Cdf-Absorbing δ(0)-Ideals of Commutative Rings

1. Introduction

Throughout the sequel, all rings are assumed to be commutative with identity $1\ne 0$. Let R denote such a ring, $\mathrm{char}\left(R\right)$ its characteristic, and $U\left(R\right)$ the set of its units. As usual, the additive monoid of nonnegative integers and the ring of integers will be denoted by $\mathbb{N}$ and $\mathbb{Z}$, respectively. For every $n\in \mathbb{N}$, the quotient of $\mathbb{Z}$ by its ideal $n\mathbb{Z}$ will be denoted by ${\mathbb{Z}}_n$. We set ${\mathbb{N}}^{*}:=\mathbb{N}\backslash \left\{0\right\}$. For any ideal I of R, its radical is $\sqrt{I}=\{x\in R:x^n\in I\mathrm{for}\mathrm{some}n\in {\mathbb{N}}^{*}\}$; in particular, $\sqrt{\left(0\right)}=\{x\in R:x^n=0\mathrm{for}\mathrm{some}n\in {\mathbb{N}}^{*}\}$ is the nilradical of R.

In 2001, Zhao [19] introduced a unified framework for prime and primary ideals by defining the notion of an expansion function of ideals of a ring R, which is a map $\delta :\mathcal{I}\left(R\right)\to \mathcal{I}\left(R\right)$, where $\mathcal{I}\left(R\right)$ stands for the set of all ideals of R, satisfying, for all ideals I, J and K of R, the conditions (i) $I\subseteq \delta \left(I\right)$ and (ii) whenever $J\subseteq K$, then $\delta \left(J\right)\subseteq \delta \left(K\right)$. Following [19], a proper ideal I of R is called $\delta$-primary if, for all elements $a,b\in R$ with $ab\in I$, one has $a\in I$ or $b\in \delta \left(I\right)$.

Various generalizations of prime ideals have been studied; see [2,4,5,14]. Recently, the concept of a square-difference factor absorbing ideal was introduced and studied in [1] as such a generalization. Following [1], a proper ideal I of R is square-difference factor absorbing (sdf-absorbing, for short) if, for all nonzero elements $a,b\in R$, the condition $a^2-b^2\in I$ implies that $a+b\in I$ or $a-b\in I$. Several characterizations of particular classes of rings were established in terms of such ideals. For instance, nonzero sdf-absorbing ideals are necessarily radical ideals, and the converse holds in characteristic 2. A generalization of this notion was given in [17], where a proper ideal of R was termed sdf-absorbing primary if, for all elements $a,b\in R$ such that $a^2-b^2\in I$, it follows that $a+b\in \sqrt{I}$ or $a-b\in I$. In [12], a generalization of both preceding notions was introduced under the term sdf-absorbing $\delta$-primary ideals.

In the cubic setting, the factorization $a^3-b^3=(a-b)(a^2+ab+b^2)$ was exploited in [15], where cdf-absorbing ideals were introduced: a proper ideal I of R is cdf-absorbing if, for all elements $a,b\in R$, $a^3-b^3\in I$ implies that $a^2+ab+b^2\in I$ or $a-b\in I$. Subsequently, Draoui [11] studied cdf-absorbing semiprimary ideals, where the condition becomes $a^2+ab+b^2\in \sqrt{I}$ or $a-b\in \sqrt{I}$.

In a different but related direction, the concept of $\delta \left(0\right)$-ideals was introduced and studied in [7], where a proper ideal I is called a $\delta \left(0\right)$-ideal of R if, for all elements $a,b\in R$ with $ab\in I$ and $a\notin \delta \left(0\right)$, then $b\in I$. This class extends the class of n-ideals of [18]. Very recently, Draoui [13] introduced the concept of sdf-absorbing $\delta \left(0\right)$-ideals: a proper ideal I of R is an sdf-absorbing $\delta \left(0\right)$-ideal if, for all nonzero $a,b\in R$, $a^2-b^2\in I$ implies that $a+b\in I$ or $a-b\in \delta \left(0\right)$. In particular, Draoui established the key equivalence that, when $a^2-b^2\in I$, the condition $a+b\in \delta \left(0\right)\iff a-b\in \delta \left(0\right)$ holds if and only if $2\in \delta \left(0\right)$ ([13] [Theorem 2.12]).

Motivated by the aforementioned classes of ideals, our objective in this paper is to introduce and study the concept of cube-difference factor absorbing $\delta \left(0\right)$-ideal (cdf-absorbing $\delta \left(0\right)$-ideal, for short). These are proper ideals I of R such that, for all $a,b\in R$, the condition $a^3-b^3\in I$ implies that $a-b\in I$ or $a^2+ab+b^2\in \delta \left(0\right)$. Our study parallels the sdf-absorbing $\delta \left(0\right)$-ideal framework of [13], transposing it from the quadratic factorization $a^2-b^2=(a+b)(a-b)$ to the cubic factorization $a^3-b^3=(a-b)(a^2+ab+b^2)$. The second factor $a^2+ab+b^2$ is the homogeneous form associated with the third cyclotomic polynomial ${\Phi }_3\left(x\right)=x^2+x+1$, which introduces fundamentally different algebraic and number-theoretic behavior compared with the sdf case.

The main contributions of this paper are as follows. In Section 2, we establish the basic properties of cdf-absorbing $\delta \left(0\right)$-ideals; we also show that, in contrast to the sdf case, a cdf-absorbing $\delta \left(0\right)$-ideal need not be a $\delta \left(0\right)$-ideal, even when $\delta \left(0\right)$ is prime (Example 5). We establish the cubic analogue of [13][Theorem 2.12]: if $3\in \delta \left(0\right)$, then $a-b\in \delta \left(0\right)$ implies $a^2+ab+b^2\in \delta \left(0\right)$, and conversely $a^2+ab+b^2\in \delta \left(0\right)$ implies $a-b\in \sqrt{\delta \left(0\right)}$ (Theorem 6), with full equivalence when $\delta \left(0\right)$ is radical. We characterize cdf-absorbing ${\delta }_0\left(0\right)$-ideals of $\mathbb{Z}$ among prime ideals (Theorem 11), connecting the theory to Eisenstein primes; we extend this to all ideals of $\mathbb{Z}$ via a squarefreeness condition (Corollary 14), and generalize further to arbitrary principal ideal domains (Theorem 13). We introduce weakly cdf-absorbing $\delta \left(0\right)$-ideals and establish annihilation results for strongly cdf-zero pairs (Theorem 18). In Section 3, we examine the transfer of this property across several ring-theoretic constructions. We investigate the transfer of the cdf-absorbing $\delta \left(0\right)$ property to polynomial rings, localizations, trivial ring extensions, product rings, and amalgamated algebras along an ideal.

We emphasize that, although our framework parallels the sdf-absorbing $\delta \left(0\right)$-ideal theory of [13], the passage from the quadratic to the cubic factorization is not a routine adaptation. Several phenomena are genuinely new: the absence of a difference-of-cubes decomposition for arbitrary products prevents a direct transposition of [13] [Theorem 2.10], and produces an essentially weaker condition (Example 5); the symmetric role of $a-b$ and $a^2+ab+b^2$ in the cdf condition gives rise to an asymmetric equivalence in Theorem 6, in contrast to the full equivalence available in the sdf setting (Remark 7); the number-theoretic content of Theorem 11 involves Eisenstein rather than Gaussian primes, with no overlap between the two characterizations; and the radical transfer property established in [13] [Theorem 2.17]for the sdf setting remains an open question in the cubic case.

2. Properties of Cdf-Absorbing $\mathit{\delta }\left(\mathbf{0}\right)$-Ideals

Recall from [19] that a map $\delta :\mathcal{I}\left(R\right)\to \mathcal{I}\left(R\right)$ is called an expansion function of ideals of a ring R if (i) $I\subseteq \delta \left(I\right)$ and (ii) $J\subseteq K\Rightarrow \delta \left(J\right)\subseteq \delta \left(K\right)$, for all ideals $I,J,K$ of R. Standard examples include the identity ${\delta }_0\left(I\right)=I$, the radical ${\delta }_1\left(I\right)=\sqrt{I}$, ${\delta }_M\left(I\right)=M$ for a quasi-local ring $(R,M)$, and ${\delta }_{+J}\left(I\right)=I+J$ or ${\delta }_{:J}\left(I\right)=(I:J)$ for a fixed ideal J; see also [7].

The main definition of this paper is as follows.

Definition 1.

Let R be a ring and $\delta$ an expansion function of its ideals. A proper ideal I of R is a cube-difference factor absorbing $\delta \left(0\right)$-ideal (cdf-absorbing $\delta \left(0\right)$-ideal, for short) if for all $a,b\in R$ such that $a^3-b^3\in I$, then $a-b\in I$ or $a^2+ab+b^2\in \delta \left(0\right)$.

Note that if $\delta \left(0\right)=R$, then every proper ideal is cdf-absorbing $\delta \left(0\right)$-ideal. Hence, throughout this paper we assume that $\delta \left(0\right)$ is proper. The following remark relates Definition 1 to several known classes of ideals.

Remark 2.

Let R be a ring, $\delta$ an expansion function of its ideals, and I a proper ideal of R.

1.

The sum-of-cubes factorization $a^3+b^3=(a+b)(a^2-ab+b^2)$ gives a symmetric variant of Definition 1: replacing b by $-b$, the ideal I is cdf-absorbing $\delta \left(0\right)$ if and only if $a^3+b^3\in I$ implies $a+b\in I$ or $a^2-ab+b^2\in \delta \left(0\right)$ for all $a,b\in R$.

2.

Every prime ideal of a ring R contained in $\delta \left(0\right)$ is a cdf-absorbing $\delta \left(0\right)$-ideal.

3.

Every $\delta \left(0\right)$-ideal I of a ring R in the sense of [7] is a cdf-absorbing $\delta \left(0\right)$-ideal. To see this, we first recall that every $\delta \left(0\right)$-ideal I satisfies the containment $I\subseteq \delta \left(0\right)$ by [7] [Proposition 2.3.7]

4.

A cdf-absorbing ideal $I\subseteq \delta \left(0\right)$ in the sense of [15] is a cdf-absorbing $\delta \left(0\right)$-ideal.

5.

In a Boolean ring, $a^3=a$ for every element, so $a^3-b^3=a-b$ and every proper ideal is trivially cdf-absorbing $\delta \left(0\right)$.

6.

If ${\delta }_1\left(0\right)\subseteq {\delta }_2\left(0\right)$, then every cdf-absorbing ${\delta }_1\left(0\right)$-ideal is also cdf-absorbing ${\delta }_2\left(0\right)$.

7.

The condition $a^2+ab+b^2\in \delta \left(0\right)$ is strictly more demanding than $a^2+ab+b^2\in \delta \left(I\right)$, since $\delta \left(0\right)\subseteq \delta \left(I\right)$ for every ideal I. Thus the cdf-absorbing $\delta \left(0\right)$ notion sits properly above the natural cubic analogue of the sdf-absorbing $\delta$-primary ideals of [12].

The examples below illustrate Definition 1 and indicate the scope of the concept.

Example 3.

1.

Let R be an integral domain. Then $\left(0\right)$ is a cdf-absorbing $\delta \left(0\right)$-ideal of R for any expansion function $\delta$.

2.

Let $R={\mathbb{Z}}_6$, $I=3{\mathbb{Z}}_6=\{0,3\}$ and $\delta ={\delta }_{+2{\mathbb{Z}}_6}$, so that $\delta \left(0\right)=2{\mathbb{Z}}_6=\{0,2,4\}$. Since $a^3\equiv a(mod6)$ for every $a\in {\mathbb{Z}}_6$, the condition $a^3-b^3\in I$ is equivalent to $a-b\in I$. An exhaustive check confirms that, besides the trivial pairs with $a=b$, the only pairs satisfying this condition are $(a,b)\in \left\{\right(0,3),$$(1,4),(2,5),(3,0),$$(4,1),(5,2)\}$; in each of them $a-b\in \{0,3\}=I$. Therefore I is a cdf-absorbing $\delta \left(0\right)$-ideal. Observe, however, that I is not a $\delta \left(0\right)$-ideal in the sense of [7]: $3·1\in I$ but $3\notin \delta \left(0\right)$, $1\notin I$. Note also that $I\neg \subseteq \delta \left(0\right)$.

3.

The next example is to illustrate that the converse of Remark 2 (2) does not hold. Let $R=\mathbb{Z}$, $I=9\mathbb{Z}$ and $\delta ={\delta }_{+3\mathbb{Z}}$. Then $\delta \left(0\right)=3\mathbb{Z}$ and $I\subseteq \delta \left(0\right)$. For $a,b\in \mathbb{Z}$ with $9\mid (a^3-b^3)$, we show $3\mid (a^2+ab+b^2)$. Using the factorization $a^3-b^3=(a-b)(a^2+ab+b^2)$, we distinguish two cases. If $3\nmid (a-b)$, then $gcd(9,a-b)=1$ (since $9=3^2$ and $3\nmid (a-b)$), so the divisibility $9\mid (a-b)(a^2+ab+b^2)$ forces $9\mid (a^2+ab+b^2)$, and in particular $3\mid (a^2+ab+b^2)$. If $3\mid (a-b)$, then $a^2+ab+b^2={(a-b)}^2+3ab\equiv 0(mod3)$. In either case $a^2+ab+b^2\in 3\mathbb{Z}=\delta \left(0\right)$, so I is a cdf-absorbing $\delta \left(0\right)$-ideal.

Under the additional assumptions that I is nonzero and $I\subseteq \delta \left(0\right)$, the elements in the definition can be sellected nonzero. The following proposition shows that we can check the same condition for just nonzero elements whenever I is itself nonzero and satisfies $I\subseteq \delta \left(0\right)$.

Proposition 4.

Let δ be an expansion function of ideals of a ring R, and let I be a nonzero ideal of R with $I\subseteq \delta \left(0\right)$. Then the following are equivalent:

1.

I is a cdf-absorbing $\delta \left(0\right)$-ideal of $R.$

2.

For all nonzero $a,b\in R$ with $a^3-b^3\in I$, one has $a-b\in I$ or $a^2+ab+b^2\in \delta \left(0\right)$.

Proof. $\left(1\right)\Rightarrow \left(2\right)$ Straightforward. $\left(2\right)\Rightarrow \left(1\right)$ Suppose that $a,b\in R$ satisfy $a^3-b^3\in I$, with at least one of a and b equal to zero. If $a=b=0$, then $a-b=0\in I$, and the conclusion is immediate. Suppose next that $a=0$ and $b\ne 0$, so that $-b^3=a^3-b^3\in I$ and hence $b^3\in I$. Since I is nonzero, we may choose some $0\ne i\in I$. As I is an ideal, $i^3=i·i^2\in I$, and therefore $b^3-i^3\in I$ with both b and i nonzero. Applying the cdf-absorbing $\delta \left(0\right)$ property to this pair yields either $b-i\in I$ or $b^2+bi+i^2\in \delta \left(0\right)$. In the former case, $b=(b-i)+i\in I\subseteq \delta \left(0\right)$, so $b^2\in \delta \left(0\right)$. In the latter case, the containment $i\in I\subseteq \delta \left(0\right)$ forces $bi,i^2\in \delta \left(0\right)$, from which $b^2\in \delta \left(0\right)$ as well. In either case, $a^2+ab+b^2=b^2\in \delta \left(0\right)$, as required. The remaining case $b=0$, $a\ne 0$ is handled symmetrically.    □

In the sdf setting, Draoui [13] [Theorem 2.10] proved that any nonzero sdf-absorbing $\delta \left(0\right)$-ideal I with $I\subseteq \delta \left(0\right)$ and $2\in U\left(R\right)$ is a $\delta \left(0\right)$-ideal. That argument depends critically on the algebraic identity that expresses any product as a difference of squares (via $a=(x+y)/2$ and $b=(x-y)/2$), an identity that admits no clean analogue in the cubic case. The example below confirms that the corresponding cubic statement fails in general, even when $\delta \left(0\right)$ is a prime ideal.

Example 5.

Let $R={\mathbb{Z}}_{12}$, $I=6{\mathbb{Z}}_{12}=\{0,6\}$ and $\delta ={\delta }_{+3{\mathbb{Z}}_{12}}$, so that $\delta \left(0\right)=3{\mathbb{Z}}_{12}=\{0,3,6,9\}$. Then $\delta \left(0\right)$ is a prime ideal (since ${\mathbb{Z}}_{12}/3{\mathbb{Z}}_{12}\cong {\mathbb{Z}}_3$) and $I\subseteq \delta \left(0\right)$. A direct computation yields the cubes modulo 12:

$$1^3\equiv 1,2^3\equiv 8,3^3\equiv 3,4^3\equiv 4,5^3\equiv 5,6^3\equiv 0,7^3\equiv 7,8^3\equiv 8,9^3\equiv 9,{10}^3\equiv 4,{11}^3\equiv 11.$$

We verify the cdf-absorbing property exhaustively. Let $a,b\in {\mathbb{Z}}_{12}$ such that $a^3-b^3\in I=\{0,6\}$. Two situations arise according to the value of $a^3-b^3$.

Case $a^3=b^3$. The pairs with $a=b$ satisfy $a-b=0\in I$ trivially. The remaining pairs with $a^3=b^3$ are exactly $(2,8)$, $(4,10)$, $(8,2)$, and $(10,4)$; in each of them $a-b\equiv 6(mod12)$, so $a-b\in I$.

Case $a^3-b^3\equiv 6(mod12)$. Inspection of the cube table shows that the pairs with this property are exactly

$$(a,b)\in \left\{\right(1,7),(3,9),(5,11),(7,1),(9,3),(11,5\left)\right\},$$

all of which satisfy $a-b\equiv 6(mod12)$, hence $a-b\in I$.

In both cases we obtain $a-b\in I$, which shows that I is a cdf-absorbing $\delta \left(0\right)$-ideal of R. However, I is not a $\delta \left(0\right)$-ideal in the sense of [7]: taking $x=2$ and $y=3$ gives $xy=6\in I$ with $x=2\notin \delta \left(0\right)$ and $y=3\notin I$.

Example 5 highlights a fundamental discrepancy between the sdf and cdf theories. The quadratic factorization $a^2-b^2=(a+b)(a-b)$ makes it possible to write any product $xy$ as a difference of squares (namely $xy=a^2-b^2$ with $a=(x+y)/2$ and $b=(x-y)/2$, provided that $2\in U\left(R\right)$). The cubic factorization $a^3-b^3=(a-b)(a^2+ab+b^2)$, in contrast, does not support an analogous decomposition. It is precisely this structural obstruction that prevents a direct transfer of [13] [Theorem 2.10]to the cdf setting.

The next theorem is the cubic counterpart of [13] [Theorem 2.12]. In the quadratic setting, the equivalence $a+b\in \delta \left(0\right)\iff a-b\in \delta \left(0\right)$ follows from the linear identity $a-b=(a+b)-2b$ whenever $2\in \delta \left(0\right)$. In the cubic setting, the analogous identity $a^2+ab+b^2={(a-b)}^2+3ab$ involves a square, and this fact introduces an inherent asymmetry between the two directions of the equivalence.

Theorem 6.

Let δ be an expansion function of ideals of a ring R and let I be a cdf-absorbing $\delta \left(0\right)$-ideal of R. Consider the following conditions.

1.

For all $a,b\in R$ with $a^3-b^3\in I$: $a-b\in \delta \left(0\right)$ implies $a^2+ab+b^2\in \delta \left(0\right)$.

2.

For all $a,b\in R$ with $a^3-b^3\in I$: $a^2+ab+b^2\in \delta \left(0\right)$ implies $a-b\in \sqrt{\delta \left(0\right)}$.

3.

$3\in \delta \left(0\right)$.

Then $\left(1\right)\iff \left(3\right)\Rightarrow \left(2\right)$. Moreover, if $\delta \left(0\right)$ is a radical ideal, then the assertions above are equivalent.

Proof. $\left(3\right)\Rightarrow \left(1\right)$: Assume $3\in \delta \left(0\right)$, so that $3ab\in \delta \left(0\right)$ for any $a,b\in R$. If $a-b\in \delta \left(0\right)$, then ${(a-b)}^2\in \delta \left(0\right)$, and consequently $a^2+ab+b^2={(a-b)}^2+3ab\in \delta \left(0\right)$. $\left(3\right)\Rightarrow \left(2\right)$: Suppose $a^2+ab+b^2\in \delta \left(0\right)$. Since $3ab\in \delta \left(0\right)$, we obtain ${(a-b)}^2=(a^2+ab+b^2)-3ab\in \delta \left(0\right)$, and therefore $a-b\in \sqrt{\delta \left(0\right)}$. $\left(1\right)\Rightarrow \left(3\right)$: Taking $a=b=1$, we have $a^3-b^3=0\in I$ and $a-b=0\in \delta \left(0\right)$. Applying (1) to this pair yields $3=a^2+ab+b^2\in \delta \left(0\right)$. Finally, if $\delta \left(0\right)=\sqrt{\delta \left(0\right)}$, then the conclusion of (2) strengthens to $a-b\in \delta \left(0\right)$, and the full equivalence follows.    □

Remark 7.

The implication $\left(2\right)\Rightarrow \left(3\right)$ does not hold in general. For instance, take $R=\mathbb{Z}$, $I=\left(0\right)$, and $\delta ={\delta }_0$ the identity expansion, so that $\delta \left(0\right)=\left(0\right)$ and $3\notin \delta \left(0\right)$. Then $I=\left(0\right)$ is a cdf-absorbing ${\delta }_0\left(0\right)$-ideal by Example 3. For $a,b\in \mathbb{Z}$ satisfying $a^3=b^3$, one necessarily has $a=b$ (as $\mathbb{Z}$ is a UFD with cube-uniqueness), so $a^2+ab+b^2=3a^2\ne 0$; the hypothesis of $\left(2\right)$ is therefore vacuous, and $\left(2\right)$ holds trivially while $\left(3\right)$ fails. In summary, the only nontrivial equivalence in Theorem 6 is $\left(1\right)\iff \left(3\right)$, and $\left(2\right)$ is a strictly weaker consequence of $\left(3\right)$.

When $\mathrm{char}\left(R\right)=3$, the cubic factorization degenerates to ${(a-b)}^3$, and this collapse establishes a close connection between the cdf-absorbing $\delta \left(0\right)$ property and the cube-radical condition discussed below.

Proposition 8.

Let δ be an expansion function of ideals of a ring R and I a proper ideal of R.

1.

If $\mathrm{char}\left(R\right)=3$ and for every $a\in R$ with $a^3\in I$, one has $a\in \delta \left(0\right)$, then I is a cdf-absorbing $\delta \left(0\right)$-ideal.

2.

If I is a cdf-absorbing $\delta \left(0\right)$-ideal, then for every $a\in R$ with $a^3\in I$, one has $a\in I$ or $a^2\in \delta \left(0\right)$.

Proof. (1) Let $a,b\in R$ with $a^3-b^3\in I$. Since $\mathrm{char}\left(R\right)=3$, the Frobenius-type identity ${(a-b)}^3=a^3-b^3$ holds, so ${(a-b)}^3\in I$; the hypothesis then yields $a-b\in \delta \left(0\right)$.

(2) Let $a\in R$ satisfy $a^3\in I$. Then $a^3-0^3\in I$ and the cdf-absorbing $\delta \left(0\right)$ property then gives either $a\in I$ or $a^2\in \delta \left(0\right)$.    □

We now turn to two structural results: the family of cdf-absorbing $\delta \left(0\right)$-ideals is closed under intersection, and -provided this family is nonempty- it contains maximal elements with respect to inclusion.

Proposition 9.

Let δ be an expansion function of ideals of a ring R and let ${\left\{I_{\alpha }\right\}}_{\alpha \in \Lambda }$ be a nonempty family of cdf-absorbing $\delta \left(0\right)$-ideals of R. Then $I={\bigcap }_{\alpha \in \Lambda }{I}_{\alpha }$, then I is a cdf-absorbing $\delta \left(0\right)$-ideal of R.

Proof. 

Let $a,b\in R$ such that $a^3-b^3\in I$ and $a-b\notin I$. Then, there exists some index $k\in \Lambda$ for which $a-b\notin I_k$. As $a^3-b^3\in I_k$ and $I_k$ is a cdf-absorbing $\delta \left(0\right)$-ideal, we conclude that $a^2+ab+b^2\in \delta \left(0\right)$, which completes the proof.    □

Theorem 10.

Let δ be an expansion function of ideals of a ring R. If R admits a cdf-absorbing $\delta \left(0\right)$-ideal, then R admits a maximal cdf-absorbing $\delta \left(0\right)$-ideal.

Proof. 

Let $\mathcal{F}$ denote the collection of all cdf-absorbing $\delta \left(0\right)$-ideals of R, partially ordered by inclusion. By hypothesis, $\mathcal{F}$ is nonempty. Let ${\left\{I_{\alpha }\right\}}_{\alpha \in \Lambda }$ be an arbitrary chain in $\mathcal{F}$, and set $U={\bigcup }_{\alpha \in \Lambda }{I}_{\alpha }$. The union U is a proper ideal of R as $1\notin I_{\alpha }$ for any $\alpha$. We claim that U belongs to $\mathcal{F}$. Suppose that $a,b\in R$ with $a^3-b^3\in U$ and $a-b\notin U$. There exists some index $k\in \Lambda$ with $a^3-b^3\in I_k$; in particular, $a-b\notin I_k$. Since $I_k$ is a cdf-absorbing $\delta \left(0\right)$-ideal, we conclude that $a^2+ab+b^2\in \delta \left(0\right)$, and so U is a cdf-absorbing $\delta \left(0\right)$-ideal. From Zorn’s Lemma, there exists a maximal element of $\mathcal{F}$.    □

We now establish an explicit characterization of the cdf-absorbing ${\delta }_0\left(0\right)$-ideals of $\mathbb{Z}$, which uncovers a number-theoretic connection with no counterpart in the semiprimary theory of [11].

Theorem 11.

Let ${\delta }_0$ be the identity expansion (so ${\delta }_0\left(0\right)=\left(0\right)$) and let p be a prime number. Then $\left(p\right)$ is a cdf-absorbing ${\delta }_0\left(0\right)$-ideal of $\mathbb{Z}$ if and only if $p=3$ or $p\equiv 2(mod3)$.

Proof. 

Suppose that $a,b\in R$ such that $a^3-b^3\in I$. Since ${\delta }_0\left(0\right)=\left(0\right)$, the identity $4(a^2+ab+b^2)={(2a+b)}^2+3b^2$ shows that $a^2+ab+b^2\ge 3b^2/4>0$ whenever $b\ne 0$, and similarly $a^2+ab+b^2>0$ when $a\ne 0$ (by symmetry). Hence the condition $a^2+ab+b^2\in {\delta }_0\left(0\right)$ is never satisfied if one of $a,b$ is nonzero, and $\left(p\right)$ is cdf-absorbing ${\delta }_0\left(0\right)$ if and only if the implication

$$p\mid (a^3-b^3)\Rightarrow p\mid (a-b)$$

holds for all $a,b\in \mathbb{Z}$. We first observe that the implication is trivially satisfied whenever $p\mid b$: indeed, $p\mid b$ yields $p\mid b^3$, and then $p\mid (a^3-b^3)$ forces $p\mid a^3$; since p is prime, we obtain $p\mid a$, and hence $p\mid (a-b)$. Hence, assume that $p\nmid b$, where we may set $\alpha =a·b^{-1}\in \mathbb{Z}/p\mathbb{Z}$. In this setting, the implication translates to the condition that every $\alpha \in \mathbb{Z}/p\mathbb{Z}$ satisfying ${\alpha }^3=1$ must satisfy $\alpha =1$, or equivalently, that the polynomial ${\Phi }_3\left(x\right)=x^2+x+1$ admits no root $\alpha \neg \equiv 1(modp)$ in $\mathbb{Z}/p\mathbb{Z}$. We now analyze the possible residue classes of p modulo 3. Case $p=2$: A direct evaluation yields ${\Phi }_3\left(0\right)\equiv 1$ and ${\Phi }_3\left(1\right)\equiv 3\equiv 1(mod2)$, so ${\Phi }_3$ has no roots in $\mathbb{Z}/2\mathbb{Z}$, and the implication holds vacuously. Since $2\equiv 2(mod3)$, this case is consistent with the residue condition in the statement. Case $p=3$: We have ${\Phi }_3\left(x\right)\equiv {(x-1)}^2(mod3)$, so the only root of ${\Phi }_3$ in $\mathbb{Z}/3\mathbb{Z}$ is $\alpha \equiv 1$. Therefore every $\alpha \in \mathbb{Z}/3\mathbb{Z}$ with ${\alpha }^3=1$ satisfies $\alpha =1$, and the implication holds. Case $p\equiv 2(mod3)$ and $p>2$: The discriminant of ${\Phi }_3$ is $1-4=-3$, so the existence of a root of ${\Phi }_3$ in $\mathbb{Z}/p\mathbb{Z}$ is equivalent to $-3$ being a quadratic residue modulo p. For an odd prime $p\ne 3$, quadratic reciprocity gives $\left({\displaystyle \frac{-3}{p}}\right)=1$ if $p\equiv 1(mod3)$ and $\left({\displaystyle \frac{-3}{p}}\right)=-1$ if $p\equiv 2(mod3)$. Hence, in the present case, $-3$ is not a quadratic residue modulo p, and ${\Phi }_3$ has no roots whatsoever in $\mathbb{Z}/p\mathbb{Z}$; the implication therefore holds vacuously. Case $p\equiv 1(mod3)$ (so $p\ne 3$): Since $3\mid (p-1)$, the multiplicative group ${(\mathbb{Z}/p\mathbb{Z})}^{\times }$ — which is cyclic of order $p-1$ — contains an element of order exactly 3, that is, an element $\alpha \in \mathbb{Z}/p\mathbb{Z}$ satisfying ${\alpha }^3=1$ and $\alpha \ne 1$. Such an $\alpha$ is a root of ${\Phi }_3$ in $\mathbb{Z}/p\mathbb{Z}$. Choosing $a=\alpha$ and $b=1$, we then have $p\mid ({\alpha }^3-1)$ but $p\nmid (\alpha -1)$, so the implication fails.    □

Remark 12.

The characterization in Theorem 11 admits the following number-theoretic interpretation: $\left(p\right)$ is a cdf-absorbing ${\delta }_0\left(0\right)$-ideal of $\mathbb{Z}$ precisely when p is inert or ramified in the ring of Eisenstein integers $\mathbb{Z}\left[\omega \right]$, where $\omega =e^{2\pi i/3}$. In the sdf theory, the analogous characterization is governed by the Gaussian primes — namely, those primes $p\equiv 3(mod4)$, which are exactly the primes inert in $\mathbb{Z}\left[i\right]$. By [11] [Example 2.3(4)], the ideal $p^n\mathbb{Z}$ is cdf-absorbing semiprimary for every prime p, regardless of its residue class modulo 3. The ${\delta }_0\left(0\right)$-property is therefore strictly more discriminating than the semiprimary property in this setting.

Theorem 11 characterizes cdf-absorbing ${\delta }_0\left(0\right)$-ideals of $\mathbb{Z}$ among prime ideals. The following proposition extends this characterization to irreducible ideals of an arbitrary principal ideal domain, and Corollary 14 then handles all ideals of $\mathbb{Z}$.

Theorem 13.

Let R be a principal ideal domain, ${\delta }_0$ the identity expansion, and $p\in R$ an irreducible element. Then $\left(p\right)$ is a cdf-absorbing ${\delta }_0\left(0\right)$-ideal of R if and only if $R/\left(p\right)$ satisfies $x^3=1\Rightarrow x=1$ for all $x\in R/\left(p\right).$

Proof. 

Since ${\delta }_0\left(0\right)=\left(0\right)$, the ideal $\left(p\right)$ is cdf-absorbing ${\delta }_0\left(0\right)$ if and only if for all $a,b\in R$ with $a^3-b^3\in \left(p\right)$, either $a-b\in \left(p\right)$ or $a^2+ab+b^2=0$ in R. $(\Leftarrow )$ Suppose $R/\left(p\right)$ satisfies $x^3=1\Rightarrow x=1$, and let $a,b\in R$ with $a^3-b^3\in \left(p\right)$. If $\overline{b}=0$ in $R/\left(p\right)$, then $p\mid b$, so $p\mid a^3$; since p is prime in the PID R, this gives $p\mid a$, hence $p\mid (a-b)$. If $\overline{b}\ne 0$, then $R/\left(p\right)$ is a field and ${(\overline{a}/\overline{b})}^3={\overline{a}}^3/{\overline{b}}^3=1$; by hypothesis $\overline{a}/\overline{b}=1$, so $\overline{a}=\overline{b}$, hence $p\mid (a-b)$. $(\Rightarrow )$ Suppose $\alpha \in R/\left(p\right)$ satisfies ${\alpha }^3=1$ and $\alpha \ne 1$. We show $\left(p\right)$ is not cdf-absorbing ${\delta }_0\left(0\right)$. Since $\alpha \ne 1$ and ${\alpha }^3=1$, we have ${\alpha }^2+\alpha +1=0$ as $R/\left(p\right)$ is a field; in particular $\mathrm{char}(R/(p\left)\right)\ne 3$ (as otherwise $x^2+x+1={(x-1)}^2$ has only the root 1). Let $a_0\in R$ be any lift of $\alpha$ and set $b=1$. We claim there exists a lift a of $\alpha$ with $a^2+a+1\ne 0$ in R. Since $a_0^2+a_0+1\equiv {\alpha }^2+\alpha +1=0(modp)$, we have $a_0^2+a_0+1\in \left(p\right)$; if $a_0^2+a_0+1\ne 0$ in R we are done. Otherwise set $a=a_0+p$; then

$$a^2+a+1={(a_0+p)}^2+(a_0+p)+1=(a_0^2+a_0+1)+p(2a_0+1+p)=p(2a_0+1+p).$$

Now $2a_0+1\equiv 2\alpha +1(modp)$, and $2\alpha +1=0$ would give $\alpha =-1/2$, so ${(-1/2)}^3=1$, i.e. $9=0$ in $R/\left(p\right)$; but the characteristic of a field is always prime, and $\mathrm{char}(R/(p\left)\right)\ne 3$ implies $\mathrm{char}(R/(p\left)\right)\nmid 9$, a contradiction. Hence $p\nmid (2a_0+1)$. To see that $2a_0+1+p\ne 0$ in R: if $2a_0+1+p=0$ then $p=-(2a_0+1)$, which gives $p\mid (2a_0+1)$, contradicting $p\nmid (2a_0+1)$. Therefore $a^2+a+1=p(2a_0+1+p)\ne 0$ in R. With this choice of a and $b=1$: $a^3-1\in \left(p\right)$ (since ${\overline{a}}^3={\alpha }^3=1$), $a-1\notin \left(p\right)$ (since $\overline{a}-1=\alpha -1\ne 0$), and $a^2+a+1\ne 0$ in R. This contradicts the cdf-absorbing property of $\left(p\right)$.    □

Corollary 14.

Let ${\delta }_0$ be the identity expansion and let $n\ge 2$ be an integer. Then $\left(n\right)$ is a cdf-absorbing ${\delta }_0\left(0\right)$-ideal of $\mathbb{Z}$ if and only if n is squarefree and every prime divisor p of n satisfies $p=3$ or $p\equiv 2(mod3)$.

Proof. 

Since ${\delta }_0\left(0\right)=\left(0\right)$ and $a^2+ab+b^2>0$ for all $a,b\in \mathbb{Z}$, the ideal $\left(n\right)$ is cdf-absorbing ${\delta }_0\left(0\right)$ if and only if $n\mid (a^3-b^3)\Rightarrow n\mid (a-b)$ for all $a,b\in \mathbb{Z}$. The squarefreeness of n is necessary: if $p^2\mid n$, write $n=p^{2}m$; taking $a\equiv p(mod{p}^2)$, $a\equiv 1(modm)$, $b\equiv 0(mod{p}^2)$, $b\equiv 1(modm)$ (when $m\ge 2$) or $a=p$, $b=p^2$ (when $m=1$) yields $n\mid (a^3-b^3)$ but $n\nmid (a-b)$. When n is squarefree, write $n=p_1\dots p_r$; by Theorem 13 and Theorem 11, each $p_i$ satisfies the cdf-absorbing condition if and only if $\mathbb{Z}/p_i\mathbb{Z}$ has no element of order 3, which holds if and only if $p_i=3$ or $p_i\equiv 2(mod3)$. The conclusion follows by coprimality of the $p_i$.    □

Having established the basic theory and the characterizations in $\mathbb{Z}$ and in arbitrary PIDs, we now turn to a natural weakening of the cdf-absorbing notion, which allows the difference $a^3-b^3$ to be nonzero but excluded from the hypothesis.

Definition 15.

Let $\delta$ be an expansion function of ideals of a ring R. A proper ideal I of R is a weakly cdf-absorbing $\delta \left(0\right)$-ideal if, for all $a,b\in R$ with $0\ne a^3-b^3\in I$, then $a-b\in I$ or $a^2+ab+b^2\in \delta \left(0\right)$.

The zero ideal is always weakly cdf-absorbing $\delta \left(0\right)$-ideal, since the condition $0\ne a^3-b^3\in \left(0\right)$ cannot be satisfied. Moreover, every cdf-absorbing $\delta \left(0\right)$-ideal is weakly cdf-absorbing $\delta \left(0\right)$. The converse fails in general, as illustrated in following:

Example 16.

Let $R={\mathbb{Z}}_{35}$, $I=\left(0\right)$ and $\delta ={\delta }_0$. Then $\delta \left(0\right)=\left(0\right)$, and I is not a cdf-absorbing ${\delta }_0\left(0\right)$-ideal: for $a=16$, $b=1$, one has $a^3-b^3\equiv 0(mod35)$ but $a-b=15\ne 0$ and $a^2+ab+b^2=273\equiv 28\ne 0(mod35)$. (Compare [11] [Example 3.2].)

Modelled on [11] [Definitions 3.3 and 3.7], the following definitions identify the obstruction that prevents the weakly version from coinciding with the strong one.

Definition 17.

Let $\delta$ be an expansion function of ideals of a ring R and let I be a weakly cdf-absorbing $\delta \left(0\right)$-ideal of R.

1.

A pair $(a,b)\in R\times R$ is a cdf-zero pair of I if $a^3-b^3=0$, $a-b\notin I$ and $a^2+ab+b^2\notin \delta \left(0\right)$.

2.

A cdf-zero pair $(a,b)$ of I (in the sense of Definition 17) is strongly cdf-zero if additionally $a^2-b^2=0$.

For strongly cdf-zero pairs, we now establish an annihilation property analogous to [11] [Lemma 3.9].

Theorem 18.

Let δ be an expansion function of ideals of a ring R with $3\in U\left(R\right)$, let I be a weakly cdf-absorbing $\delta \left(0\right)$-ideal with $I\subseteq \delta \left(0\right)$, and let $(x,y)$ be a strongly cdf-zero pair of I. Then $(x-y)i^2=0$ for every $i\in I$.

Proof. 

Assume, for contradiction, that $(x-y)i^2\ne 0$ for some $i\in I$. Since 3 is a unit in R, we also have $3(x-y)i^2\ne 0$. Expanding the cube and using the assumptions $x^3=y^3$ and $x^2=y^2$, we compute

$${(x+i)}^3-{(y+i)}^3=(x^3-y^3)+3i(x^2-y^2)+3i^2(x-y)=3i^2(x-y).$$

Since $0\ne 3i^2(x-y)={(x+i)}^3-{(y+i)}^3\in I$, so the weakly cdf-absorbing $\delta \left(0\right)$ property applies and yields either $(x+i)-(y+i)=x-y\in I$, or ${(x+i)}^2+(x+i)(y+i)+{(y+i)}^2=(x^2+xy+y^2)+3(x+y)i+3i^2\in \delta \left(0\right)$.

Since $i\in I\subseteq \delta \left(0\right)$, every term involving i lies in $\delta \left(0\right)$, and $x^2+xy+y^2\in \delta \left(0\right)$. Both alternatives — namely $x-y\in I$ and $x^2+xy+y^2\in \delta \left(0\right)$ which contradict the assumption that $(x,y)$ is a cdf-zero pair of I. Thus, $(x-y)i^2=0$.    □

3. Cdf-Absorbing $\mathit{\delta }\left(\mathbf{0}\right)$-Ideals in Ring-Theoretic Constructions

In this section, we examine how the cdf-absorbing $\delta \left(0\right)$ property behaves under several fundamental ring-theoretic constructions. We begin with polynomial ring extensions. For an ideal I of a ring R, denote by $(I,X)=I·R\left[X\right]+XR\left[X\right]$ of $R\left[X\right]$ generated by I and the indeterminate X. For every ideal K of $R\left[X\right]$, we set $J=\left\{P\right(0):P(X)\in K\}$ and define $\tilde{\delta }\left(K\right):=\delta \left(J\right)+XR\left[X\right]$. As shown in [13] [Lemma 3.1], $\tilde{\delta }$ is an expansion function of ideals of $R\left[X\right]$, and it satisfies $\tilde{\delta }\left((I,X)\right)=(\delta \left(I\right),X)$.

The following theorem characterizes the cdf-absorbing $\tilde{\delta }\left(0\right)$-ideals of $R\left[X\right]$ of the form $(I,X)$.

Theorem 19.

Let δ be an expansion function of ideals of a ring R and let I be a proper ideal of R. Then I is a cdf-absorbing $\delta \left(0\right)$-ideal of R if and only if $(I,X)$ is a cdf-absorbing $\tilde{\delta }\left(0\right)$-ideal of $R\left[X\right]$. In particular, $\left(0\right)$ is a cdf-absorbing $\delta \left(0\right)$-ideal in R if and only if $\left(X\right)$ is a cdf-absorbing $\tilde{\delta }\left(0\right)$-ideal of $R\left[X\right]$.

Proof. $(\Rightarrow )$: Let $f,g\in R\left[X\right]$ such that $f^3-g^3\in (I,X)$, and write $f=a+XM$ and $g=b+XN$ with $a,b\in R$ and $M,N\in R\left[X\right]$. Evaluating at $X=0$ yields $a^3-b^3\in I$. Hence, we conclude either $a-b\in I$ or $a^2+ab+b^2\in \delta \left(0\right)$. In the former case, $f-g=(a-b)+X(M-N)\in (I,X)$; in the latter, $f^2+fg+g^2=(a^2+ab+b^2)+X(\dots )\in (\delta \left(0\right),X)=\tilde{\delta }\left(0\right)$, where the omitted term lies in $XR\left[X\right]$.

$(\Leftarrow )$: Let $a,b\in R$ with $a^3-b^3\in I\subseteq (I,X)$. By hypothesis, either $a-b\in (I,X)$ or $a^2+ab+b^2\in \tilde{\delta }\left(0\right)$. Intersecting with R in each case, we conclude that $a-b\in (I,X)\cap R=I$ or $a^2+ab+b^2\in \tilde{\delta }\left(0\right)\cap R=\delta \left(0\right)$, as required.    □

Example 20.

Take $R=\mathbb{Z}$, $\delta ={\delta }_{+2\mathbb{Z}}$ (so that $\delta \left(0\right)=2\mathbb{Z}$), and $I=2\mathbb{Z}$. Then, I is a cdf-absorbing $\delta \left(0\right)$-ideal of $\mathbb{Z}$ by Example 3(2). Theorem 19 therefore gives that $(2\mathbb{Z},X)=(2,X)$ is a cdf-absorbing $\tilde{\delta }\left(0\right)$-ideal of $\mathbb{Z}\left[X\right]$, where $\tilde{\delta }\left(0\right)=(\delta \left(0\right),X)=(2,X)$.

To see both directions directly: if $f,g\in \mathbb{Z}\left[X\right]$ with $f^3-g^3\in (2,X)$, then evaluating at $X=0$ gives $f{\left(0\right)}^3-g{\left(0\right)}^3\in 2\mathbb{Z}$. Since $a^3\equiv a(mod2)$ for all $a\in \mathbb{Z}$ (the cube of any integer has the same parity as the integer itself), we get $f\left(0\right)-g\left(0\right)\in 2\mathbb{Z}$, and hence $f-g=\left(f\right(0)-g(0\left)\right)+X(\dots )\in (2,X)$, where the remaining terms are divisible by X. Conversely, if $(2,X)$ is a cdf-absorbing $\tilde{\delta }\left(0\right)$-ideal, then restricting to constant polynomials $f=a$ and $g=b$ immediately recovers the cdf-absorbing $\delta \left(0\right)$ property of $I=2\mathbb{Z}$ in $\mathbb{Z}$.

Our next result demonstrates that the cdf-absorbing $\delta \left(0\right)$ property is preserved under localization with respect to a multiplicatively closed set of non-zero-divisors.

Theorem 21.

Let S be a multiplicatively closed subset of a ring R and δ be an expansion function of ideals of R. Suppose that for every ideal I disjoint from S, $\delta \left(I\right)\cap S=\varnothing$. Define ${\delta }_S\left(S^{-1}J\right)=S^{-1}\delta \left(J\right)$. Let I be an ideal of R with $I\cap S=\varnothing$.

1.

If I is a cdf-absorbing $\delta \left(0\right)$-ideal of R, then $S^{-1}I$ is a cdf-absorbing ${\delta }_S\left(0\right)$-ideal of $S^{-1}R$.

2.

If additionally $N_I\cap S=N_{\delta \left(0\right)}\cap S=\varnothing$ (where $N_J=\{x\in R:xr\in J\mathrm{for}\mathrm{some}r\in R\backslash J\}$), the converse also holds.

Proof. (1) Let ${\displaystyle \frac{a}{s}},{\displaystyle \frac{b}{t}}\in S^{-1}R$ with ${\left({\displaystyle \frac{a}{s}}\right)}^3-{\left({\displaystyle \frac{b}{t}}\right)}^3\in S^{-1}I$. Then, there exists $u\in S$ satisfying $u(t^3{a}^3-s^3{b}^3)\in I$. Setting $c=uat$ and $d=ubs$, we compute $c^3-d^3=u^3(t^3{a}^3-s^3{b}^3)\in I$, The cdf-absorbing $\delta \left(0\right)$ property of I now yields either $c-d\in I$ or $c^2+cd+d^2\in \delta \left(0\right)$. In the first case,

$${\displaystyle \frac{a}{s}}-{\displaystyle \frac{b}{t}}={\displaystyle \frac{c-d}{ust}}\in S^{-1}I;$$

in the second,

$${\left({\displaystyle \frac{a}{s}}\right)}^2+{\displaystyle \frac{ab}{st}}+{\left({\displaystyle \frac{b}{t}}\right)}^2={\displaystyle \frac{c^2+cd+d^2}{{\left(ust\right)}^2}}\in S^{-1}\delta \left(0\right)={\delta }_S\left(0\right).$$

(2) Let $a,b\in R$ with $a^3-b^3\in I$. Applying the cdf-absorbing ${\delta }_S\left(0\right)$ property of $S^{-1}I$ to ${\displaystyle \frac{a^3-b^3}{1}}\in S^{-1}I$, we find elements $r,s\in S$ such that $r(a-b)\in I$ or $s(a^2+ab+b^2)\in \delta \left(0\right)$. In the first case, the assumption $N_I\cap S=\varnothing$ gives $r\notin N_I$, which together with $r(a-b)\in I$ forces $a-b\in I$. Similarly, in the second case, the assumption $N_{\delta \left(0\right)}\cap S=\varnothing$ forces $a^2+ab+b^2\in \delta \left(0\right)$.    □

Example 22.

Let $R=\mathbb{Z}$, $S=\mathbb{Z}\backslash 2\mathbb{Z}$ (the set of odd integers), so that $S^{-1}R={\mathbb{Z}}_{\left(2\right)}$ is the localization of $\mathbb{Z}$ at the prime 2. Take $\delta ={\delta }_{+2\mathbb{Z}}$ (so that $\delta \left(0\right)=2\mathbb{Z}$) and $I=2\mathbb{Z}$. Note that $I\cap S=\varnothing$. Since $I=2\mathbb{Z}$ is a cdf-absorbing $\delta \left(0\right)$-ideal of $\mathbb{Z}$ by Example 3(2), Theorem 21(1) gives that $S^{-1}I=2{\mathbb{Z}}_{\left(2\right)}$ is a cdf-absorbing ${\delta }_S\left(0\right)$-ideal of ${\mathbb{Z}}_{\left(2\right)}$, where ${\delta }_S\left(0\right)=S^{-1}\delta \left(0\right)=2{\mathbb{Z}}_{\left(2\right)}$. Note that $2{\mathbb{Z}}_{\left(2\right)}$ is the unique maximal ideal of the local ring ${\mathbb{Z}}_{\left(2\right)}$. For the converse, $N_I=\{x\in \mathbb{Z}:xr\in 2\mathbb{Z}\mathrm{for}\mathrm{some}r\in \mathbb{Z}\backslash 2\mathbb{Z}\}=2\mathbb{Z}$, so $N_I\cap S=2\mathbb{Z}\cap (\mathbb{Z}\backslash 2\mathbb{Z})=N_{\delta \left(0\right)}\cap S=\varnothing$. The conditions of Theorem 21(2) are therefore satisfied, and the cdf-absorbing ${\delta }_S\left(0\right)$ property of $2{\mathbb{Z}}_{\left(2\right)}$ implies the cdf-absorbing $\delta \left(0\right)$ property of $2\mathbb{Z}$ in $\mathbb{Z}$, as confirmed by Example 3(2).

The next theorem describes the behavior of the cdf-absorbing $\delta \left(0\right)$ property under ring homomorphisms, in both the pullback and pushforward directions.

Theorem 23.

Let $f:R\to T$ be a unital ring homomorphism, let $\delta ,\gamma$ be expansion functions of ideals of a ring R and T respectively, and let ${\delta }^{\prime },{\gamma }^{\prime }$ be expansion functions of ideals of T and R respectively, with $f\left(\delta \left(0\right)\right)\subseteq {\delta }^{\prime }\left(0\right)$ and $f^{-1}\left(\gamma \left(0\right)\right)\subseteq {\gamma }^{\prime }\left(0\right)$.

1.

If J is a cdf-absorbing $\gamma \left(0\right)$-ideal of T, then $f^{-1}\left(J\right)$ is a cdf-absorbing ${\gamma }^{\prime }\left(0\right)$-ideal of R.

2.

If f is surjective, $ker\left(f\right)\subseteq I$, and I is a cdf-absorbing $\delta \left(0\right)$-ideal of R, then $f\left(I\right)$ is a cdf-absorbing ${\delta }^{\prime }\left(0\right)$-ideal of T.

Proof. (1) Let $a,b\in R$ such that $a^3-b^3\in f^{-1}\left(J\right)$. Then, $f{\left(a\right)}^3-f{\left(b\right)}^3=f(a^3-b^3)\in J$. Hence, we obtain either $f\left(a\right)-f\left(b\right)\in J$ or $f{\left(a\right)}^2+f\left(a\right)f\left(b\right)+f{\left(b\right)}^2\in \gamma \left(0\right)$. In the first case, $a-b\in f^{-1}\left(J\right)$; in the second, $a^2+ab+b^2\in f^{-1}\left(\gamma \left(0\right)\right)\subseteq {\gamma }^{\prime }\left(0\right)$. (2) The assumptions $ker\left(f\right)\subseteq I$ and the properness of I imply that $f\left(I\right)$ is a proper ideal of T. Let $x,y\in T$ such that $x^3-y^3\in f\left(I\right)$, and pick $a,b\in R$ with $f\left(a\right)=x$ and $f\left(b\right)=y$, which exist by the surjectivity of f. Then $f(a^3-b^3)\in f\left(I\right)$, and so $a^3-b^3=i+k$ for some $i\in I$ and $k\in ker\left(f\right)$. The containment $ker\left(f\right)\subseteq I$ places k in I, whence $a^3-b^3\in I$. Applying the cdf-absorbing $\delta \left(0\right)$ property of I yields $a-b\in I$ or $a^2+ab+b^2\in \delta \left(0\right)$, which in turn translate to $x-y\in f\left(I\right)$ or $x^2+xy+y^2\in f\left(\delta \left(0\right)\right)\subseteq {\delta }^{\prime }\left(0\right)$, respectively.    □

Specializing the preceding theorem to canonical projections yields the following version for quotient rings.

Corollary 24.

Let δ be an expansion function of ideals of a ring R and let $J\subseteq I$ be proper ideals of R.

1.

If I is a cdf-absorbing $\delta \left(0\right)$-ideal of R containing J, then $I/J$ is a cdf-absorbing ${\delta }_q\left(0\right)$-ideal of $R/J$, where ${\delta }_q(K/J)=\delta \left(K\right)/J$.

2.

If $J⊊I\subseteq \delta \left(0\right)$, ${\delta }_q\left(0\right)=\delta \left(0\right)/J$, and $I/J$ is a cdf-absorbing ${\delta }_q\left(0\right)$-ideal of $R/J$, then I is a cdf-absorbing $\delta \left(0\right)$-ideal of R.

Proof. 

Consider the canonical projection $\pi :R\to R/J$, whose kernel is $ker\left(\pi \right)=J\subseteq I$. (1) The assertion follows from Theorem 23 (2) applied to $\pi$. (2) Since $I/J\subseteq \delta \left(0\right)/J={\delta }_q\left(0\right)$, we may apply Theorem 23 (1b) to $\pi$, which yields the desired conclusion.    □

Example 25.

Take $R=\mathbb{Z}$, $J=6\mathbb{Z}$, $I=2\mathbb{Z}$, and $\delta ={\delta }_{+2\mathbb{Z}}$ (so that $\delta \left(0\right)=2\mathbb{Z}$). Then I is a cdf-absorbing $\delta \left(0\right)$-ideal of $\mathbb{Z}$ by Example 3(2). The quotient ring is $R/J={\mathbb{Z}}_6$, and the induced expansion function satisfies ${\delta }_q\left(0\right)=\delta \left(0\right)/J=2\mathbb{Z}/6\mathbb{Z}=\{0,2,4\}$ in ${\mathbb{Z}}_6$.

Corollary 24(1) gives that $I/J=2\mathbb{Z}/6\mathbb{Z}=\{0,2,4\}$ is a cdf-absorbing ${\delta }_q\left(0\right)$-ideal of ${\mathbb{Z}}_6$. One can check this directly: since $a^3\equiv a(mod6)$ for every $a\in {\mathbb{Z}}_6$ (as $a^3-a=a(a-1)(a+1)$ is always divisible by 6), the condition $a^3-b^3\in \{0,2,4\}$ reduces to $a-b\in \{0,2,4\}$, which is precisely $a-b\in I/J$. For the converse part, note that $J⊊I$ and $I/J\subseteq {\delta }_q\left(0\right)$, so Corollary 24(2) applies and recovers the cdf-absorbing $\delta \left(0\right)$ property of $I=2\mathbb{Z}$ in $\mathbb{Z}$, consistent with Example 3(2).

We now describe the behavior of cdf-absorbing $\delta \left(0\right)$-ideals under finite direct products of rings.

Theorem 26.

Let ${\delta }_i$ be an expansion function of ideals of a ring $R_i$ and let $I_i$ be a proper ideal of $R_i$ for $i=1,2$. Set $R=R_1\times R_2$, $I=I_1\times I_2$, and define the product expansion function $\delta ={\delta }_1\times {\delta }_2$ on ideals of R by $\delta (K_1\times K_2)={\delta }_1\left(K_1\right)\times {\delta }_2\left(K_2\right)$, so that $\delta \left(0\right)={\delta }_1\left(0\right)\times {\delta }_2\left(0\right)$.

1.

If I is a cdf-absorbing $\delta \left(0\right)$-ideal of R, then $I_1$ and $I_2$ are cdf-absorbing ${\delta }_1\left(0\right)$- and ${\delta }_2\left(0\right)$-ideals of $R_1$ and $R_2$, respectively.

2.

If $3\in {\delta }_i\left(0\right)$, and $I_i$ is a cdf-absorbing ${\delta }_i\left(0\right)$-ideal of $R_i$ for $i=1,2$, then I is a cdf-absorbing $\delta \left(0\right)$-ideal of R.

Proof. (1) Let $a_1,b_1\in R_1$ with $a_1^3-b_1^3\in I_1$. Setting $a=(a_1,0)$ and $b=(b_1,0)$, we have $a^3-b^3=(a_1^3-b_1^3,0)\in I$. The cdf-absorbing $\delta \left(0\right)$ property of I then yields either $(a_1-b_1,0)\in I$ or $(a_1^2+a_1{b}_1+b_1^2,0)\in \delta \left(0\right)$, which project onto $a_1-b_1\in I_1$ or $a_1^2+a_1{b}_1+b_1^2\in {\delta }_1\left(0\right)$, respectively. Hence $I_1$ is a cdf-absorbing ${\delta }_1\left(0\right)$-ideal, and a symmetric argument applies to $I_2$. (2) Let $(a_1,a_2),(b_1,b_2)\in R$ with $(a_1^3-b_1^3,a_2^3-b_2^3)\in I$. Since each $I_i$ is a cdf- ${\delta }_i\left(0\right)$-ideal, we have for $i=1,2$:

$$a_i-b_i\in I_i\mathrm{or}{a}_i^2+a_i{b}_i+b_i^2\in {\delta }_i\left(0\right).$$

We proceed by cases according to which alternative holds in each factor. If $a_i-b_i\in I_i$ for both i, then $(a_1-b_1,a_2-b_2)\in I$, as required. Suppose now that $a_1-b_1\in I_1$ while $a_2^2+a_2{b}_2+b_2^2\in {\delta }_2\left(0\right)$; we show that $a_1^2+a_1{b}_1+b_1^2\in {\delta }_1\left(0\right)$, which then yields $(a_1^2+a_1{b}_1+b_1^2,a_2^2+a_2{b}_2+b_2^2)\in \delta \left(0\right)$. Then, $a_1-b_1\in I_1\subseteq {\delta }_1\left(0\right)$ combined with $3\in {\delta }_1\left(0\right)$ allows us to invoke Theorem 6 (1), which produces $a_1^2+a_1{b}_1+b_1^2\in {\delta }_1\left(0\right)$. The symmetric situation $a_2-b_2\in I_2$ and $a_1^2+a_1{b}_1+b_1^2\in {\delta }_1\left(0\right)$ is handled identically. Finally, if $a_i^2+a_i{b}_i+b_i^2\in {\delta }_i\left(0\right)$ for both i, then $(a_1^2+a_1{b}_1+b_1^2,a_2^2+a_2{b}_2+b_2^2)\in \delta \left(0\right)$ directly.    □

Remark 27.

The hypothesis $3\in {\delta }_i\left(0\right)$ in Theorem 26 (2) cannot be omitted. To see this, take $R_1=R_2=\mathbb{Z}$, $I_1=I_2=4\mathbb{Z}$, and ${\delta }_1={\delta }_2={\delta }_{+4\mathbb{Z}}$, so that ${\delta }_i\left(0\right)=4\mathbb{Z}=I_i$, and $3\notin {\delta }_i\left(0\right)$. A direct verification shows that $4\mathbb{Z}$ is a cdf-absorbing ${\delta }_{+4\mathbb{Z}}\left(0\right)$-ideal of $\mathbb{Z}$. However, taking $(a_1,a_2)=(2,1)$ and $(b_1,b_2)=(0,1)$ in $R=\mathbb{Z}\times \mathbb{Z}$, one computes $(a_1^3-b_1^3,a_2^3-b_2^3)=(8,0)\in I$, while

$$a_1-b_1=2\notin 4\mathbb{Z},a_2^2+a_2{b}_2+b_2^2=3\notin 4\mathbb{Z}.$$

Hence $(a_1-b_1,a_2-b_2)\notin I$ and $(a_1^2+a_1{b}_1+b_1^2,a_2^2+a_2{b}_2+b_2^2)\notin \delta \left(0\right)$, so $I=4\mathbb{Z}\times 4\mathbb{Z}$ fails to be a cdf-absorbing $\delta \left(0\right)$-ideal of R.

We now turn to trivial ring extensions. Let A be a ring, E an A-module, and $\delta$ an expansion function of ideals of A. Recall that the trivial ring extension $A⋉E$ is the ring whose underlying set is $A\times E$, equipped with the multiplication $(a,e)(b,f)=(ab,af+be)$; we refer the reader to [3,16] for background material. Following [7] [Corollary 2.39], we associate to $\delta$ the expansion function ${\delta }_{\propto }$ on ideals of $A⋉E$ defined by ${\delta }_{\propto }(I⋉N):=\delta \left(I\right)⋉E$ for every ideal $I⋉N$; in particular, ${\delta }_{\propto }\left(0\right)=\delta \left(0\right)⋉E$.

Theorem 28.

Let A be a ring, E an A-module, δ an expansion function of ideals of A, I a proper ideal of A, and N a submodule of E with $IE\subseteq N$.

1.

If $I⋉N$ is a cdf-absorbing ${\delta }_{\propto }\left(0\right)$-ideal of $A⋉E$, then I is a cdf-absorbing $\delta \left(0\right)$-ideal of A.

2.

If I is a cdf-absorbing $\delta \left(0\right)$-ideal of A, then $I⋉E$ is a cdf-absorbing ${\delta }_{\propto }\left(0\right)$-ideal of $A⋉E$.

Proof. (1) Let $a,b\in A$ with $a^3-b^3\in I$. Then ${(a,0)}^3-{(b,0)}^3=(a^3-b^3,0)\in I⋉N$. The cdf-absorbing ${\delta }_{\propto }\left(0\right)$ property therefore yields either $(a-b,0)\in I⋉N$ (giving $a-b\in I$) or $(a^2+ab+b^2,0)\in {\delta }_{\propto }\left(0\right)=\delta \left(0\right)⋉E$ (giving $a^2+ab+b^2\in \delta \left(0\right)$). (2) Let $(a,m),(b,n)\in A⋉E$ satisfying ${(a,m)}^3-{(b,n)}^3=(a^3-b^3,3a^{2}m-3b^{2}n)\in I⋉E$. Projecting onto the first coordinate gives $a^3-b^3\in I$, which imply either $a-b\in I$ or $a^2+ab+b^2\in \delta \left(0\right)$. In the first case, $(a-b,m-n)\in I⋉E$; in the second, $(a^2+ab+b^2,2am+an+bm+2bn)\in \delta \left(0\right)⋉E={\delta }_{\propto }\left(0\right)$, the second coordinate belonging to ${\delta }_{\propto }\left(0\right)$ automatically since its module component is all of E.    □

Under additional hypotheses on A and E, the cdf-absorbing $\delta \left(0\right)$ property of $I⋉N$ forces N to be a $\delta \left(0\right)$-submodule of E, in the sense of [7] [Definition 2.36].

Theorem 29.

Let A be a ring, E an A-module, δ an expansion function of ideals of A with $\delta \left(0\right)$ a radical ideal, I a proper ideal of A with $I\subseteq \delta \left(0\right)$ and $3\in U\left(A\right)$, and N a submodule of E with $IE\subseteq N$. If $I⋉N$ is a cdf-absorbing ${\delta }_{\propto }\left(0\right)$-ideal of $A⋉E$, then N is a $\delta \left(0\right)$-submodule of E.

Proof. 

Let $am\in N$ with $a\in A$ and $m\in E$. Then,

$${(a,0)}^3-{(a,m)}^3=(0,-3a^{2}m)\in I⋉N,$$

This yields either

$$(a,0)-(a,m)=(0,-m)\in I⋉N\mathrm{or}{(a,0)}^2+(a,0)(a,m)+{(a,m)}^2=(3a^2,3am)\in {\delta }_{\propto }\left(0\right).$$

The first alternative gives $m\in N$, as desired. In the second, we obtain $3a^2\in \delta \left(0\right)$; since 3 is a unit in A, it follows that $a^2\in \delta \left(0\right)$, and the radicality of $\delta \left(0\right)$ then yields $a\in \delta \left(0\right)$.    □

Finally, we examine amalgamated algebras along an ideal. Let $f:A\to B$ be a unital ring homomorphism and J an ideal of B. The amalgamated algebra of A and B along J with respect to f is

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

a construction introduced and studied in [8,9,10]. Let ${\pi }_A:A{\bowtie }^{f}J\to A$ denote the canonical projection $(a,f(a)+j)\mapsto a$. For every ideal K of $A{\bowtie }^{f}J$, we set $\widehat{\delta }\left(K\right):=\delta \left({\pi }_A\left(K\right)\right){\bowtie }^{f}J$, which defines an expansion function of ideals of $A{\bowtie }^{f}J$ with $\widehat{\delta }\left(0\right)=\delta \left(0\right){\bowtie }^{f}J$.

Theorem 30.

Let δ be an expansion function of ideals of A, let $f:A\to B$ be a unital ring homomorphism, J an ideal of B, and let I be a proper ideal of A.

1.

If $I{\bowtie }^{f}J$ is a cdf-absorbing $\widehat{\delta }\left(0\right)$-ideal of $A{\bowtie }^{f}J$, then I is a cdf-absorbing $\delta \left(0\right)$-ideal of A.

2.

If I is a cdf-absorbing $\delta \left(0\right)$-ideal of A, then $I{\bowtie }^{f}J$ is a cdf-absorbing $\widehat{\delta }\left(0\right)$-ideal of $A{\bowtie }^{f}J$. In the special case, I is a cdf-absorbing $\delta \left(0\right)$-ideal of A if and only if $I{\bowtie }^f\left(0\right)$ is a cdf-absorbing $\widehat{\delta }\left(0\right)$-ideal of $A{\bowtie }^f\left(0\right)$.

Proof. (1) Let $a,b\in A$ with $a^3-b^3\in I$. Then $(a,f(a\left)\right)$, $(b,f\left(b\right))\in A{\bowtie }^{f}J$ such that

$${(a,f\left(a\right))}^3-{(b,f\left(b\right))}^3=(a^3-b^3,f(a^3-b^3))\in I{\bowtie }^{f}J.$$

The cdf-absorbing $\widehat{\delta }\left(0\right)$ property therefore yields either $(a-b,f(a-b))\in I{\bowtie }^{f}J$ or $(a^2+ab+b^2,f(a^2+ab+b^2))\in \widehat{\delta }\left(0\right)$. Projecting onto the first and second coordinates, we obtain $a-b\in I$ or $a^2+ab+b^2\in \delta \left(0\right)$, respectively. (2) Let $x=(a,f\left(a\right)+j_1),y=(b,f\left(b\right)+j_2)\in A{\bowtie }^{f}J$ with $x^3-y^3\in I{\bowtie }^{f}J$. Comparing first coordinates gives $a^3-b^3\in I$, we have either $a-b\in I$ or $a^2+ab+b^2\in \delta \left(0\right)$. In the former case,

$$x-y=(a-b,f(a-b)+(j_1-j_2))\in I{\bowtie }^{f}J;$$

in the latter, $x^2+xy+y^2\in \widehat{\delta }\left(0\right)$, since its first coordinate $a^2+ab+b^2$ belongs to $\delta \left(0\right)$. When $J=\left(0\right)$, the map $a\mapsto (a,f(a\left)\right)$ is a ring isomorphism from A onto $A{\bowtie }^f\left(0\right)$, under which I corresponds precisely to $I{\bowtie }^f\left(0\right)$ and $\delta \left(0\right)$ corresponds to $\widehat{\delta }\left(0\right)$. The equivalence follows at once from this correspondence.    □