On Anderson-Badawi conjectures

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.

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:

• 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.

• 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.

• 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.

• 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 $x_1^2x_2^2x_3^2(x_1+x_2+x_3)\in I$.

• Let $x_1$, $x_2$, $x_3\in Rad(I)$ then $x_1^2x_2x_3\in I$, since $x_1^2x_2x_3(x_2+x_3)\in I$.

• 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 $x\in Rad(I)$.

• Let $x_1$, $x_2\in Rad(I)$ then $x_1^4x_2^4\in I$, since $x_1^4(x_1+x_2)x_2^4\in I$.

• 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.

• 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 $x_1^4x_2^2\in I$.

• 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 $x_1^4(x_1+x_2)x_2\in I$.

• Let $x_1$, $x_2$, $x_3\in Rad(I)$ then $(x_1x_2x_3)^2\in I$. It suffices to remark that $x_1^2x_2^2x_3^2(x_1+x_2+x_3)\in I$.

• 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 $x_1^3x_2x_3(x_1+x_2+x_3)\in I$.

• 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.

• 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 $x_1^2x_2x_3x_4(x_2+x_3+x_4)\in I$.

• 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 $n\in \{3, 4, 5\}$.

• (1) If $P^n$ is a P-primary ideal of R and $P^n\subset P^{n-1}$, then $\omega _R(P^n)=n$.

• (2) If P is a maximal ideal of R and $P^n\subset P^{n-1}$, then $\omega _R(P^n)=n$.

• (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:

• (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.

• (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 $x_1^{n-2}(x_1+x_2)x_2^{n-2}$.

• (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) $(I:J)=\{x\in R\; |\; xJ\subseteq I\}$.

• (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:

• $\hat{x_i}$ the product $x_1\ldots x_{i-1}x_{i+1}\ldots x_n$.

• $\hat{I_i}$ 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.

• (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

• “$1)\Rightarrow 2)$” 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 $a\in \cup _{1\le i \le n}(I:\hat{x_i})$.

• $``2)\Rightarrow 1)$” 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

• (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:

• (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.

• $2)\Rightarrow 3)$ This follows from the last proposition, since R is a U-ring.

• $3)\Rightarrow 4)$ 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.

• $4)\Rightarrow 5)$ 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.

• $5)\Rightarrow 1)$ 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.

• (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, $y\in R{\setminus } I$.

  • If x, $y\in P{\setminus } I$, it’s clear by Badawi (2007 Theorem 2.5).

  • If x, $y\in R{\setminus } P$, it’s clear by the last remark.

  • if $x\in R{\setminus } P$ and $y\in P{\setminus } I$, we have $(I:x)\subset P\subset (I:y)$ by the last remark and Badawi (2007 Theorem 2.5).

• (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 $(I:y)\subseteq P_1\cup P_2\subseteq (I:x)$.

  • If x, $y\in R{\setminus } Rad(I)$:

    • if x, $y\in P_1{\setminus } P_2$, we conclude that $(I:x)=(I:y)=P_2$.

    • 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]]). $\square$