_2^4\in I\) and since \(x_1^4x_2^3(x_1+x_2)\in I\), then we get the result.
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Let \(x_1\), \(x_2\in Rad(I)\) then \(x_1^3x_2^3\in I\) and \(x_1^4x_2^2\in I\). In fact, since \(x_1^4x_2^3\in I\) and I is a 5-absorbing ideal we have either \(x_1^4x_2^2\in I\) or \(x_1^3x_2^3\in I\). Suppose that \(x_1^4x_2^2\in I\), since \(x_1^3(x_1+x_2)x_2^3\in I\) and \(x_1^3x_2^4\in I\), we prove that \(x_1^3x_2^3\in I\). Suppose that \(x_1^3x_2^3\in I\) and since \(x_1^4x_2^2(x_1+x_2)\in I\), we conclude that \(x_1^4x_2^2\in I\).
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Let \(x_1\), \(x_2\in Rad(I)\) then \(x_1^3x_2^2\in I\) and \(x_1^4x_2\in I\). In fact, we have \(x_1^4x_2^2\in I\) so either \(x_1^3x_2^2\in I\) or \(x_1^4x_2\in I\). If \(x_1^4x_2\in I\) we prove that \(x_1^3x_2^2\in I\) since \(x_1^3x_2^3\in I\) and \(x_1^3(x_1+x_2)x_2^2\in I\). If \(x_1^3x_2^2\in I\), we prove that \(x_1^4x_2\in I\) since \(x_1^4(x_1+x_2)x_2\in I\).
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Let \(x_1\), \(x_2\), \(x_3\in Rad(I)\) then \((x_1x_2x_3)^2\in I\). It suffices to remark that \(x_1^2x_2^2x_3^2(x_1+x_2+x_3)\in I\).
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Let \(x_1\), \(x_2\), \(x_3\in Rad(I)\) then \(x_1^3x_2x_3\in I\). In fact, it is clear since \(x_1^3x_2x_3(x_2+x_3)\in I\) and \(x_1^3x_2x_3(x_1+x_2+x_3)\in I\).
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Let \(x_1\), \(x_2\), \(x_3\in Rad(I)\) then \(x_1^2x_2^2x_3\in I\) since \(x_1^2x_2^2x_3(x_1+x_2+x_3)\in I\), then either \(x_1^2x_2^2x_3+x_1x_2^2x_3^2\in I\) (1’) or \(x_1^2x_2^2x_3+x_1^2x_2x_3^2\in I\) (2’) or \(x_1^2x_2^2x_3\in I\). If (1’) is true, since \(x_1^2x_2^2x_3^2\in I\) then either \(x_1x_2^2x_3^2\in I\) or \(x_1^2x_2x_3^2\in I\) or \(x_1^2x_2^2x_3\in I\). If \(x_1x_2^2x_3^2\in I\), we get the result. If \(x_1^2x_2x_3^2\in I\), since \(x_1x_2^2x_3^2(x_1+x_2+x_3^2)\in I\), we conclude.
If (2’) is true, since \(x_1^2x_2^2x_3^2\in I\) then either \(x_1x_2^2x_3^2\in I\) or \(x_1^2x_2x_3^2\in I\) or \(x_1^2x_2^2x_3\in I\). If \(x_1^2x_2x_3^2\in I\), we get the result. If \(x_1x_2^2x_3^2\in I\), since \(x_1^2x_2x_3^2(x_1+x_2+x_3^2)\in I\), we conclude.
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Let \(x_1\), \(x_2\), \(x_3\),\(x_4\in Rad(I)\) then \(x_1^2x_2x_3x_4\in I\). It is clear since \(x_1^2x_2x_3x_4(x_1+x_2+x_3+x_4)\in I\) and \(x_1^2x_2x_3x_4(x_2+x_3+x_4)\in I\).
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Let \(x_1\), \(x_2\), \(x_3\),\(x_4\in Rad(I)\) then \(x_1x_2x_3x_4\in I\). In fact, remark that \(x_1x_2x_3x_4(x_1+x_2+x_3+x_4)\in I\).
\(\square \)
Notation (Anderson and Badawi 2011) If I is an n-absorbing ideal of R for some positive integer n, then define \(\omega _R(I)=min\{n|\; I\; is\; an\; n-absorbing\; ideal \;of\; R\}\).
Applying Anderson and Badawi (2011, Theorem 6.3), we obtain the following result:
Corollary 2.1
Let P be a prime ideal of a ring R and \(n\in \{3, 4, 5\}\).
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(1)
If \(P^n\) is a P-primary ideal of R and \(P^n\subset P^{n-1}\), then \(\omega _R(P^n)=n\).
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(2)
If P is a maximal ideal of R and \(P^n\subset P^{n-1}\), then \(\omega _R(P^n)=n\).
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(3)
Let I be a P-primary ideal of a ring R. If \(P^n\subseteq I\) and \(P^{n-1}\not \subset I\), then \(\omega _R(I)=n\).
Remark that in the case where \(n\ge 6\), we can prove the following results:
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(1)
Let \(x_1\), \(x_2\in Rad(I)\) then \(x_1x_2^{n-1}\in I\). In fact, since \( x_1(x_1^{n-1}+x_2)x_2^{n-1}\in I\), we conclude that either \(x_1x_2^{n-1}\in I\) or \(x_1^{n-1}x_2^{n-1}\in I\).
Now, for each \(1\le k\le n-1\), we suppose that \(x_1^{n-k}x_2^{n-1}\in I\) and we prove that \(x_1^{n-k-1}x_2^{n-1}\in I\).
Since \(x_1(x_1^{n-k-1}+x_2)x_2^{n-1}\in I\), we conclude that either \(x_1^{n-k-1}x_2^{n-1}\in I\) or \(x_1x_2^{n-1}\in I\) or \(x_1^{n-k}x_2^{n-2}+x_1x_2^{n-1}\in I\). As \(x_1^{n-k}x_2^{n-1}\in I\), then either \(x_1^{n-k-1}x_2^{n-1}\in I\) or \(x_1^{n-k}x_2^{n-2}\in I\). So the result is clear.
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(2)
Let \(x_1\), \(x_2\in Rad(I)\) then \(x_1^{n-2}x_2^{n-2}\in I\). In fact, it is clear since \(x_1^{n-2}(x_1+x_2)x_2^{n-2}\).
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(3)
Let \(x_1\), \(x_2\in Rad(I)\) then \(x_1^{n-2}x_2^{n-3}\in I\). In fact, it is clear since \(x_1^{n-2}(x_1+x_2)x_2^{n-3}\in I\) and \(x_1^{n-2}x_2^{n-2}\in I\).
In the next step, we prove that Conjecture 1 holds for U-rings.
Definition 2.1
Let R be a commutative ring, I, J two ideals of R and \(a\in R\). We define:
-
(1)
\((I:J)=\{x\in R\; |\; xJ\subseteq I\}\).
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(2)
\((I:a)=\{x\in R\; | \;ax\in I\}\).
Notation Let R be a commutative ring, \(n\in \mathbb {N}^*\), \(x_1,\ldots ,\) \(x_n\in R\) and \(I_1\),\(\ldots ,I_n\) be n ideals of R. For \(i\in \{1,\ldots ,n\}\), we denote by:
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\(\hat{x_i}\) the product \(x_1\ldots x_{i-1}x_{i+1}\ldots x_n\).
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\(\hat{I_i}\) the product \(I_1\ldots I_{i-1}I_{i+1}\ldots I_n\).
Proposition 2.1
Let I be a proper ideal of a commutative ring R and \(n\in \mathbb {N}^*\). The following conditions are equivalent:
-
(1)
I is an n-absorbing ideal of R.
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(2)
For every elements \(x_1,\ldots ,\) \(x_n\in R\) with \(x_1\ldots x_n\not \in I\), \((I:x_1\ldots x_n)\subseteq \cup _{1\le i\le n}(I:\hat{x_i})\)
Proof
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“\(1)\Rightarrow 2)\)” Let \(a\in (I:x_1\ldots x_n)\) then \(ax_1\ldots x_n\in I\). Since I is an n-absorbing ideal and \(x_1\ldots x_n\not \in I\), we conclude that \(a\hat{x_i}\in I\) for some i with \(1\le i\le n\). Thus \(a\in \cup _{1\le i \le n}(I:\hat{x_i})\).
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\(``2)\Rightarrow 1)\)” Let \(x_1,\ldots ,\) \(x_{n+1}\in R\) such that \(x_1\ldots x_{n+1}\in I\), then \(x_1\in (I:x_2\ldots x_{n+1})\). If \(x_2\ldots x_{n+1}\in I\) then we are done. Hence we may assume that \(x_2\ldots x_{n+1}\not \in I\) and so by (1), \((I:x_2\ldots x_{n+1})\subseteq \cup _{2\le i\le n+1}(I:\hat{x_i})\). So \(x_1\in (I:\hat{x_i})\) for some i with \(2\le i\le n+1\). \(\square \)
Definition 2.2
(Quartararo and Butts 1975) A commutative ring R is said to be a U-ring provided R has the property that an ideal contained in a finite union of ideals must be contained in one of those ideals.
Example 2.1
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(1)
Every Pr\(\ddot{u}\)fer domain is a U-ring (Quartararo and Butts 1975, Corollary 1.6).
-
(2)
Let D be an integral domain with quotiont field K. If D is a U-ring and \(D\subseteq R\subseteq K\), then R is a U-domain. If D / P is finite for all maximal ideals P of D, then D is a U-domain if and only if D is a Pr\(\ddot{u}\)fer domain (Quartararo and Butts 1975).
Recall that a proper ideal I of a ring R is a strongly n-absorbing ideal if whenever \(I_1\ldots I_{n+1}\subseteq I\) for ideals \(I_1,\ldots ,\) \(I_{n+1}\) of R, then the product of some n of the \(I_i\)’s is contained in I.
Theorem 2.4
Let R be a U-ring and \(n\ge 3\). The following conditions are equivalent:
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(1)
I is a strongly n-absorbing ideal.
-
(2)
I is an n-absorbing ideal.
-
(3)
For every \(x_1\), \(x_2,\ldots ,\) \(x_n\in R\) such that \(x_1\ldots x_n\not \in I\), \((I:x_1\ldots x_n)=(I:\hat{x_i})\) for some \(1\le i\le n\).
-
(4)
For every t ideals \(I_1,\ldots ,\) \(I_t\), \(1\le t\le n-1\), and for every elements \(x_1,\ldots ,\) \(x_{n-t}\) such that \(x_1\ldots x_{n-t}I_1\ldots I_t\not \subseteq I\), \((I:x_1\ldots x_{n-t}I_1\ldots I_t)=(I:\hat{x_i}I_1\ldots I_t)\) for some \(1\le i\le n-t\) or \((I:x_1\ldots x_{n-t}I_1\ldots I_t)=(I:x_1\ldots x_{n-t}\hat{I_j})\) for some \(1\le j\le t\).
-
(5)
For every ideals \(I_1,\ldots \) \(I_n\) of R with \(I_1\ldots I_n\not \subseteq I\), \((I:I_1\ldots I_n)=(I:\hat{I_i})\), for some \(1\le i\le n\).
Proof
-
\(1)\Rightarrow 2)\) It is clear.
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\(2)\Rightarrow 3)\) This follows from the last proposition, since R is a U-ring.
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\(3)\Rightarrow 4)\) We prove the result by induction on \(t\in \{1,\ldots ,n-1\}\). For \(t=1\) consider \(x_1,\ldots ,\) \(x_{n-1}\in R\) and an ideal \(I_1\) of R such that \(x_1\ldots x_{n-1}I_1\not \subseteq I\).
Let \(a\in (I:x_1\ldots x_{n-1}I_1)\). Then \(I_1\subseteq (I:ax_1\ldots x_{n-1})\). If \(ax_1\ldots x_{n-1}\in I\), then \(a\in (I:x_1\ldots x_{n-1})\). If \(ax_1\ldots x_{n-1}\not \in I\), then by 3), either \((I:ax_1\ldots x_{n-1})=(I:x_1\ldots x_{n-1})\) or \((I:ax_1\ldots x_{n-1})=(I:a\hat{x_i})\) for some \(1\le i\le n-1\). Since \(I_1\not \subset (I:x_1\ldots x_{n-1})\), we conclude that \(I_1\subseteq (I:a\hat{x_i})\) for some \(1\le i\le n-1\), and thus \(a\in (I:\hat{x_i}I_1)\). Hence \((I:x_1\ldots x_{n-1}I_1)\subseteq (I:x_1\ldots x_{n-1})\cup \cup _{1\le i\le n-1}(I:\hat{x_i}I_1)\). Since R is a U-ring, then either \((I:x_1\ldots x_{n-1}I_1)\subseteq (I:x_1\ldots x_{n-1})\) or \((I:x_1\ldots x_{n-1}I_1)\subseteq (I:\hat{x_i}I_1)\). The other inclusions are evident. Now, suppose that \(t> 1\) and assume that the claim holds for \(t-1\). Let \(x_1,\ldots ,\) \(x_{n-t}\) be elements of R and let \(I_1,\ldots ,\) \(I_t\) be ideals of R such that \(x_1\ldots x_{n-t}I_1\ldots I_t\not \subseteq I\).
Consider an element \(a\in (I:x_1\ldots x_{n-t}I_1\ldots I_t)\). Thus \(I_t\subseteq (I:ax_1\ldots x_{n-t}I_1\ldots I_{t-1})\). If \(ax_1\ldots x_{n-t}I_1\ldots I_{t-1}\subseteq I\), then \(a\in (I:x_1\ldots x_{n-t}I_1\ldots I_{t-1})\). If \(ax_1\ldots x_{n-t}I_1\ldots I_{t-1}\not \subseteq I\), then by the induction hypothesis, either \((I:ax_1\ldots x_{n-t}I_1\ldots I_{t-1})=(I:x_1\ldots x_{n-t}I_1\ldots I_{t-1})\) or \((I:ax_1\ldots x_{n-t}I_1\ldots I_{t-1})=(I:a\hat{x_i}I_1\ldots I_{t-1})\) for some \(1\le i\le n-t\) or
\((I:ax_1\ldots x_{n-t}I_1\ldots I_{t-1})=(I:ax_1\ldots x_{n-t}I_1\ldots I_{j-1}I_{j+1}\ldots I_{t-1})\) for some \(1\le j\le t-1\). Since \(x_1\ldots x_{n-t}I_1\ldots I_t\not \subseteq I\), then the first case is removed. Consequently, either \((I:ax_1\ldots x_{n-t}I_1\ldots I_{t-1})= (I:a\hat{x_i}I_1\ldots I_{t-1})\) for some \(1\le i\le n-t\) or
\((I:ax_1\ldots x_{n-t}I_1\ldots I_{t-1})= (I:ax_1\ldots x_{n-t}I_1\ldots I_{j-1}I_{j+1}\ldots I_t)\) for some \(1\le j\le t-1\).
Hence \((I:x_1\ldots x_{n-t}I_1\ldots I_t)\subseteq {\cup }_{1\le i\le n-1}(I:\hat{x_i}I_1\ldots I_t)\cup \underset{1\le j\le t}{\cup }(I:x_1\ldots x_{n-t}\hat{I_j})\). Now, since R is a U-ring, \((I:x_1\ldots x_{n-t}I_1\ldots I_t)\) is included in \((I:\hat{x_i}I_1\ldots I_t)\) for some \(1\le i\le n-t\) or \((I:x_1\ldots x_{n-t}\hat{I_j})\) for some \(1\le j\le t\). The other inclusions are evident.
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\(4)\Rightarrow 5)\) Let \(I_1,\ldots ,\) \(I_n\) be ideals of R such that \(I_1\ldots I_n\not \subseteq I\). Suppose that \(a\in (I:I_1\ldots I_n)\). Then \(I_n\subseteq (I:aI_1\ldots I_{n-1})\). If \(aI_1\ldots I_{n-1}\subseteq I\), then \(a\in (I:I_1\ldots I_{n-1})\). If \(aI_1\ldots I_{n-1}\not \subseteq I\), then by 4), we have either \((I:aI_1\ldots I_{n-1})=(I:a\hat{I_j})\) for some \(1\le j\le n-1\) or \((I:aI_1\ldots I_{n-1})=(I:I_1\ldots I_{n-1})\).
By hypothesis, the second case does not hold. The first case implies that \(a\in (I:I_1\ldots I_{j-1}I_{j+1}\ldots I_n)\) for some \(1\le j\le n-1\). Hence \((I:I_1\ldots I_n)\subseteq (I:I_1\ldots I_{n-1})\cup \underset{1\le j\le n-1}{\cup }(I:\hat{I_j})=\underset{1\le i\le n}{\cup }(I:\hat{I_j})\). Since R is a U-ring, we conclude that \((I:I_1\ldots I_n)\subseteq (I:\hat{I_j})\) for some \(1\le j\le n\). The other inclusions are evident.
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\(5)\Rightarrow 1)\) Let \(I_1,\ldots ,\) \(I_{n+1}\) be ideals of R such that \(I_1\ldots I_{n+1}\subseteq I\). Then \(I_1\subseteq (I:I_2\ldots I_{n+1})\). If \(I_2\ldots I_{n+1}\subseteq I\), that is clear. If \(I_2\ldots I_{n+1}\not \subseteq I\), then by 5), \((I:I_2\ldots I_{n+1})=(I:I_2\ldots I_{j-1}I_{j+1}\ldots I_{n+1})\) for some \(2\le j\le n+1\). So \(I_1\hat{I_j}\subseteq I\) for some \(2\le j\le n+1\). \(\square \)
Example 2.2
Let R be a Pr\(\ddot{u}\)fer domain, I a proper ideal of R and \(n\ge 3\). Using Anderson and Badawi (2011, Theorem 5.7), we conclude that I is a strongly n-absorbing ideal of R if and only if I is a product of prime ideals of R.
Badawi (2007) proved that if I is a 2-absorbing ideal of a commutative ring R, then either \((I:x)\subseteq (I:y)\) or \((I:y)\subseteq (I:x)\) for each x, \(y\in Rad(I){\setminus } I\). It is natural to ask if this result can be generalized for each x, \(y\in R{\setminus } I\). The answer is given by the next theorem. Recall, from Badawi (2007), that if I a 2-absorbing ideal, then one of the following statements must hold:
-
(1)
\(Rad(I)=P\) is a prime ideal of R and \(P^2\subseteq I\).
-
(2)
\(Rad(I)=P_1\cap P_2\), \(P_1P_2\subseteq I\) and \(Rad(I)^2\subseteq I\) where \(P_1\), \(P_2\) are the only distinct prime ideals of R that are minimal over I.
Theorem 2.5
Let I be a 2-absorbing ideal of a commutative ring R.
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(1)
If \(Rad(I)=P\) is a prime ideal of R, then either \((I:x)\subseteq (I:y)\) or \((I:y)\subseteq (I:x)\), for every x, \(y\in R{\setminus } I\).
-
(2)
If \(Rad(I)=P_1\cap P_2\), where \(P_1\), \(P_2\) are the only distinct prime ideals of R that are minimal over I and \(I\ne Rad(I)\), then either \((I:x)\subseteq (I:y)\) or \((I:y)\subseteq (I:x)\) for every x, \(y\in R{\setminus } I\) except if \(x\in P_1{\setminus } P_2\) and \(y\in P_2{\setminus } P_1\), in which case \((I:x)=P_2\) and \((I:y)=P_1\).
Proof
-
(1)
Let I be a 2-absorbing ideal of R such that \(Rad(I)=P\) is a prime ideal of R. First, remark that:
-
(a)
For each \(x\in R{\setminus } P\), \((I:x)\subseteq P\). In fact, let \(y\in R\) such that \(yx\in I\). Since P is a prime ideal and \(x\not \in P\) we conclude that \(y\in P\).
-
(b)
Let x, \(y\in R{\setminus } P\) then (I : x) and (I : y) are linearly ordered. Otherwise, let \(z_1\in (I:x){\setminus } (I:y)\) and \(z_2\in (I:y){\setminus } (I:x)\). Then \(x(z_1+z_2)y\in I\). Since I is a 2-absorbing ideal, we have \(x(z_1+z_2)\in I\) or \((z_1+z_2)y\in I\) or \(xy\in I\) which is impossible.
Now, let x, \(y\in R{\setminus } I\).
-
(a)
-
(2)
Let I be a 2-absorbing ideal such that \(Rad(I)=P_1\cap P_2\) and \(x\in R{\setminus } Rad(I)\). Then \((I:x)\subseteq P_1\cup P_2\). In fact, let \(z\in (I:x)\), so \(zx\in I\subseteq P_1\cap P_2\). Since \(x\not \in Rad(I)\), we have \(x\not \in P_1\) or \(x\not \in P_2\). So we conclude that \(z\in P_1\) or \(z\in P_2\).
Remark that if \(x\in P_1{\setminus } P_2\), then \((I:x)=P_2\). In fact, let \(z\in (I:x)\) then \(xz\in I\subseteq P_1\cap P_2\subseteq P_2\). As \(x\not \in P_2\) then \(z\in P_2\). So \((I:x)\subseteq P_2\). Conversely, let \(z\in P_2\) then \(xz\in P_1P_2\subseteq I\). So \(z\in (I:x)\).
Similarly, if \(x\in P_2{\setminus } P_1\) then \((I:x)=P_1\).
Now let x, \(y\in R{\setminus } I\).
If x, \(y\in Rad(I){\setminus } I\), then (I : x) and (I : y) are linearly ordered by Badawi (2007 Theorem 2.6).
If not, we have the following cases:
-
If \(x\in Rad(I){\setminus } I\) and \(y\in R{\setminus } Rad(I)\), we have \((I:y)\subseteq P_1\cup P_2\subseteq (I:x)\).
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If x, \(y\in R{\setminus } Rad(I)\):
-
if x, \(y\in P_1{\setminus } P_2\), we conclude that \((I:x)=(I:y)=P_2\).
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if x, \(y\in P_2{\setminus } P_1\), in this case we have \((I:x)=(I:y)=P_1\).
-
if x, \(y\in R{\setminus } (P_1\cup P_2)\), we assume that (I : x) and (I : y) are not linearly ordered. Then there exist \(z_1\in (I:x){\setminus } (I:y)\) and \(z_2\in (I:y){\setminus } (I:x)\). So \(x(z_1+z_2)y\in I\) and no product of two elements is in I which is a contradiction.
-
if \(x\in P_1{\setminus } P_2\) and \(y\in P_2{\setminus } P_1\), we have \((I:x)=P_2\) and \((I:y)=P_1\) and it is clear that (I : x) and (I : y) are not linearly ordered in this case.
-
-
\(\square \)
Recall that a 2-absorbing ideal is a generalization of a prime ideal and there are many characterization of a commutative ring with their set of prime ideals , so one can ask if we have a similar result for a commutative ring such that every nonzero proper ideal of R is a 2-absorbing ideal. The following proposition gives an answer.
Proposition 2.2
Let R be a commutative ring. If every nonzero proper ideal of R is a 2-absorbing ideal then R is an SFT ring.
Proof
By Badawi (2007 Theorem 3.4), R is a zero-dimensional ring and we have three cases.
Case 1: R is quasi-local with maximal ideal \(M=Nil(R)\ne \{0\}\) such that \(M^2\subseteq xR\) for each nonzero \(x\in M\). To prove that R is an SFT ring it suffices to prove that M is an SFT ideal of R. Since \(M\ne (0)\), then there is a nonzero element \(y\in M\). Thus \(F=(y)\) is a principal ideal of R such that \(x^2\in F\) for each \(x\in M\). So we conclude that M is an SFT ideal.
Case 2: R has exactly two distinct maximal ideals, say \(\{M_1, M_2\}\). So either R is isomorphic to \(D=R/M_1 \oplus R/M_2\) or \(Nil(R)^2=\{0\}\) and \(Nil(R)=\omega R\) for each nonzero \(\omega \in Nil(R)\). In the first situation, R is isomorphic to an SFT ring so R is an SFT ring. In the second situation, we have \(R\cong R/M_1^2 \oplus R/M_2\), by Badawi (2007 Lemma 3.3). The ring \(R/M_1^2\) is SFT. In fact, let J be an ideal of \(R/M_1^2\), then there exists an ideal I of R such that \(M_1^2\subseteq I\subseteq M_1\) and \(J=I/M_1^2\). It is easy to see that \(J\subseteq Nil(R/M_1^2)=M_1/M_1^2\) and for each \(\bar{x}\in J\), we have \(\bar{x}^2=\bar{0}\). Then by Hizem and Benhissi (2011, Proposition 2.1) \(R/M_1^2\) is an SFT ring.
Case 3: We suppose that R is isomorphic to \(F_1\oplus F_2\oplus F_3\), where \(F_1\), \(F_2\) and \(F_3\) are fields. It is clear in this case that R is an SFT ring. \(\square \)
Example 2.3
-
(1)
Let \(R=\mathbb {Z}+6X\mathbb {Z}[X]\) and \(P=6X\mathbb {Z}[X]\). First observe that \(P^2\) is not a 2-absorbing ideal of R. In fact, let \(f_1=6X^2\), \(f_2=2\) and \(f_3=3\) in R, then it is clear to see that \(f_1f_2f_3\in P^2\) but \(f_1f_2\not \in I\), \(f_2f_3\not \in I\) and also \(f_1f_3\not \in I\). So R is not an SFT ring.
-
(2)
Let D be a valuation domain with Krull dimension \(n\ge 1\), K the quotient field of D and X an indeterminate. Set \(R=D+XK[[X]]\), by [4, Example 3.12], R is not a 2-absorbing ring so R is not an SFT ring.
Next, we give some classes of rings in which Conjecture 3 holds. Recall that Conjecture 3 is true if \(n=2\) and we can easily prove that if I is a 2-absorbing ideal of R then I[[X]] is also a 2-absorbing ideal of the power series ring R[[X]]. In fact, we prove that either \(Rad(I[[X]])=P[[X]]\), with P a prime ideal of R or \(Rad(I[[X]])=P_1[[X]]\cap P_2[[X]]\), with \(P_1\) and \(P_2\) are two prime ideals of R. By Badawi (2007 Theorems 2.8 and 2.9), we conclude that I[[X]] is a 2-absorbing ideal since \(I[[X]]_{f}\) is a prime ideal of R[[X]] for each \(f\in Rad(I[[X]]){\setminus } I[[X]]\).
Nasehpour (2016) proves that for a Pr\(\ddot{u}\)fer domain R and \(n\ge 3\), an ideal I is n-absorbing if and only if I[X] is n-absorbing. In the following, we generalize this result in the case of a Gaussian U-ring.
Remark also that in a Pr\(\ddot{u}\)fer domain, we can prove the last result in the power series ring. In fact, let I be an n-absorbing ideal then \(I[[X]]=P_1^{n_1}[[X]]\ldots P_k^{n_k}[[X]]\), where \(P_1,\ldots ,\) \(P_k\) are the minimal prime ideals over I and \(n_1,\ldots ,\) \(n_k\) positive integer such that \(n_1+\cdots +n_k=n\). By Fields (1971, Corollary 4) and Anderson and Badawi (2011, Theorems 3.1 and 2.1) we conclude that I[[X]] is an n-absorbing ideal of R[[X]].
Recall that a commutative ring R is said to be a \(Gaussian ring \) (respectively \(P-Gaussian \)) if \(C(fg)=C(f)C(g)\) for every polynomials f and g in R[X] (respectively f and g in R[[X]]).
Theorem 2.6
Let R be a Gaussian ring (respectively a Noetherian Gaussian ring). If R is a U-ring, then I is an n-absorbing ideal of R if and only if I[X] (respectively I[[X]]) is an n-absorbing ideal of R[X] (respectively R[[X]]). Moreover, \(\omega _R(I)=\omega _{R[X]}(I[X])\) (respectively \(\omega _R(I)=\omega _{R[[X]]}(I[[X]])\)).
Proof
We prove the result in the case of polynomial rings.
“\(\Leftarrow \)” It follows from Anderson and Badawi (2011, Corollary 4.3).
“\(\Rightarrow \)” Suppose that I is an n-absorbing ideal of R and let \(f_1\), \(f_2,\ldots ,\) \(f_{n+1}\in R[X]\) such that \(f_1\ldots f_{n+1}\in I[X]\).
Since R is a Gaussian ring, we conclude that \(C(f_1)\cdots C(f_{n+1})=C(f_1\cdots f_{n+1})\subseteq I\). As I is a strongly n-absorbing ideal of R, by Theorem 2.2, hence \(\hat{C(f_i)}\subseteq I\) for some \(1\le i\le n+1\), thus \(\hat{f_i}\in I[X]\).
The same proof works also in the case of power series rings as a Noetherian Gaussian ring is P-Gaussian (Tsang 1965).
Recall that a commutative ring R is said to be a pseudo-valuation domain (PVD) if every prime ideal of R is strongly prime.
Theorem 2.7
Let R be a pseudo-valuation domain with associated valuation domain V and let I be an ideal of R such that Rad(I) is not maximal. Then I is an n-absorbing ideal of R if and only if I[X] (respectively I[[X]]) is an n-absorbing ideal of R[X] (respectively of R[[X]]). Moreover, \(\omega _R(I)=\omega _{R[X]}(I[X])\) (respectively \(\omega _R(I)=\omega _{R[[X]]}(I[[X]])\)).
Proof
Let I be an n-absorbing ideal of R. Then there are at most n prime ideal of R minimal over I. Since Rad(I) is the intersection of all the prime ideals minimal over I and the prime ideals are comparable in a PVD, we conclude that \(Rad(I)=P\) for some prime ideal minimal over I.
Recall that a PVD is a divided ring, so I is a P-primary ideal of R by Anderson and Badawi (2011, Theorem 3.2). As Rad(I) is not maximal then I is also a P-primary ideal of V by Anderson and Dobbs (1980, Proposition 3.13).
We show that \(P^n\subseteq I\). Let \(x_1,\ldots ,\) \(x_n\in P\), then there is an \(x\in P\) such that \((x_1,\ldots ,x_n)_V=xV\) since V is a valuation domain.
Hence \(x_1\ldots x_n=x^nb\) for some \(b\in V\). As \(x\in P=Rad(I)\) and I is n-absorbing then \(x^n\in I\) and so \(x^nb\in I\). Then I[X] is an n-absorbing ideal of R[X] by Anderson and Badawi (2011, Theorem 3.1) (respectively, by Fields (1971, Corollary 4)), I[[X]] is P[[X]]-primary since \(P^n[[X]]\subseteq I[[X]]\), so I[[X]] is an n-absorbing ideal of R[[X]]). \(\square \)
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Sihem, S., Sana, H. On Anderson-Badawi conjectures. Beitr Algebra Geom 58, 775–785 (2017). https://doi.org/10.1007/s13366-017-0343-9
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DOI: https://doi.org/10.1007/s13366-017-0343-9