A New Generalization of the class of r-Ideals in Commutative Rings

1. Introduction

Throughout, all rings are assumed to be commutative with identity and all modules are unital. For a ring R, $Zd\left(R\right)$, $Reg\left(R\right)$ and $Nil\left(R\right)$ denote the sets of zero-divisors, regular elements and the nilradical of R, respectively. For an element $a\in R$, ${Ann}_R\left(a\right)=\{r\in R:ar=0\}$ denotes the annihilator of a.

In 2015, R. Mohamadian [1] introduced the concept of r-ideals of commutative rings. A proper ideal I of a ring R is called an r-ideal if whenever $a,b\in R$ such that $ab\in I$ and ${Ann}_R\left(a\right)=0$, then $b\in I$. Prime and r-ideals are not comparable in general; however, it is verified that every maximal r-ideal in a ring is a prime ideal, while every minimal prime ideal is an r-ideal. In 2017, Tekir, Koç and Oral [2] introduced the concept of n-ideals as a special kind of r-ideals by considering the set of nilpotent elements instead of zero divisors. In [3], Yetkin Celikel and Khashan generalized n-ideals by defining and studying the class of semi n-ideals. A proper ideal I of R is called a semi n-ideal if for $a\in R$, $a^2\in I$ and $a\notin \sqrt{0}$ imply $a\in I$. Later, several other generalizations of n-ideals and r-ideals have been introduced; see, for example, [4,5].

Over the last decades, many generalizations of prime ideals and submodules have been built (see, for example, [6,7,8,9]). Among them, the one originally proposed by Anderson, Badawi and Coykendall offers a new perspective. According to their paper [10], a proper ideal I of R is called a square-difference factor absorbing ideal ($sdf$-absorbing ideal) of R if for nonzero $a,b\in R$, whenever $a^2-b^2\in I$, then $a+b\in I$ or $a-b\in I$. They gave many properties of $sdf$-absorbing ideals, such as showing that a nonzero $sdf$-absorbing ideal is a radical ideal, which leads to a characterization of rings in which every nonzero proper ideal is $sdf$-absorbing. Moreover, they determine the $sdf$-absorbing ideals in a PID, direct product of two rings, polynomial rings, idealizations, amalgamation rings and $D+M$ constructions.

A natural extension of this notion, which involves the radical of the ideal, was proposed by Khashan, Yetkin Celikel, and Tekir [11]. They defined an ideal I of a ring R to be $sdf$-absorbing primary if for all $a,b\in R$, $a^2-b^2\in I$ implies $a+b\in \sqrt{I}$ or $a-b\in I$. A fundamental structural result established in [11] states that if I is an $sdf$-absorbing primary ideal, then its radical $\sqrt{I}$ is necessarily $sdf$-absorbing. Building on this concept in module theory, the notion of square-difference factor absorbing submodules ($sdf$-absorbing submodules) was recently introduced in [12]. A proper submodule N of M is called an $sdf$-absorbing submodule of M if for $m\in M$ and $a,b\in R\setminus {Ann}_R\left(m\right)$, whenever $(a^2-b^2)m\in N$, then $(a+b)m\in N$ or $(a-b)m\in N$.

Recently, Abouhalaka and Kim [13] introduced a new class of ideals which is a generalization of n-ideals using the square-difference factor concept. A proper ideal I of R is a Nil-$sdf$-absorbing ideal of R if for all $a,b\in R$, $a^2-b^2\in I$ implies $a-b\in I$ or $a+b\in Nil\left(R\right)$. They establish permanence properties under localization and homomorphisms, and provide characterizations in trivial ring extensions and in pullback settings, including amalgamated algebras and amalgamated modules along an ideal.

Inspired by the above developments, we introduce and systematically study a new class of ideals, called r-$sdf$-absorbing ideals, which naturally unifies and extends both the classical notions of r-ideals and $sdf$-absorbing ideals. A proper ideal I of a ring R is said to be r-$sdf$-absorbing if whenever $a^2-b^2\in I$ for $a,b\in R$, then either $a+b$ is a zero divisor or $a-b\in I$. This definition replaces the nilpotent condition in Nil-$sdf$-absorbing ideals by the weaker zero-divisor condition, thereby providing a unified framework that encompasses several important ideal classes.

The paper is organized as follows. In Section 2, we establish the fundamental properties of r-$sdf$-absorbing ideals. We begin by giving the definition and providing the basic examples. We fully characterize these ideals in the ring of integers $\mathbb{Z}$, showing that the only proper r-$sdf$-absorbing ideals are $\left(0\right)$ and $2\mathbb{Z}$ (Proposition 1). We then prove that in any ring of characteristic 2, every radical ideal is r-$sdf$-absorbing (Proposition 2). Next, we provide a complete characterization of when the ideal $\left(x\right)$ in a polynomial ring $A\left[x\right]$ over a characteristic 2 ring is r-$sdf$-absorbing, namely that A must be reduced (Proposition 3). We also show (Proposition 4) that when 2 is a unit, the notions of r-ideals and r-$sdf$-absorbing ideals coincide. Several examples are given to illustrate the proper inclusions among the classes of r-ideals, n-ideals, $sdf$-absorbing ideals, Nil-$sdf$-absorbing ideals, and r-$sdf$-absorbing ideals. We also give a complete description of rings in which every proper ideal is r-$sdf$-absorbing, showing that this is equivalent to every regular element being a unit (Theorem 1). In analogy with the r-ideal case, we determine precisely when the zero ideal is the only r-$sdf$-absorbing ideal (Proposition 6). We also prove that the intersection of any family of r-$sdf$-absorbing ideals is again r-$sdf$-absorbing, while the sum and the product need not be.

In Section 3, we investigate the transfer of the r-$sdf$-absorbing property under several standard ring-theoretic constructions. We provide sufficient conditions for the image and preimage of such an ideal under surjective ring homomorphisms to inherit the property (Theorem 2), and we give examples showing that the extra hypotheses are essential. We then establish analogous results for quotient rings (Corollary 1) , localizations (Theorem 3), direct products (Theorem 4), idealizations (Theorem 6), and amalgamations of rings (Theorem 7). In each case, we obtain sharp criteria with necessary and sufficient conditions. In particular, for direct products, we show that the property is well-behaved under full products and under extensions of the form $I_1\times R_2$, but requires an additional regularity condition for the descent from a product to its components. For idealizations, we prove that $I(+)M$ is r-$sdf$-absorbing if and only if I is r-$sdf$-absorbing. For amalgamations, we establish the lifting of the property and provide the precise zero-divisor conditions under which the descent holds.

Our results demonstrate that r-$sdf$-absorbing ideals form a rich and natural extension of the existing theory, with deep connections to zero-divisor structures and ideal-theoretic generalizations of prime ideals.

2. Fundamental Properties of r-$sdf$-Absorbing Ideals

In this section, we establish the foundational properties of r-$sdf$-absorbing ideals. We begin with the definition and provide basic examples, including a full characterization of these ideals in $\mathbb{Z}$. We then explore their behavior in rings of characteristic 2, determine when the ideal $\left(x\right)$ in a polynomial ring is r-$sdf$-absorbing, and show that the notions of r-ideals and r-$sdf$-absorbing ideals coincide when 2 is a unit. We also examine the relationships between r-$sdf$-absorbing ideals and related classes, classify rings in which every proper ideal is r-$sdf$-absorbing, and establish closure properties under intersections.

Definition 1.

Let R be a ring and let I be a proper ideal of R. Then I is called anr-$sdf$-absorbing idealif for $a,b\in R$, whenever $a^2-b^2\in I$, then $a+b\in Zd\left(R\right)$ or $a-b\in I$.

By symmetry, replacing b with $-b$ in the definition shows that I is an r-$sdf$-absorbing ideal of R if and only if, for any $a,b\in R$, whenever $a^2-b^2\in I$, then either $a-b\in Zd\left(R\right)$ or $a+b\in I$.

It is clear that every r-ideal (and so every n-ideal) of a ring R is an r-$sdf$-absorbing ideal. However, the converse is false in general. For example, the ideal $2\mathbb{Z}$ is not an r-ideal of $\mathbb{Z}$ because $2\mathbb{Z}⊈Zd\left(\mathbb{Z}\right)=\left\{0\right\}$. On the other hand, $2\mathbb{Z}$ is r-$sdf$-absorbing in $\mathbb{Z}$. Indeed, if $a,b\in \mathbb{Z}$ and $a^2-b^2\in 2\mathbb{Z}$, then $a^2\equiv b^2(mod2)$. Since $x^2\equiv x(mod2)$ for every integer x, we get $a\equiv b(mod2)$, and hence $a-b\in 2\mathbb{Z}$. Thus, $2\mathbb{Z}$ is r-$sdf$-absorbing.

Having seen that $2\mathbb{Z}$ provides a counterexample to the converse, it is natural to ask what all r-$sdf$-absorbing ideals of $\mathbb{Z}$ look like. The following proposition answers this question completely.

Proposition 1.

For a non-negative integer n and a proper ideal $I=n\mathbb{Z}$ of $\mathbb{Z}$, I is r-$sdf$-absorbing if and only if $n=0$ or $n=2$.

Proof. 

If $n=0$, then $I=\left\{0\right\}$ is an r-ideal and so r-$sdf$-absorbing in $\mathbb{Z}$. Next, if $n=2$, then $I=2\mathbb{Z}$ is also r-$sdf$-absorbing as shown above. Conversely, assume $I=n\mathbb{Z}$ is a proper r-$sdf$-absorbing ideal. Suppose $n\ge 3$ and choose $a=n-1$, $b=1$. Then $a^2-b^2={(n-1)}^2-1=n(n-2)\in I$. However, $a+b=n$ is not a zero divisor in $\mathbb{Z}$ and $a-b=n-2\notin I$. Consequently, for $n\ge 3$, $I=n\mathbb{Z}$ is not an r-$sdf$-absorbing ideal of $\mathbb{Z}$. □

The following result provides a general sufficient condition for an ideal to be r-$sdf$-absorbing, based only on the characteristic of the ring and the radicality of the ideal.

Proposition 2.

Let R be a ring of characteristic 2, and let I be a proper radical ideal of R. Then I is an r-$sdf$-absorbing ideal of R.

Proof. 

Let $a,b\in R$ and suppose that $a^2-b^2\in I.$ Since $char\left(R\right)=2$, then ${(a-b)}^2=a^2+b^2=a^2-b^2\in I$. Since I is a radical ideal, then $a-b\in I$. Therefore, I is r-$sdf$-absorbing. □

The following example illustrates the proposition in a concrete setting and shows that, despite being r-$sdf$-absorbing, such ideals need not be r-ideals.

Example 1.

Let $R={\mathbb{F}}_2\left[x\right]$ and $I=\left(x\right).$ First, we show that I is r-$sdf$-absorbing in R. Indeed, take $a,b\in R$ such that $a^2-b^2\in I$. Since $char\left({\mathbb{F}}_2\right)=2$, we have $a^2-b^2=a^2+b^2={(a+b)}^2={(a-b)}^2.$ Thus, ${(a-b)}^2\in \left(x\right)$ and so $a-b\in \left(x\right)=I$ because $\left(x\right)$ is a prime ideal in ${\mathbb{F}}_2\left[x\right]$. Hence, I is r-$sdf$-absorbing in R. However, I is not an r-ideal since it contains the regular element x of R.

The characteristic 2 hypothesis in Proposition 2 is essential and cannot be discarded. For instance, in ${\mathbb{F}}_3\left[x\right]$, the radical ideal $\left(x\right)$ fails to be r-$sdf$-absorbing. Take $a=1$ and $b=2+x$. Then

$$a^2-b^2=1-{(2+x)}^2=1-(4+4x+x^2)=-x-x^2=x(-1-x)\in \left(x\right).$$

However, $a+b=x\notin Zd\left({\mathbb{F}}_3\left[x\right]\right)$ and $a-b=-1-x\notin \left(x\right)$. Thus, $\left(x\right)$ is not r-$sdf$-absorbing in ${\mathbb{F}}_3\left[x\right]$.

The following result sharpens the preceding example by providing a complete characterization for the ideal $\left(x\right)$ in the polynomial ring $A\left[x\right]$ over a ring of characteristic 2.

Proposition 3.

Let A be a ring with $char\left(A\right)=2$. Set $R=A\left[x\right]$ and $I=\left(x\right).$ Then I is an r-$sdf$-absorbing ideal of R if and only if A is reduced.

Proof.(⇐) Assume A is reduced. Let $a,b\in R$ and suppose $a^2-b^2\in I=\left(x\right).$ Since $char\left(A\right)=2$, $a^2-b^2=a^2+b^2={(a+b)}^2={(a-b)}^2.$ Put $c=a-b=c_0+c_{1}x+c_2{x}^2+\dots +c_n{x}^n\in A\left[x\right]$. Then $c^2\in \left(x\right)$, which means the constant term $c_0^2$ of $c^2$ is zero. Since A is reduced, this implies $c_0=0$ and hence $c=a-b\in \left(x\right)=I$. Thus I is r-$sdf$-absorbing.

(⇒) Conversely, suppose A is not reduced. Then there exists a nonzero element $u\in A$ such that $u^2=0$. Choose $a=u+x$ and $b=0$. Then $a^2-b^2={(u+x)}^2=u^2+x^2=x^2\in \left(x\right)=I.$ We show that $a+b=u+x$ is not a zero divisor in $A\left[x\right]$. By McCoy’s theorem, if $u+x$ were a zero divisor, there would exist a nonzero $r\in A$ such that $r(u+x)=0$. Comparing coefficients gives $r=0$, a contradiction. Hence $a+b\notin Zd\left(R\right)$. On the other hand, $a-b=u+x$ has constant term $u\ne 0$, so $a-b\notin \left(x\right)=I$. Therefore I fails the r-$sdf$-absorbing condition. This completes the proof. □

Nevertheless, the next proposition shows that the two notions of r-$sdf$-absorbing ideals and r-ideals are equivalent in any ring where 2 is a unit.

Proposition 4.

Let R be a ring such that $2\in U\left(R\right)$. Then an ideal I of R is r-$sdf$-absorbing if and only if I is an r-ideal of R.

Proof. 

Suppose that I is an r-$sdf$-absorbing ideal. Let $a,b\in R$ with $ab\in I$ and $a\notin Zd\left(R\right)$. Write $x={\displaystyle \frac{a+b}{2}}$ and $y={\displaystyle \frac{a-b}{2}}$. Then a straightforward calculation gives

$$x^2-y^2={\displaystyle \frac{{(a+b)}^2-{(a-b)}^2}{4}}=ab\in I.$$

Since I is r-$sdf$-absorbing and $x+y=a\notin Zd\left(R\right)$, it follows that $x-y=b\in I$. Thus I is an r-ideal. The converse is straightforward. □

Since $Nil\left(R\right)\subseteq Zd\left(R\right)$, it is clear that every Nil-$sdf$-absorbing ideal is r-$sdf$-absorbing. However, the converse is false in general, as the following two examples demonstrate.

Example 2.

Let $R=\mathbb{Z}\times \mathbb{Z}$ and $I=2\mathbb{Z}\times \left\{0\right\}$. Then I is r-$sdf$-absorbing but not Nil-$sdf$-absorbing. Indeed, take $a=(a_1,a_2)$, $b=(b_1,b_2)\in R$ with $a^2-b^2\in I$. Then $a_1^2-b_1^2\in 2\mathbb{Z}$ and $a_2^2-b_2^2=0$. From the first condition, $a_1^2\equiv b_1^2(mod2)$ implies $a_1\equiv b_1(mod2)$ and so $a_1-b_1\in 2\mathbb{Z}$. From the second condition, $a_2^2=b_2^2$ in the domain $\mathbb{Z}$, either $a_2=b_2$ or $a_2=-b_2$. If $a_2=b_2$, then $a-b=(a_1-b_1,0)\in I$. If $a_2=-b_2$, then $a+b=(a_1+b_1,0)$ is a zero divisor in $R=\mathbb{Z}\times \mathbb{Z}$. Thus, I is r-$sdf$-absorbing in R. However, I is not Nil-$sdf$-absorbing. Take $a=(1,1)$ and $b=(1,-1)$. Then $a^2-b^2=(0,0)\in I$, but $a-b=(0,2)\notin I$ and $a+b=(2,0)\notin Nil(\mathbb{Z}\times \mathbb{Z})=\left\{\right(0,0\left)\right\}$.

Example 3.

Let K be any field with $char\left(K\right)\ne 2$, and set $R={\mathbb{F}}_2\left[t\right]\times K$, $I=\left(t\right)\times \left\{0\right\}.$ Then I is r-$sdf$-absorbing but not Nil-sdf-absorbing in R. To see that I is r-$sdf$-absorbing, let $a=(a_1,a_2)$, $b=(b_1,b_2)\in R$ with $a^2-b^2\in I$. Then $a_1^2-b_1^2\in \left(t\right)$ and $a_2^2-b_2^2=0$. Since $char\left({\mathbb{F}}_2\right)=2$, we have $a_1^2-b_1^2={(a_1-b_1)}^2$. As $\left(t\right)$ is prime, this gives $a_1-b_1\in \left(t\right)$. From $a_2^2=b_2^2$ in the field K, we get $a_2=b_2$ or $a_2=-b_2$. If $a_2=b_2$, then $a-b=(a_1-b_1,0)\in I$. If $a_2=-b_2$, then $a+b=(a_1+b_1,0)$, which is a zero divisor in R. Thus I is r-$sdf$-absorbing. To show that I is not Nil-sdf-absorbing, choose $a=(1,1)$ and $b=(1+t,-1).$ Then

$$a^2-b^2=(1,1)-({(1+t)}^2,{(-1)}^2)=(1-(1+t^2),0)=(t^2,0)\in I.$$

However, $a-b=(1-(1+t),1-(-1\left)\right)=(t,2)\notin I,$ because $char\left(K\right)\ne 2$. Also, $a+b=(1+1+t,1+(-1\left)\right)=(t,0)\notin Nil\left(R\right)$ since $Nil\left(R\right)=\left\{\right(0,0\left)\right\}$. Hence neither $a-b\in I$ nor $a+b\in Nil\left(R\right)$ holds, so I is not Nil-$sdf$-absorbing.

We can also observe that neither of the ideals in the above examples is an r-ideal. Indeed, in $R=\mathbb{Z}\times \mathbb{Z}$, taking $a=(2,1)$ and $b=(1,0)$ gives $ab=(2,0)\in I$, while $a\notin Zd\left(R\right)$ and $b\notin I$. Similarly, in $R={\mathbb{F}}_2\left[t\right]\times K$, the elements $a=(t,1)$ and $b=(1,0)$ yield $ab=(t,0)\in I$, but $a\notin Zd\left(R\right)$ and $b\notin I$. Hence both ideals fail to be r-ideals.

Remark 1.

By ([10], Example 2.8(a)), the sdf-absorbing ideals of $\mathbb{Z}$ are precisely the prime ideals and the ideals of the form $2q\mathbb{Z}$, where q is an odd prime. Thus by Proposition 1, for $n=p$, $p\ge 3$ or $n=2q\mathbb{Z}$, where q is an odd prime, $I=n\mathbb{Z}$ is an sdf-absorbing ideal of $\mathbb{Z}$ that is not r-$sdf$-absorbing. This establishes that sdf-absorbing does not imply r-$sdf$-absorbing. To show that the reverse implication also fails, consider the ideal $I=\left\{0\right\}$ in $R={\mathbb{Z}}_{18}$. It is shown in ([13], Example 2.7) that I is a Nil-$sdf$-absorbing (and so r-$sdf$-absorbing) ideal of R that is not sdf-absorbing. Therefore, the notions of sdf-absorbing and r-$sdf$-absorbing ideals are genuinely distinct.

Although the notions of sdf-absorbing and r-$sdf$-absorbing ideals are independent in general, the following proposition establishes precise conditions under which they imply one another, based on the relative position of the ideal with respect to the set of zero divisors.

Proposition 5.

Let I be a proper ideal of R. Then the following hold:

1.

If I is an r-$sdf$-absorbing ideal of R and $Zd\left(R\right)\subseteq I$, then I is an sdf-absorbing ideal of R.

2.

If I is an sdf-absorbing ideal of R and $I\subseteq Zd\left(R\right)$, then I is an r-$sdf$-absorbing ideal of R.

Proof. (1) Let $a,b\in R$ with $a^2-b^2\in I$. Since I is r-$sdf$-absorbing, we have $a-b\in I$ or $a+b\in Zd\left(R\right)$. If $a-b\in I$, we are done. Otherwise, $a+b\in Zd\left(R\right)\subseteq I$, so $a+b\in I$. Hence I is sdf-absorbing.

(2) Let $a,b\in R$ and suppose that $a^2-b^2\in I$. Since I is sdf-absorbing, we have $a-b\in I$ or $a+b\in I$. If $a-b\in I$, there is nothing to prove. Otherwise, $a+b\in I\subseteq Zd\left(R\right)$. Thus I is r-$sdf$-absorbing. □

We now give a complete description of rings in which every proper ideal is r-$sdf$-absorbing. The following theorem shows that this property is equivalent to a simple condition on the regular elements of the ring.

Theorem 1.

Let R be a ring. Then the following are equivalent.

1.

Every proper ideal of R is an r-ideal.

2.

Every proper ideal of R is an r-$sdf$-absorbing ideal.

3.

Every regular element of R is a unit.

Proof.$\left(1\right)⟹\left(2\right)$ Clear.

$\left(2\right)⟹\left(3\right)$ Suppose that every proper ideal of R is r-$sdf$-absorbing, and let $u\in R$ be a regular element. We prove that u is a unit. If u is not a unit, then its square $u^2$ is also not a unit and so the ideal $I=\left(u^2\right)$ is proper. Since I is r-$sdf$-absorbing, applying the definition with $a=u$ and $b=0$ gives $u^2-0=u^2\in I$, and therefore either $u\in Zd\left(R\right)$ or $u\in \left(u^2\right)$. But u is regular, so we must have $u\in \left(u^2\right)$. Thus, there exists $r\in R$ such that $u=r{u}^2$, whence $u(1-ru)=0$. Since u is regular, cancellation yields $1=ru$, proving that u is a unit. Therefore every regular element of R is a unit.

$\left(3\right)⟹\left(1\right)$ Assume that every regular element of R is a unit. Let I be any proper ideal of R, and let $a,b\in R$ with $ab\in I$. If a is regular, then by assumption a is a unit which yields $b=a^{-1}\left(ab\right)\in I$. If a is not regular, then by definition $a\in Zd\left(R\right)$. Hence, in all cases, $ab\in I$ implies that either $a\in Zd\left(R\right)$ or $b\in I$. Therefore, I is an r-ideal. □

In complete analogy with [1] for r-ideals, we now determine precisely when the zero ideal is the only proper r-$sdf$-absorbing ideal of a ring. The result is as follows.

Proposition 6.

The zero ideal is the only r-$sdf$-absorbing ideal of a ring R if and only if R is a domain and either 2 is a unit of R or R is a field of characteristic 2.

Proof. 

Suppose that $\left(0\right)$ is the only r-$sdf$-absorbing ideal of R. Since every r-ideal is r-$sdf$-absorbing, it follows that $\left(0\right)$ is also the only r-ideal of R. Hence, by [1], R is a domain. Assume, toward a contradiction, that 2 is not a unit and that R is not a field of characteristic 2. If $char\left(R\right)=2$, then R is a domain which is not a field. Hence R possesses a nonzero maximal ideal M. Thus, by Proposition 2, M is a nonzero r-$sdf$-absorbing ideal of R, contradicting the hypothesis. Next, suppose that $char\left(R\right)\ne 2$ so that $2\ne 0$. Since 2 is not a unit, the principal ideal $\left(2\right)$ is a nonzero proper ideal of R. Thus, there exists a prime ideal $P\ne \left(0\right)$ containing $\left(2\right)$. We claim that P is r-$sdf$-absorbing. Let $a,b\in R$ satisfy $a^2-b^2\in P$. Since P is prime, either $a-b\in P$ or $a+b\in P$. In the latter case, $a-b=(a+b)-2b\in P$ because $2\in P$. Thus, in either case, $a-b\in P$, proving that P is an r-$sdf$-absorbing ideal. This contradicts the assumption that $\left(0\right)$ is the only r-$sdf$-absorbing ideal. Therefore, either 2 is a unit of R or R is a field of characteristic 2. Conversely, assume that R is a domain and either 2 is a unit or R is a field of characteristic 2. If R is a field of characteristic 2, then $\left(0\right)$ is clearly the only r-$sdf$-absorbing ideal. Suppose now that 2 is a unit, and let I be a nonzero proper ideal of R. Choose $0\ne x\in I$ and let $a=1+x$, $b=x-1$. Then

$$a^2-b^2={(1+x)}^2-{(x-1)}^2=4x\in I.$$

On the other hand, $a+b=2x$ is a nonzero element of the domain R, and hence is regular. Moreover, $a-b=2\notin I$ since 2 is a unit and I is proper. Therefore, I does not satisfy the defining property of an r-$sdf$-absorbing ideal. As I was arbitrary, $\left(0\right)$ is the only r-$sdf$-absorbing ideal of R. □

Analogous to [1, Proposition 2.7], we have the following result for r-$sdf$-absorbing ideals.

Proposition 7.

Let $P_1,\dots ,P_n$ be pairwise incomparable prime ideals of a ring R. Suppose that for each i, there exists a regular element $y_i\in \left({\prod }_{j\ne i}{P}_j\right)\setminus P_i$. If the intersection $I={\bigcap }_{i=1}^n{P}_i$ is an r-$sdf$-absorbing ideal of R, then each $P_i$ is an r-$sdf$-absorbing ideal of R.

Proof. 

Fix an index i. Let $a,b\in R$ with $a^2-b^2\in P_i$. Choose a regular element $y\in \left({\prod }_{j\ne i}{P}_j\right)\setminus P_i$, which exists by hypothesis. Since $y\in {\prod }_{j\ne i}{P}_j$, we have $y^2\in P_j$ for every $j\ne i$. Also, $y^2(a^2-b^2)\in P_i$ because $a^2-b^2\in P_i$. Hence

$${\left(ya\right)}^2-{\left(yb\right)}^2=y^2(a^2-b^2)\in {\displaystyle \bigcap _{j=1}^n}{P}_j=I.$$

Since I is r-$sdf$-absorbing, either $y(a+b)\in Zd\left(R\right)$ or $y(a-b)\in I$. If $y(a-b)\in I$, then $y(a-b)\in P_i$. As $P_i$ is prime and $y\notin P_i$, we conclude $a-b\in P_i$. If $y(a+b)\in Zd\left(R\right)$, then since y is regular, there exists a nonzero z such that $y(a+b)z=0$. Thus $(a+b)\left(yz\right)=0$ with $yz\ne 0$, proving that $a+b\in Zd\left(R\right)$. In either case, the condition for $P_i$ to be r-$sdf$-absorbing is satisfied. Therefore, each $P_i$ is r-$sdf$-absorbing. □

The following example demonstrates that the regularity condition in the above proposition cannot be omitted.

Example 4.

Let $R=\mathbb{Z}\times {\mathbb{F}}_2$ and consider the two prime ideals $P_1=3\mathbb{Z}\times {\mathbb{F}}_2$ and $P_2=\mathbb{Z}\times \left\{0\right\}$ of R. These primes are incomparable and their intersection is $I=P_1\cap P_2=3\mathbb{Z}\times \left\{0\right\}$. First, observe that I is an r-$sdf$-absorbing ideal. Indeed, let $a=(a_1,a_2)$ and $b=(b_1,b_2)$ in R with $a^2-b^2\in I$. Since $a_2^2-b_2^2=0$ in ${\mathbb{F}}_2$ then clearly $a_2=b_2$. Hence, $a+b=(a_1+b_1,0)\in Zd\left(R\right)$. Therefore the r-$sdf$-absorbing condition is satisfied, so I is indeed r-$sdf$-absorbing. We now show that $P_1$ fails to be r-$sdf$-absorbing. Take $a=(1,1)$ and $b=(2,0).$ Then $a^2-b^2=(-3,1)$$\in P_1$. However, $a+b=(3,1)\notin Zd\left(R\right)$ and $a-b=(-1,1)$ $\notin P_1$. Thus $P_1$ is not r-$sdf$-absorbing. We note that the set $P_2\setminus P_1$ consists of elements of the form $(n,0)$ where $n\in \mathbb{Z}$ is not divisible by 3. Every such element is clearly a zero divisor in R. Therefore, there is no regular element in $P_2\setminus P_1$. This example demonstrates that the conclusion of the proposition fails if the regularity condition is omitted.

We now establish an elementary but important closure property of r-$sdf$-absorbing ideals.

Proposition 8.

Let ${\left\{I_{\alpha }\right\}}_{\alpha \in \Lambda }$ be a family of r-$sdf$-absorbing ideals of a ring R. Then their intersection

$$I=\bigcap _{\alpha \in \Lambda }{I}_{\alpha }$$

is an r-$sdf$-absorbing ideal of R.

Proof. 

Let $a,b\in R$ such that $a^2-b^2\in I$. Then $a^2-b^2\in I_{\alpha }$ for every $\alpha \in \Lambda$. Suppose, for contradiction, that $a+b\notin Zd\left(R\right)$. Since each $I_{\alpha }$ is r-$sdf$-absorbing, the definition forces the second alternative to hold for every $\alpha$. Thus $a-b\in I_{\alpha }$ for all $\alpha$, and consequently $a-b\in {\bigcap }_{\alpha \in \Lambda }{I}_{\alpha }=I$. Hence, whenever $a^2-b^2\in I$, either $a+b\in Zd\left(R\right)$ or $a-b\in I$. Therefore, I is r-$sdf$-absorbing. □

Remark 2.

While the class of r-$sdf$-absorbing ideals is closed under arbitrary intersections, it is not closed under products and Sums. Indeed, in the ring $\mathbb{Z}$, the ideal $2\mathbb{Z}$ is r-$sdf$-absorbing, as shown earlier. However, its square ${\left(2\mathbb{Z}\right)}^2=4\mathbb{Z}$ is not r-$sdf$-absorbing. This follows immediately from Proposition 1, which characterizes the r-$sdf$-absorbing ideals of $\mathbb{Z}$ as precisely $\left\{0\right\}$, $2\mathbb{Z}$.

Before presenting the next example, we briefly recall some notation concerning rings of continuous functions. Let $C\left(\mathbb{R}\right)$ denote the ring of all real-valued continuous functions on $\mathbb{R}$. For $f\in C\left(\mathbb{R}\right)$, let

$$Z\left(f\right)=\{x\in \mathbb{R}:f(x)=0\}$$

denote the zero-set of f. For a subset $A\subseteq \mathbb{R}$, define

$$M_A=\{f\in C\left(\mathbb{R}\right):A\subseteq Z\left(f\right)\}.$$

It is well known that $M_A$ is an ideal of $C\left(\mathbb{R}\right)$. We also recall that a function $f\in C\left(\mathbb{R}\right)$ is regular if and only if the interior of $Z\left(f\right)$ is empty.

Example 5.

The sum of two r-$sdf$-absorbing ideals need not be r-$sdf$-absorbing. Let

$$I=M_{[0,1)}\mathrm{and}J=M_{(-1,0]}$$

be ideals of the ring $C\left(\mathbb{R}\right)$. By [1], both I and J are r-ideals and hence r-$sdf$-absorbing ideals. Moreover, the function $x\in C\left(\mathbb{R}\right)$, defined by $x\left(t\right)=t$, belongs to $I+J$ and is a regular element, so $I+J$ is not an r-ideal. Now let $a=1+x$ and $b=x-1.$ Then $a^2-b^2={(1+x)}^2-{(x-1)}^2=4x\in I+J.$ Furthermore, $a+b=2x$ is regular, since 2 is a unit in $C\left(\mathbb{R}\right)$ and x is regular, while $a-b=2\notin I+J$ because $I+J$ is a proper ideal and 2 is a unit. Hence, $I+J$ is not an r-$sdf$-absorbing ideal.

3. Behavior of r-$sdf$-Absorbing Ideals Under Extensions

In this section, we study the behavior of r-sdf-absorbing ideals under the standard ring-theoretic constructions, namely homomorphic images, localizations, direct products, idealizations, and amalgamations of rings. These transfer results provide a systematic understanding of how the r-sdf-absorbing property is preserved, reflected, or lost when passing between a ring and its extensions.

We now investigate how r-$sdf$-absorbing ideals behave under surjective ring homomorphisms. The following theorem gives sufficient conditions for the image and the preimage of such an ideal to inherit the same property.

Theorem 2.

Let $f:R\to S$ be a ring epimorphism. Then the following hold.

1.

If I is an r-$sdf$-absorbing ideal of R such that $Ker\left(f\right)\subseteq I$ and $f(Zd(R\left)\right)\subseteq Zd\left(S\right)$, then $f\left(I\right)$ is an r-$sdf$-absorbing ideal of S.

2.

If J is an r-$sdf$-absorbing ideal of S such that $f^{-1}(Zd\left(S\right))\subseteq Zd\left(R\right)$, then $f^{-1}\left(J\right)$ is an r-$sdf$-absorbing ideal of R.

Proof. (1) Set $J=f\left(I\right)$. We first show that J is proper. Suppose, to the contrary, that $J=S$. Since f is surjective, for every $r\in R$, there exists $i\in I$ such that $f\left(r\right)=f\left(i\right)$. Hence $r-i\in Ker\left(f\right)\subseteq I$, so $r=(r-i)+i\in I$. Thus, $I=R$, contradicting the properness of I. Therefore J is a proper ideal of S. Now let $u,v\in S$ such that $u^2-v^2\in J$. Choose $a,b\in R$ with $f\left(a\right)=u$ and $f\left(b\right)=v$. Then

$$f(a^2-b^2)=f{\left(a\right)}^2-f{\left(b\right)}^2=u^2-v^2\in f\left(I\right).$$

Again since $Ker\left(f\right)\subseteq I$, we conclude $a^2-b^2\in I$. Since I is r-$sdf$-absorbing, either $a-b\in I$ or $a+b\in Zd\left(R\right)$. If $a-b\in I$, then $u-v=f(a-b)\in f\left(I\right)=J$. If $a+b\in Zd\left(R\right)$, then using the hypothesis $f(Zd(R\left)\right)\subseteq Zd\left(S\right)$, we obtain $u+v=f(a+b)\in Zd\left(S\right)$. Thus, J is an r-$sdf$-absorbing ideal of S.

(2) Set $I=f^{-1}\left(J\right)$. Since f is surjective and J is proper, I is a proper ideal of R. Let $a,b\in R$ such that $a^2-b^2\in I$. Then

$$f{\left(a\right)}^2-f{\left(b\right)}^2=f(a^2-b^2)\in J.$$

Since J is r-$sdf$-absorbing, either $f\left(a\right)-f\left(b\right)\in J$ or $f\left(a\right)+f\left(b\right)\in Zd\left(S\right)$. If $f\left(a\right)-f\left(b\right)\in J$, then $a-b\in f^{-1}\left(J\right)=I$. If $f\left(a\right)+f\left(b\right)\in Zd\left(S\right)$, then applying the hypothesis $f^{-1}(Zd\left(S\right))\subseteq Zd\left(R\right)$ yields $a+b\in f^{-1}(Zd\left(S\right))\subseteq Zd\left(R\right).$ Hence $I=f^{-1}\left(J\right)$ is r-$sdf$-absorbing in R. □

The following examples demonstrate that the extra conditions in the above theorem cannot be omitted. Firstly, we show that the condition $Ker\left(f\right)\subseteq I$ in part (1) of the theorem cannot be discarded.

Example 6.

Let $R={\mathbb{Z}}_2\times \mathbb{Z}$ and $S=\mathbb{Z}$. Define the epimorphism $f:R\to S$ by $f(x,y)=y$. Then $Ker\left(f\right)={\mathbb{Z}}_2\times \left\{0\right\}$. Consider the ideal $I=\left\{0\right\}\times 3\mathbb{Z}$ of R. We first show that I is r-$sdf$-absorbing. Let $a=(a_1,a_2)$, $b=(b_1,b_2)\in R$ such that $a^2-b^2\in I$. Then

$$a^2-b^2=(a_1^2-b_1^2,a_2^2-b_2^2)=(a_1-b_1,a_2^2-b_2^2).$$

Thus, $a_1-b_1=0$, so $a_1=b_1$ and so $a+b=(0,a_2+b_2)\in Zd\left(R\right)$. Hence, I is r-$sdf$-absorbing. However, $Ker\left(f\right)={\mathbb{Z}}_2\times \left\{0\right\}$ is not contained in I. Moreover, $f\left(I\right)=3\mathbb{Z}$ which is not an r-$sdf$-absorbing ideal of $\mathbb{Z}$. Thus the conclusion fails when $Ker\left(f\right)⊈I$.

Next, we show that the condition $f^{-1}(Zd\left(S\right))\subseteq Zd\left(R\right)$ in part (2) of the theorem also cannot be discarded.

Example 7.

Let $R=\mathbb{Z}\left[x\right]$ and $S=\mathbb{Z}$. Define the epimorphism $f:R\to S$ by $f\left(p\right(x\left)\right)=p\left(0\right)$. Then $Ker\left(f\right)=\left(x\right)$. Moreover, we have $f^{-1}(Zd\left(S\right))=f^{-1}\left(0\right)=\left(x\right).$ Since the element $x\in \left(x\right)$ is regular in $\mathbb{Z}\left[x\right]$, then $f^{-1}(Zd\left(S\right))⊈Zd\left(R\right)$. Now, let $J=\left\{0\right\}$ which is an r-$sdf$-absorbing ideal of S. Then $f^{-1}\left(J\right)=f^{-1}\left(0\right)=\left(x\right)$ which is not r-$sdf$-absorbing in $\mathbb{Z}\left[x\right]$. Indeed, take $a=1$ and $b=x-1.$ Then

$$a^2-b^2=1-{(x-1)}^2=-x^2+2x=x(2-x)\in \left(x\right).$$

However, $a+b=1+x-1=x$ is regular in $\mathbb{Z}\left[x\right]$ and $a-b=1-(x-1)=2-x\notin \left(x\right)$. Thus, $\left(x\right)$ fails the r-$sdf$-absorbing condition. Therefore, without the hypothesis $f^{-1}(Zd\left(S\right))\subseteq Zd\left(R\right)$, the preimage of an r-$sdf$-absorbing ideal need not be r-$sdf$-absorbing.

Applying the epimorphism theorem to the natural surjection $\pi :R\to R/J$ yields the following corollary.

Corollary 1.

Let $J\subseteq I$ be ideals of R, and let $\pi :R\to R/J$ be the canonical surjection.

1.

If I is an r-$sdf$-absorbing ideal of R such that $\pi (Zd(R\left)\right)\subseteq Zd(R/J)$, then $I/J$ is an r-$sdf$-absorbing ideal of $R/J$.

2.

If $I/J$ is an r-$sdf$-absorbing ideal of $R/J$ such that ${\pi }^{-1}(Zd(R/J))\subseteq Zd\left(R\right)$, then I is an r-$sdf$-absorbing ideal of R.

We now investigate the behavior of r-$sdf$-absorbing ideals under localization. The following theorem provides a sufficient condition ensuring that the localization of such an ideal retains the desired property. For a subset A of a ring R, by $Z_A\left(R\right)$, we mean the set of all $r\in R$ such that $rb\in A$ for some $b\in R\setminus A.$

Theorem 3.

Let S be a multiplicatively closed subset of a ring R and I be an ideal of R disjoint with S.

1.

If I is an r-$sdf$-absorbing ideal of R and $S^{-1}Zd\left(R\right)\subseteq Zd\left(S^{-1}R\right)$, then $S^{-1}I$ is an r-$sdf$-absorbing ideal of $S^{-1}R$.

2.

If $S\cap Z_I\left(R\right)=S\cap Z_{Zd\left(R\right)}\left(R\right)=\varnothing$ and $S^{-1}I$ is an r-$sdf$-absorbing ideal of $S^{-1}R,$ then I is an r-$sdf$-absorbing ideal of $R.$

Proof. (1) Let ${\displaystyle \frac{a}{s}},{\displaystyle \frac{b}{t}}\in S^{-1}R$ and assume that

$${\left({\displaystyle \frac{a}{s}}\right)}^2-{\left({\displaystyle \frac{b}{t}}\right)}^2={\displaystyle \frac{a^2{t}^2-b^2{s}^2}{s^2{t}^2}}\in S^{-1}I.$$

Then there exists $u\in S$ such that $u(a^2{t}^2-b^2{s}^2)\in I$. Set $x=uat$ and $y=ubs$ so that $x^2-y^2=u^2(a^2{t}^2-b^2{s}^2)\in I$. Because I is r-$sdf$-absorbing, either $x-y\in I$ or $x+y\in Zd\left(R\right)$. If $x-y\in I$, then

$${\displaystyle \frac{a}{s}}-{\displaystyle \frac{b}{t}}={\displaystyle \frac{at-bs}{st}}={\displaystyle \frac{u(at-bs)}{ust}}={\displaystyle \frac{x-y}{ust}}\in S^{-1}I.$$

If $x+y\in Zd\left(R\right)$, then

$${\displaystyle \frac{a}{s}}+{\displaystyle \frac{b}{t}}={\displaystyle \frac{at+bs}{st}}={\displaystyle \frac{u(at+bs)}{ust}}={\displaystyle \frac{x+y}{ust}}\in S^{-1}Zd\left(R\right)\subseteq Zd\left(S^{-1}R\right).$$

Thus, in all cases, $S^{-1}I$ is an r-$sdf$-absorbing ideal of $S^{-1}R$.

(2) Note that $Zd\left(S^{-1}R\right)\subseteq S^{-1}(Zd\left(R\right))$ is always true. Let $a,b\in R$ with $a^2-b^2\in I.$ Then ${\left({\displaystyle \frac{a}{1}}\right)}^2-{\left({\displaystyle \frac{b}{1}}\right)}^2\in S^{-1}I$ which yields that $\left({\displaystyle \frac{a}{1}}+{\displaystyle \frac{b}{1}}\right)\in Zd\left(S^{-1}R\right)\subseteq S^{-1}(Zd\left(R\right))$ or $\left({\displaystyle \frac{a}{1}}-{\displaystyle \frac{b}{1}}\right)\in S^{-1}I.$ Hence, $u(a+b)\in Zd\left(R\right)$ for some $u\in S$ or $v(a-b)\in I$ for some $v\in S.$ By our assumption $S\cap Z_I\left(R\right)=S\cap Z_{Zd}\left(R\right)=\varnothing ,$ we conclude either $a+b\in Zd\left(R\right)$ or $a-b\in I$, as required. □

The extra hypothesis $S^{-1}Zd\left(R\right)\subseteq Zd\left(S^{-1}R\right)$ in (1) of the above theorem is essential and cannot be omitted. The following example demonstrates this.

Example 8.

Let $R={\mathbb{Z}}_2\times \mathbb{Z}$ and consider the multiplicatively closed subset $S=\left\{\right(1,1),(0,1\left)\right\}$ of R. We showed previously that the ideal $I={\mathbb{Z}}_2\times 3\mathbb{Z}$ is r-$sdf$-absorbing in R and moreover, $S\cap I=\varnothing$. It is straightforward to verify that the map $\phi :S^{-1}R⟶\mathbb{Z}$ defined by ${\displaystyle \frac{(\overline{a},n)}{(\overline{b},1)}}⟼n$ is a well-defined ring isomorphism. Consequently, $S^{-1}R\cong \mathbb{Z}$ and under this isomorphism, $S^{-1}I=S^{-1}({\mathbb{Z}}_2\times 3\mathbb{Z})\cong 3\mathbb{Z}.$ Since $3\mathbb{Z}$ is not an r-$sdf$-absorbing ideal of $\mathbb{Z}$, it follows that $S^{-1}I$ is not r-$sdf$-absorbing. Note that the inclusion $S^{-1}Zd\left(R\right)\subseteq Zd\left(S^{-1}R\right)$ fails. Indeed, in R, the element $(0,1)$ is a zero divisor, and its image in $S^{-1}R\cong \mathbb{Z}$ is the integer 1, which is regular in $\mathbb{Z}$. Thus, $S^{-1}Zd\left(R\right)⊈Zd\left(S^{-1}R\right)$.

The following theorem establishes the behavior of r-$sdf$-absorbing ideals under direct products.

Theorem 4.

Let $I_1$ and $I_2$ be proper ideals of rings $R_1$ and $R_2$, respectively. Then the following statements hold.

1.

If $I_1$ is an r-$sdf$-absorbing ideal of $R_1$ and $I_2$ is an r-$sdf$-absorbing ideal of $R_2$, then $I=I_1\times I_2$ is an r-$sdf$-absorbing ideal of $R_1\times R_2$.

2.

If $I=I_1\times I_2$ is an r-$sdf$-absorbing ideal of $R_1\times R_2$ and $2u\in reg\left(R_2\right)$ for some $u\in R_2$, then $I_1$ is an r-$sdf$-absorbing ideal of $R_1$.

3.

If $I=I_1\times I_2$ is an r-$sdf$-absorbing ideal of $R_1\times R_2$ and $2u\in reg\left(R_1\right)$ for some $u\in R_1$, then $I_2$ is an r-$sdf$-absorbing ideal of $R_2$.

4.

$I_1\times R_2$ is an r-$sdf$-absorbing ideal of $R_1\times R_2$ if and only if $I_1$ is an r-$sdf$-absorbing ideal of $R_1$.

5.

$R_1\times I_2$ is an r-$sdf$-absorbing ideal of $R_1\times R_2$ if and only if $I_2$ is an r-$sdf$-absorbing ideal of $R_2$.

Proof. (1) Let $a=(a_1,a_2)$, $b=(b_1,b_2)\in R_1\times R_2$ such that

$$a^2-b^2=(a_1^2-b_1^2,a_2^2-b_2^2)\in I_1\times I_2.$$

Suppose that $a+b=(a_1+b_1,a_2+b_2)\notin Zd(R_1\times R_2)$. Then $a_1+b_1\notin Zd\left(R_1\right)$ and $a_2+b_2\notin Zd\left(R_2\right)$. Since $I_1$ and $I_2$ are r-$sdf$-absorbing, we obtain $a_1-b_1\in I_1$ and $a_2-b_2\in I_2$. Hence $a-b=(a_1-b_1,a_2-b_2)\in I_1\times I_2$. Therefore, $I_1\times I_2$ is r-$sdf$-absorbing.

(2) Assume $I=I_1\times I_2$ is r-$sdf$-absorbing and let $u\in R_2$ such that $2u$ is regular in $R_2$. We prove that $I_1$ is r-$sdf$-absorbing. Take $a_1,b_1\in R_1$ with $a_1^2-b_1^2\in I_1$. Set $a=(a_1,u)$ and $b=(b_1,u)$. Then

$$a^2-b^2=(a_1^2-b_1^2,u^2-u^2)=(a_1^2-b_1^2,0)\in I_1\times I_2=I.$$

Since I is r-$sdf$-absorbing, either $a+b=(a_1+b_1,2u)\in Zd(R_1\times R_2)$ or $a-b\in I$. If $a+b\in Zd(R_1\times R_2)$, then $2u$ being regular in $R_2$ implies $a_1+b_1\notin Zd\left(R_1\right)$. If $a-b\in I$, then $a_1-b_1\in I_1$. Hence, $I_1$ is r-$sdf$-absorbing.

(3) This is symmetric to (2), obtained by interchanging the roles of $R_1$ and $R_2$.

(4) Suppose first that $I_1\times R_2$ is r-$sdf$-absorbing. Let $a_1,b_1\in R_1$ with $a_1^2-b_1^2\in I_1$. Set $a=(a_1,0)$ and $b=(b_1,1).$ Then

$$a^2-b^2=(a_1^2-b_1^2,-1)\in I_1\times R_2.$$

If $a_1+b_1\notin Zd\left(R_1\right)$, then $a+b=(a_1+b_1,1)\notin Zd(R_1\times R_2)$. Hence, $a-b=(a_1-b_1,-1)\in I_1\times R_2$, forcing $a_1-b_1\in I_1$. Thus $I_1$ is r-$sdf$-absorbing.

Conversely, suppose $I_1$ is r-$sdf$-absorbing. Let $a=(a_1,a_2)$, $b=(b_1,b_2)\in R_1\times R_2$ such that

$$a^2-b^2=(a_1^2-b_1^2,a_2^2-b_2^2)\in I_1\times R_2.$$

Then $a_1^2-b_1^2\in I_1$. If $a+b=(a_1+b_1,a_2+b_2)\notin Zd(R_1\times R_2)$, then $a_1+b_1\notin Zd\left(R_1\right)$. Since $I_1$ is r-$sdf$-absorbing, we get $a_1-b_1\in I_1$. Hence $a-b=(a_1-b_1,a_2-b_2)\in I_1\times R_2$. Therefore $I_1\times R_2$ is r-$sdf$-absorbing.

(5) This is symmetric to (4), interchanging $R_1$ with $R_2$ and $I_1$ with $I_2$. □

The following example shows that the condition $2u\in reg\left(R_2\right)$ for some $u\in R_2$ is indispensable in the second statement of the above theorem.

Example 9.

Let $R_1=\mathbb{Z}$, $R_2={\mathbb{Z}}_2,$ and consider the ideals $I_1=3\mathbb{Z}$ and $I_2=\left\{0\right\}$ of $R_1$ and $R_2$ respectively. As we showed previously, the ideal $I=I_1\times I_2=3\mathbb{Z}\times \left\{0\right\}$ is an r-$sdf$-absorbing ideal of $R_1\times R_2$. However, the component $I_1=3\mathbb{Z}$ is not an r-$sdf$-absorbing ideal of $\mathbb{Z}$ by Proposition 1. We note that since $char\left({\mathbb{Z}}_2\right)=2$, we have $2u=0\notin reg\left({\mathbb{Z}}_2\right)$. This demonstrates that the condition $2u\in reg\left(R_2\right)$ for some $u\in R_2$ cannot be omitted.

The following result extends the direct product theorem to finitely many factors. Since the proof is obtained by a straightforward induction on the number of factors using the two-factor case, it is omitted.

Theorem 5.

Let $R_1,\dots ,R_n$ be rings and let $I_i$ be a proper ideal of $R_i$ for each $i=1,\dots ,n$. Put $R={\prod }_{i=1}^n{R}_i$ and $I={\prod }_{i=1}^n{I}_i.$ Then the following statements hold.

1.

If each $I_i$ is an r-sdf-absorbing ideal of $R_i$, then I is an r-sdf-absorbing ideal of R.

2.

If I is an r-sdf-absorbing ideal of R and, for a fixed index $j\in \{1,\dots ,n\}$, there exists an element $u_i\in R_i$ such that $2u_i\in reg\left(R_i\right)$ for every $i\ne j$, then $I_j$ is an r-sdf-absorbing ideal of $R_j$.

3.

For any index $j\in \{1,\dots ,n\}$, the ideal $I_j\times {\prod }_{i\ne j}{R}_i$ is an r-sdf-absorbing ideal of R if and only if $I_j$ is an r-sdf-absorbing ideal of $R_j$.

We recall that for a ring R and an R-module M, the idealization ring $R(+)M$ has underlying set $R\times M$ with multiplication

$$(r_1,m_1)(r_2,m_2)=(r_1{r}_2,r_1{m}_2+r_2{m}_1).$$

For an ideal I of R and a submodule N of M, the set $I(+)N=\left\{\right(i,n):i\in I,n\in N\}$ is an ideal of $R(+)M$ if and only if $IM\subseteq N$.

Theorem 6.

Let I be a proper ideal of a ring R and N be a submodule of an R-module M with $IM\subseteq N$. Then the following hold.

1.

I is an r-$sdf$-absorbing ideal of R if and only if $I(+)M$ is an r-$sdf$-absorbing ideal of $R(+)M$.

2.

If $I(+)N$ is an r-$sdf$-absorbing ideal of $R(+)M$, then I is an r-$sdf$-absorbing ideal of R.

Proof. 

We shall use the following standard fact about zero divisors in the idealization ring.

For any $(r,m)\in R(+)M$, we have

$$(r,m)\in Zd\left(R\right(+\left)M\right)⟺r\in Zd\left(R\right).$$

Indeed, if $r\in Zd\left(R\right)$, there exists $0\ne s\in R$ such that $rs=0$. Then $(s,0)\ne (0,0)$ and $(r,m)(s,0)=(rs,sm)=(0,0)$, so $(r,m)$ is a zero divisor. Conversely, if $(r,m)\in Zd\left(R\right(+\left)M\right)$, then there exists $(s,n)\ne (0,0)$ such that $(r,m)(s,n)=(rs,rn+sm)=(0,0)$. If $s\ne 0$, then $rs=0$ implies $r\in Zd\left(R\right)$. If $s=0$, then $n\ne 0$ and $rn=0$, again implies $r\in Zd\left(R\right)$. Hence the equivalence holds.

(1) Suppose first that I is r-$sdf$-absorbing in R. Let $A=(a_1,m_1)$ and $B=(a_2,m_2)$ be arbitrary elements of $R(+)M$ such that $(a_1^2-a_2^2,2a_1{m}_1-2a_2{m}_2)=A^2-B^2\in I(+)M$. Then $a_1^2-a_2^2\in I$ and since I is r-$sdf$-absorbing, either $a_1-a_2\in I$ or $a_1+a_2\in Zd\left(R\right)$. If $a_1-a_2\in I$, then $A-B=(a_1-a_2,m_1-m_2)\in I(+)M.$ If $a_1+a_2\in Zd\left(R\right)$, then by the above zero-divisor equivalence, $A+B=(a_1+a_2,m_1+m_2)\in Zd\left(R(+)M\right).$ Thus $I(+)M$ is r-$sdf$-absorbing in $R(+)M$.

Conversely, suppose $I(+)M$ is r-$sdf$-absorbing. Let $a,b\in R$ such that $a^2-b^2\in I$ and consider $A=(a,0),$$B=(b,0).$ Then $A^2-B^2=(a^2-b^2,0)\in I(+)M.$ Since $I(+)M$ is r-$sdf$-absorbing, either $A-B=(a-b,0)\in I(+)M$ or $A+B=(a+b,0)\in Zd\left(R\right(+\left)M\right).$ Thus, either $a-b\in I$ or $a+b\in Zd\left(R\right)$ and I is r-$sdf$-absorbing in R.

(2) Assume that $I(+)N$ is r-$sdf$-absorbing in $R(+)M$. Let $a,b\in R$ with $a^2-b^2\in I$. As before, set $A=(a,0),$$B=(b,0).$ Then $A^2-B^2=(a^2-b^2,0)\in I(+)N$ and so $A-B=(a-b,0)\in I(+)N$ or $A+B=(a+b,0)\in Zd\left(R\right(+\left)M\right)$. Therefore, $a-b\in I$ or $a+b\in Zd\left(R\right)$ and I is r-$sdf$-absorbing in R. □

Remark 3.

If I is an r-$sdf$-absorbing ideal of R and N is a proper submodule of M, then $I(+)N$ need not be r-$sdf$-absorbing in $R(+)M$. For example, take $R=\mathbb{Z}$, $M=\mathbb{Z}$, $I=2\mathbb{Z}$, and $N=2\mathbb{Z}$. Since $IM=2\mathbb{Z}\subseteq N=2\mathbb{Z}$, the ideal $I(+)N$ is well-defined. We claim that $2\mathbb{Z}(+)2\mathbb{Z}$ is not r-$sdf$-absorbing in $\mathbb{Z}(+)\mathbb{Z}$. Consider $A=(1,2)$ and $B=(1,-1).$ Then

$$A^2-B^2=(1,4)-(1,-2)=(0,6)\in 2\mathbb{Z}(+)2\mathbb{Z}$$

However, $A+B=(2,1)\notin Zd\left(\mathbb{Z}\right(+\left)\mathbb{Z}\right)$ since 2 is regular in $\mathbb{Z}$. Also, $A-B=(0,3)$$\notin 2\mathbb{Z}(+)2\mathbb{Z}$. Therefore, $I(+)N$ is not r-$sdf$-absorbing.

We recall the amalgamation construction. Let $f:R\to S$ be a ring homomorphism and let J be an ideal of S. The amalgamation of R and S along J with respect to f is the subring of $R\times S$ defined by

$$R{\bowtie }^{f}J=\{(a,f\left(a\right)+j):a\in R,j\in J\}.$$

If f is the identity on R, we obtain the amalgamated duplication $R\bowtie J$. For more related definitions and several properties of this kind of rings, one can see [14,15]. For an ideal I of R and an ideal K of $f\left(R\right)+J$, the sets

$$I{\bowtie }^{f}J=\{(i,f\left(i\right)+j):i\in I,j\in J\}$$

and

$${\widehat{K}}^f=\{(a,f\left(a\right)+j):a\in R,j\in J,f\left(a\right)+j\in K\}$$

are ideals of $R{\bowtie }^{f}J$,16].

A fundamental property of the amalgamation ring is the following characterization of its zero divisors:

$$(x,f\left(x\right)+j)\in Zd\left(R{\bowtie }^{f}J\right)⟺x\in Zd\left(R\right)\mathrm{or}f\left(x\right)+j\in Zd(f\left(R\right)+J).$$

The following theorem establishes the behavior of r-$sdf$-absorbing ideals under amalgamation.

Theorem 7.

Let $R,S,f,J,I$ and K be as above. Then the following hold.

1.

If I is an r-$sdf$-absorbing ideal of R, then $I{\bowtie }^{f}J$ is an r-$sdf$-absorbing ideal of $R{\bowtie }^{f}J$. The converse is true if $f^{-1}(Zd(f\left(R\right)+J))\subseteq Zd\left(R\right)$.

2.

If K is an r-$sdf$-absorbing ideal of $f\left(R\right)+J$, then ${\widehat{K}}^f$ is an r-$sdf$-absorbing ideal of $R{\bowtie }^{f}J$. The converse is true if $f(Zd(R\left)\right)+J\subseteq Zd\left(f\right(R)+J)$.

Proof. (1) Assume I is r-$sdf$-absorbing in R. Let $A=(a,f\left(a\right)+j_1)$, $B=(b,f\left(b\right)+j_2)$ be elements of $R{\bowtie }^{f}J$ such that $A^2-B^2\in I{\bowtie }^{f}J.$ Then $a^2-b^2\in I$ and so $a-b\in I$ or $a+b\in Zd\left(R\right)$ as I is r-$sdf$-absorbing in R. If $a-b\in I$, then

$$A-B=(a-b,f(a-b)+j_1-j_2)\in I{\bowtie }^{f}J.$$

If $a+b\in Zd\left(R\right)$, then by the zero-divisor characterization of $R{\bowtie }^{f}J$,

$$A+B=(a+b,f(a+b)+j_1+j_2)\in Zd\left(R{\bowtie }^{f}J\right).$$

Thus $I{\bowtie }^{f}J$ is r-$sdf$-absorbing in $R{\bowtie }^{f}J$.

Conversely, assume $I{\bowtie }^{f}J$ is r-$sdf$-absorbing and $f^{-1}(Zd(f\left(R\right)+J))\subseteq Zd\left(R\right)$. Let $a,b\in R$ such that $a^2-b^2\in I$ and consider $A=(a,f(a\left)\right),$$B=(b,f(b\left)\right).$ Then

$$A^2-B^2=(a^2-b^2,f{\left(a\right)}^2-f{\left(b\right)}^2)\in I{\bowtie }^{f}J.$$

Since $I{\bowtie }^{f}J$ is r-$sdf$-absorbing, either $A-B\in I{\bowtie }^{f}J$ (which gives $a-b\in I$) or $A+B\in Zd\left(R{\bowtie }^{f}J\right)$. In the second case, again by the zero-divisor characterization, $a+b\in Zd\left(R\right)$ or $f(a+b)\in Zd\left(f\right(R)+J).$ The extra hypothesis $f^{-1}(Zd(f\left(R\right)+J))\subseteq Zd\left(R\right)$ implies that both alternatives force $a+b\in Zd\left(R\right)$. Hence I is r-$sdf$-absorbing in R.

(2) Assume K is r-$sdf$-absorbing in $f\left(R\right)+J$. Let $A=(a,f\left(a\right)+j_1)$ and $B=(b,f\left(b\right)+j_2)$ be elements of $R{\bowtie }^{f}J$ such that $A^2-B^2\in {\widehat{K}}^f.$ Then ${(f\left(a\right)+j_1)}^2-{(f\left(b\right)+j_2)}^2\in K$. Since K is r-$sdf$-absorbing, either $(f\left(a\right)-f\left(b\right)+j_1-j_2)\in K$ or $(f\left(a\right)+f\left(b\right)+j_1+j_2)\in Zd(f\left(R\right)+J).$ In the first case, $A-B\in {\widehat{K}}^f$. In the second case, by the zero-divisor characterization,

$$A+B=(a+b,f(a+b)+j_1+j_2)\in Zd\left(R{\bowtie }^{f}J\right).$$

Thus, ${\widehat{K}}^f$ is r-$sdf$-absorbing.

Conversely, assume ${\widehat{K}}^f$ is r-$sdf$-absorbing and that $f(Zd(R\left)\right)+J\subseteq Zd\left(f\right(R)+J)$. Let $u=f\left(a\right)+j_1$ and $v=f\left(b\right)+j_2$ be elements of $f\left(R\right)+J$ such that $u^2-v^2\in K$. Consider $A=(a,u),$$B=(b,v)\in R{\bowtie }^{f}J.$ Then

$$A^2-B^2=(a^2-b^2,u^2-v^2)\in {\widehat{K}}^f.$$

Since ${\widehat{K}}^f$ is r-$sdf$-absorbing, either $A-B\in {\widehat{K}}^f$ (which gives $u-v\in K$) or $A+B\in Zd\left(R{\bowtie }^{f}J\right)$. In the second case, by the zero-divisor characterization, $a+b\in Zd\left(R\right)$ or $u+v\in Zd\left(f\right(R)+J).$ If $a+b\in Zd\left(R\right)$, then by the hypothesis $f(Zd(R\left)\right)+J\subseteq Zd\left(f\right(R)+J)$, we get $u+v=f(a+b)+j_1+j_2\in Zd(f\left(R\right)+J)$. Thus, in all cases, K is r-$sdf$-absorbing in $f\left(R\right)+J$. □

Corollary 2.

Let I, J, and K be ideals of a ring R. Then the following hold.

1.

$I\bowtie J$ is an r-$sdf$-absorbing ideal of $R\bowtie J$ if and only if I is an r-$sdf$-absorbing ideal of R.

2.

$\widehat{K}$ is an r-$sdf$-absorbing ideal of $R\bowtie J$ if and only if K is an r-$sdf$-absorbing ideal of R.

Proof. 

The proof is straightforward and follows immediately by applying the corresponding parts of the theorem with $S=R$ and $f={\mathrm{id}}_R$, noting that the zero-divisor conditions required in the general setting are automatically fulfilled in this special case. □

The following example show that the extra conditions in the converses of (1) in Theorem 7 cannot be omitted.

Example 10.

Let $R=\mathbb{Z}$, $S={\mathbb{Z}}_4$, and let $f:R\to S$ be the canonical epimorphism given by $f\left(n\right)=\overline{n}(mod4)$. Put $J=\overline{2}{\mathbb{Z}}_4=\{\overline{0},\overline{2}\}$. Then $f\left(R\right)+J={\mathbb{Z}}_4$, and the set of zero divisors of ${\mathbb{Z}}_4$ is precisely J. We observe that the condition $f^{-1}(Zd(f\left(R\right)+J))\subseteq Zd\left(R\right)$ is violated, since $f^{-1}\left(J\right)$ is the set of even integers, whereas $Zd\left(\mathbb{Z}\right)=\left\{0\right\}$. Now consider the ideal $I=4\mathbb{Z}$ of R. It is well known that I is not an r-$sdf$-absorbing ideal of $\mathbb{Z}$. We claim that $I{\bowtie }^{f}J$ is an r-$sdf$-absorbing ideal of the $R{\bowtie }^{f}J$. Let $A=(a,\overline{a}+j_1)$ and $B=(b,\overline{b}+j_2)$ be arbitrary elements such that $A^2-B^2\in I{\bowtie }^{f}J$. Then the first coordinate gives $a^2-b^2\in 4\mathbb{Z}$, so a and b have the same parity. Hence $a+b$ is even, which implies that $\overline{a+b}\in J$. Now $A+B=(a+b,\overline{a+b}+j_1+j_2)$. Since the second coordinate is in J, which is precisely the zero divisors of $f\left(R\right)+J$, it follows that $A+B\in Zd\left(R{\bowtie }^{f}J\right)$. Thus, $I{\bowtie }^{f}J$ is r-$sdf$-absorbing in $R{\bowtie }^{f}J$ while I is not r-$sdf$-absorbing in R.