p-Difference Factor Absorbing Ideals of Commutative Rings

1. Introduction

The notion of prime ideals plays a central role in commutative ring theory and has long served as a source of inspiration for various generalizations. Among the more recent ones, Anderson, Badawi, and Coykendall introduced in [1] the class of square-difference factor absorbing ideals, or sdf-absorbing ideals for short. A proper ideal I of a commutative ring R is called an sdf-absorbing ideal if, whenever $a^2-b^2\in I$ for nonzero $a,b\in R$, at least one of $a+b$ or $a-b$ must lie in I as well. The idea behind this definition is transparent: the identity $a^2-b^2=(a+b)(a-b)$ shows that membership of a difference of squares in I should force one of its factors into I, much in the spirit of primeness. Building on this, Farshadifar [4] introduced cubes-difference factor absorbing ideals (cdf-absorbing ideals) by replacing squares with cubes, now using the factorization $a^3-b^3=(a-b)(a^2+ab+b^2)$, and established a parallel theory with results closely mirroring those of [1].

The present paper continues this line of investigation by taking the natural next step: we replace the exponent with an arbitrary prime p. The starting point is the factorization

$$a^p-b^p=(a-b){\Phi }_p(a,b),$$

where ${\Phi }_p(a,b)=a^{p-1}+a^{p-2}b+\dots +b^{p-1}$ is the classical p-th cyclotomic-type factor. We call a proper ideal I of R a p-difference factor absorbing ideal, or p-dfa ideal, if whenever $a^p-b^p\in I$ for $a,b\in R$, then either $a-b\in I$ or ${\Phi }_p(a,b)\in I$. When $p=2$, this recovers (with the restriction $a,b\ne 0$ removed) the notion of an sdf-absorbing ideal from [1], and when $p=3$, it recovers (again without the nonzero restriction) the cdf-absorbing ideals of [4], since ${\Phi }_2(a,b)=a+b$ and ${\Phi }_3(a,b)=a^2+ab+b^2$. In this way, the present notion unifies and extends both of these earlier concepts.

A simple but important observation is that ${\Phi }_p(a,a)=p·a^{p-1}$ for all $a\in R$. This means that the prime p itself enters the picture in a very concrete way, and indeed the status of p in R — whether it is zero, a unit, or a zero-divisor — will play a decisive role throughout the paper, just as the element 2 did in [1] and 3 did in [4].

The main contributions of this paper are the following. We introduce and study ${*}_p$-prime ideals (Definition 2), a new class that sits between prime ideals and p-dfa ideals, and we demonstrate that the implication “nonzero p-dfa and $p\in U\left(R\right)$ implies prime” — which holds for $p=2$ by [1, Theorem 2.6] — fails for every prime $p\ge 3$ (Example 1). For principal ideal domains of characteristic zero, we obtain a complete characterization of when a prime-power ideal $\left({\pi }^k\right)$ is p-dfa when $p\notin \left(\pi \right)$ (Theorem 3), and for those with finite residue fields we determine exactly when a product $\left(\pi \tau \right)$ of two distinct irreducibles is p-dfa in terms of the injectivity of the p-power map on the residue fields (Theorem 4); the integers $\mathbb{Z}$ are recovered as a special case. We further study p-dfa ideals in polynomial rings, direct products, idealizations, and amalgamation rings.

We work throughout with commutative rings having a nonzero identity, and all ring homomorphisms are assumed to send 1 to 1. We use standard notation: $U\left(R\right)$ for the group of units and $\mathrm{char}\left(R\right)$ for the characteristic of R. Recall that a proper ideal I is a radical ideal if $I=\sqrt{I}=\{r\in R\mid r^n\in I\mathrm{for}\mathrm{some}n\ge 1\}$, and a prime ideal if $ab\in I$ implies $a\in I$ or $b\in I$. Every prime ideal is radical, and every radical ideal is an intersection of primes. As usual, $\mathbb{Z}$, $\mathbb{Q}$, ${\mathbb{Z}}_n$, and ${\mathbb{F}}_q$ stand for the integers, rationals, integers modulo n, and the finite field with q elements. For any background material not covered here, we refer to [5,6,7].

The paper is organized as follows. Section 2 develops the basic theory: basic properties, the equivalence with a system-of-equations condition (Proposition 1), ${*}_p$-prime ideals, a result on von Neumann regular rings (Corollary 1), and the behavior under localization and homomorphisms. Section 3 treats p-dfa ideals in PIDs, culminating in Theorems 3 and 4. Section 4 studies p-dfa ideals in polynomial rings, direct products, idealizations, and amalgamation rings.

2. Basic Properties of p-dfa Ideals

We begin with the central definition of the paper.

Definition 1.

Let p be a prime number and let R be a commutative ring. A proper ideal I of R is called ap-difference factor absorbing ideal(p-dfa ideal) of R if whenever $a^p-b^p\in I$ for $a,b\in R$, then $a-b\in I$ or ${\Phi }_p(a,b)\in I$.

Since every prime ideal P of R satisfies $xy\in P$ implies $x\in P$ or $y\in P$, and since $a^p-b^p=(a-b){\Phi }_p(a,b)$, it is immediate that every prime ideal is a p-dfa ideal. The converse does not hold in general, as Example 1 below illustrates.

Remark 1.

A straightforward computation shows that ${\Phi }_p(a,a)=p·a^{p-1}$ for every $a\in R$. In particular, setting $a=b=1$ gives ${\Phi }_p(1,1)=p$, a fact that will be used repeatedly in what follows.

The next lemma records a basic consequence of the definition, analogous to [4, Proposition 2.4].

Lemma 1.

Let I be a p-dfa ideal of R and let $a\in R$. If $a^p\in I$, then $a\in I$ or $a^{p-1}\in I$.

Proof. 

Suppose $a^p\in I$. Then $a^p-0^p=a^p\in I$, and since I is a p-dfa ideal, either $a-0=a\in I$ or ${\Phi }_p(a,0)=a^{p-1}\in I$, as claimed.    □

The following lemma characterizes when the prime p belongs to a p-dfa ideal I in terms of an equivalent condition on ${\Phi }_p$ and the characteristic of the quotient ring. We note that the converse of condition (a) does not hold in general; see Remark 2 below.

Lemma 2.

Let I be a p-dfa ideal of R. The following statements are equivalent.

(a)

Whenever $a^p-b^p\in I$ for $a,b\in R$ and $a-b\in I$, we have ${\Phi }_p(a,b)\in I$.

(b)

$p\in I$.

(c)

$\mathrm{char}(R/I)=p$.

Proof. (a) ⇒ (b): Set $a=b=1$. Then $a^p-b^p=0\in I$ and $a-b=0\in I$, so by (a) we get ${\Phi }_p(1,1)=p\in I$.

(b) ⇒ (a): Assume $p\in I$, $a^p-b^p\in I$, and $a-b\in I$. Then $a\equiv b(modI)$, so each term satisfies $a^{p-1-k}{b}^k\equiv b^{p-1-k}·b^k=b^{p-1}(modI)$, and thus

$${\Phi }_p(a,b)={\displaystyle \sum _{k=0}^{p-1}}{a}^{p-1-k}{b}^k\equiv {\displaystyle \sum _{k=0}^{p-1}}{b}^{p-1}=p·b^{p-1}\in I,$$

since $p\in I$.

(b) ⇔ (c): This is clear since $p\in I$ if and only if the image of p in $R/I$ is zero, that is, $\mathrm{char}(R/I)=p$.    □

Remark 2.

The converse of Lemma 2(a) does not hold in general: it is not true that ${\Phi }_p(a,b)\in I$ implies $a-b\in I$, even when $p\in I$ and $a^p-b^p\in I$. A concrete counterexample is given by $R={\mathbb{F}}_3\left[x\right]/\left(x^2\right)$, $I=\left\{0\right\}$, and $p=3$. Since $\mathrm{char}\left(R\right)=3$, we have $p=3=0\in I$. To see that I is a 3-dfa ideal of R, note that for any $a=\alpha +\beta x\in R$ one computes $a^3={\alpha }^3=\alpha$ (using $x^2=0$ and Fermat’s little theorem in ${\mathbb{F}}_3$), so $a^3-b^3=0$ if and only if $\alpha =\gamma$, where $b=\gamma +\delta x$. In that case ${\Phi }_3(a,b)=3{\alpha }^2+\left(\mathrm{terms}\mathrm{in}x\right)\equiv 0+0·x=0\in I$, confirming that I is a 3-dfa ideal. Taking $a=x$ and $b=0$, we get $a^3=0\in I$, so $a^3-b^3=0\in I$. Moreover, ${\Phi }_3(x,0)=x^2=0\in I$. However, $a-b=x\ne 0$, so $a-b\notin I$.

The following lemma is central to much of what follows. When the characteristic of R equals p, the p-dfa condition takes a particularly clean form.

Lemma 3.

Let R be a commutative ring with $\mathrm{char}\left(R\right)=p$. Then every radical ideal of R is a p-dfa ideal of R.

Proof. 

Let I be a radical ideal of R and suppose $a^p-b^p\in I$ for some $a,b\in R$. Since $\mathrm{char}\left(R\right)=p$, the Frobenius endomorphism (the identity ${(x+y)}^p=x^p+y^p$ in characteristic p) gives ${(a-b)}^p=a^p-b^p\in I$. Because I is a radical ideal, it follows that $a-b\in I$. Hence I is a p-dfa ideal.    □

Recall that a commutative ring R is called von Neumann regular if for every $a\in R$ there exists $x\in R$ with $a=axa$. Von Neumann regular rings are reduced and of Krull dimension zero; in particular, every ideal of a von Neumann regular ring is a radical ideal (see, e.g., [5]). Combined with Lemma 3, this immediately gives the following.

Corollary 1.

Let R be a von Neumann regular ring with $\mathrm{char}\left(R\right)=p$. Then every proper ideal of R is a p-dfa ideal of R.

One of the notable features of sdf-absorbing ideals is that every nonzero sdf-absorbing ideal with $2\in U\left(R\right)$ is a prime ideal [1, Theorem 2.6]. A natural question is whether an analogous result holds for p-dfa ideals when $p\in U\left(R\right)$. The following example shows that this fails already for $p=3$, marking a fundamental difference between the case $p=2$ and all larger primes.

Example 1.

Let $p=3$ and $R={\mathbb{Z}}_8$. Since $3·3=9\equiv 1(mod8)$, we have $3\in U\left({\mathbb{Z}}_8\right)$. Consider the ideal $I=4{\mathbb{Z}}_8=\{0,4\}$, which is nonzero. A direct computation shows that I is a 3-dfa ideal of ${\mathbb{Z}}_8$: one checks that whenever $a^3-b^3\in I$, either $a-b\in I$ or ${\Phi }_3(a,b)=a^2+ab+b^2\in I$. However, I is not a prime ideal of ${\mathbb{Z}}_8$, since $2·2=4\in I$ while $2\notin I$. The same ideal serves as a counterexample for all primes $p\ge 3$: one verifies that $4{\mathbb{Z}}_8$ is a p-dfa ideal of ${\mathbb{Z}}_8$ for every prime p with $p\in U\left({\mathbb{Z}}_8\right)$, yet it is never prime.

Remark 3.

Example 1 shows that for primes $p\ge 3$, the implication “I is a nonzero p-dfa ideal and $p\in U\left(R\right)$ implies I is prime” fails in general. This is a genuine departure from the case $p=2$ established in [1]. We note that for $p=3$, Farshadifar [4] obtained a related but weaker result: when $3\in U\left(R\right)$, a cdf-absorbing ideal is equivalent to a *-prime ideal (in the sense of [4]), a notion weaker than primeness. The question of what additional conditions beyond $p\in U\left(R\right)$ force a p-dfa ideal to be prime remains open and is a direction for further investigation.

The following result gives a useful equivalent reformulation of the p-dfa condition in terms of a system of equations, analogous to [1, Theorem 2.7] and [4, Theorem 2.13].

Proposition 1.

Let I be a proper ideal of a commutative ring R. Then the following statements are equivalent.

(a)

I is a p-dfa ideal of R.

(b)

For all $x,y\in R\setminus I$ with $xy\in I$, there are no $c,d\in R$ satisfying both $c-d=x$ and ${\Phi }_p(c,d)=y$ simultaneously.

Proof. (a) ⇒ (b): Suppose I is a p-dfa ideal of R, and let $xy\in I$ for some $x,y\in R\setminus I$. Assume for contradiction that there exist $c,d\in R$ with $c-d=x$ and ${\Phi }_p(c,d)=y$. Then

$$c^p-d^p=(c-d){\Phi }_p(c,d)=xy\in I,$$

so since I is a p-dfa ideal, either $c-d\in I$ or ${\Phi }_p(c,d)\in I$. But $c-d=x\notin I$ and ${\Phi }_p(c,d)=y\notin I$, a contradiction. Hence no such $c,d$ can exist.

(b) ⇒ (a): Suppose $c^p-d^p\in I$ for some $c,d\in R$. Define $x=c-d$ and $y={\Phi }_p(c,d)$, so that $xy=c^p-d^p\in I$. Assume for contradiction that $x\notin I$ and $y\notin I$. Then c and d themselves satisfy the system $c-d=x$ and ${\Phi }_p(c,d)=y$, which directly contradicts (b). Hence we must have $x=c-d\in I$ or $y={\Phi }_p(c,d)\in I$, showing that I is a p-dfa ideal of R.    □

We now introduce a new class of ideals related to p-dfa ideals, which generalizes the notion of *-prime ideals introduced in [4] for the case $p=3$.

Definition 2.

A proper ideal I of R is called a${*}_p$-prime idealof R if whenever $x·{\Phi }_p(a,b)\in I$ for some $a,b,x\in R$, then $x\in I$ or ${\Phi }_p(a,b)\in I$.

Note that every prime ideal of R is a ${*}_p$-prime ideal, since primeness immediately forces $x\in I$ or ${\Phi }_p(a,b)\in I$ whenever $x·{\Phi }_p(a,b)\in I$. The converse, however, need not hold in general, as the following example illustrates.

Example 2.

Consider $R={\mathbb{Z}}_4$ and $p=3$. The zero ideal $I=\left\{0\right\}$ is not a prime ideal of ${\mathbb{Z}}_4$, since $2·2=0\in I$ but $2\notin I$. However, I is a ${*}_3$-prime ideal of ${\mathbb{Z}}_4$. Indeed, a direct computation shows that ${\Phi }_3(a,b)=a^2+ab+b^2$ takes only the values $0,1,3$ in ${\mathbb{Z}}_4$ — it never equals 2. Therefore, if $x·{\Phi }_3(a,b)=0$ in ${\mathbb{Z}}_4$, then either $x=0\in I$ or ${\Phi }_3(a,b)=0\in I$, confirming that I is a ${*}_3$-prime ideal of ${\mathbb{Z}}_4$.

Lemma 4.

Let I be a proper ideal of R.

(a)

If I is a ${*}_p$-prime ideal of R, then I is a p-dfa ideal of R.

(b)

Every prime ideal of R is a ${*}_p$-prime ideal of R.

Proof. (a) Suppose I is a ${*}_p$-prime ideal of R and let $a^p-b^p\in I$ for some $a,b\in R$. Since $a^p-b^p=(a-b){\Phi }_p(a,b)\in I$, we may apply the ${*}_p$-prime condition with $x=a-b$ to conclude that $a-b\in I$ or ${\Phi }_p(a,b)\in I$. Hence I is a p-dfa ideal of R.

(b) Let P be a prime ideal of R and suppose $x·{\Phi }_p(a,b)\in P$ for some $a,b,x\in R$. Since P is prime, either $x\in P$ or ${\Phi }_p(a,b)\in P$. Hence P is a ${*}_p$-prime ideal of R.    □

Remark 4.

In [1, Theorem 2.2], it is shown that every nonzero sdf-absorbing ideal is a radical ideal. One might naturally expect an analogous statement to hold for p-dfa ideals in general; however, the argument breaks down for primes $p\ge 3$. The proof for $p=2$ hinges on the fact that if $a^2-i^2\in I$ for some nonzero $i\in I$, then either $a+i$ or $a-i$ lies in I, and in either case one immediately deduces $a\in I$ since $i\in I$. For general p, however, when $a^p-i^p\in I$ and it is ${\Phi }_p(a,i)\in I$ that holds rather than $a-i\in I$, one cannot in general conclude that $a\in I$: the expression ${\Phi }_p(a,i)=a^{p-1}+a^{p-2}i+\dots +i^{p-1}$ is a sum of mixed terms, and knowing it lies in I gives no direct handle on a alone. Whether every nonzero p-dfa ideal is necessarily a radical ideal for $p\ge 3$ remains an open question.

The following example shows that the zero ideal need not be a p-dfa ideal, even in local rings.

Example 3.

Let $p\ge 3$ be prime and consider the ring ${\mathbb{Z}}_{p^2}$. Take $a=1$ and $b=1+p$ in ${\mathbb{Z}}_{p^2}$. By the binomial theorem,

$${(1+p)}^p={\displaystyle \sum _{k=0}^p}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{k}}\right)p^k\equiv 1+p·p=1+p^2\equiv 1(mod{p}^2),$$

so $a^p-b^p=1-1=0\in \left\{0\right\}$. On the other hand, $a-b=-p\ne 0$ in ${\mathbb{Z}}_{p^2}$, and

$${\Phi }_p(1,1+p)={\displaystyle \sum _{k=0}^{p-1}}{(1+p)}^k\equiv {\displaystyle \sum _{k=0}^{p-1}}(1+kp)=p+p·{\displaystyle \frac{p(p-1)}{2}}\equiv p(mod{p}^2),$$

which is nonzero in ${\mathbb{Z}}_{p^2}$ (as $p\ge 3$ implies $p<p^2$). Thus $\left\{0\right\}$ is not a p-dfa ideal of ${\mathbb{Z}}_{p^2}$ for any prime $p\ge 3$. For $p=2$ the same conclusion holds — $\left\{0\right\}$ is not a 2-dfa ideal of ${\mathbb{Z}}_4$ — as witnessed by $a=0$, $b=2$: then $a^2-b^2=-4\equiv 0(mod4)$, yet $a-b=-2\neg \equiv 0$ and ${\Phi }_2(0,2)=0+2=2\neg \equiv 0(mod4)$. This contrasts sharply with the sdf case, where $\left\{0\right\}$ is an sdf-absorbing ideal of ${\mathbb{Z}}_4$ (see [1, Remark 2.3(a)]); the difference arises because the sdf definition requires $a,b\ne 0$.

We next establish that the p-dfa property is preserved under localization and ring homomorphisms, in direct analogy with the results of [1] and [4].

Theorem 1.

Let I be a p-dfa ideal of R, and let S be a multiplicatively closed subset of R with $I\cap S=\varnothing$. Then $S^{-1}I$ is a p-dfa ideal of $S^{-1}R$.

Proof. 

Suppose ${(a/s)}^p-{(b/t)}^p\in S^{-1}I$ for some $a/s,b/t\in S^{-1}R$. Then there exists $u\in S$ such that $u(a^p{t}^p-b^p{s}^p)\in I$. Since I is an ideal and $u^{p-1}\in R$, we get

$${\left(uat\right)}^p-{\left(ubs\right)}^p=u^p(a^p{t}^p-b^p{s}^p)=u^{p-1}·u(a^p{t}^p-b^p{s}^p)\in I.$$

Since I is a p-dfa ideal of R, either $uat-ubs\in I$ or ${\Phi }_p(uat,ubs)\in I$. In the first case, $u(at-bs)\in I$ and $ust\in S$, so $a/s-b/t=(at-bs)/st=u(at-bs)/\left(ust\right)\in S^{-1}I$. In the second case, since

$${\Phi }_p(uat,ubs)=u^{p-1}{\left(st\right)}^{p-1}{\Phi }_p(a/s,b/t),$$

and $u^{p-1}{\left(st\right)}^{p-1}\in S$, we get ${\Phi }_p(a/s,b/t)\in S^{-1}I$. Hence $S^{-1}I$ is a p-dfa ideal of $S^{-1}R$.    □

Theorem 2.

Let $f:R\to T$ be a homomorphism of commutative rings.

(a)

If f is injective and J is a p-dfa ideal of T, then $f^{-1}\left(J\right)$ is a p-dfa ideal of R.

(b)

If f is surjective and I is a p-dfa ideal of R with $ker\left(f\right)\subseteq I$, then $f\left(I\right)$ is a p-dfa ideal of T.

Proof. (a) Suppose $a^p-b^p\in f^{-1}\left(J\right)$ for some $a,b\in R$. Then $f{\left(a\right)}^p-f{\left(b\right)}^p=f(a^p-b^p)\in J$. Since J is a p-dfa ideal, either $f\left(a\right)-f\left(b\right)\in J$ or ${\Phi }_p(f\left(a\right),f\left(b\right))\in J$. Since f is a ring homomorphism, $f(a-b)=f\left(a\right)-f\left(b\right)$ and $f\left({\Phi }_p(a,b)\right)={\Phi }_p(f\left(a\right),f\left(b\right))$. Since f is injective, it follows that $a-b\in f^{-1}\left(J\right)$ or ${\Phi }_p(a,b)\in f^{-1}\left(J\right)$. Hence $f^{-1}\left(J\right)$ is a p-dfa ideal of R.

(b) Let $x^p-y^p\in f\left(I\right)$ for some $x,y\in T$. Since f is surjective, write $x=f\left(a\right)$ and $y=f\left(b\right)$ for some $a,b\in R$. Then $f(a^p-b^p)=f{\left(a\right)}^p-f{\left(b\right)}^p=x^p-y^p\in f\left(I\right)$, so there exists $c\in I$ with $f(a^p-b^p)=f\left(c\right)$. Thus $a^p-b^p-c\in ker\left(f\right)\subseteq I$, giving $a^p-b^p=c+(a^p-b^p-c)\in I$. Since I is a p-dfa ideal, either $a-b\in I$ or ${\Phi }_p(a,b)\in I$, and applying f gives $x-y=f(a-b)\in f\left(I\right)$ or ${\Phi }_p(x,y)=f\left({\Phi }_p(a,b)\right)\in f\left(I\right)$. Hence $f\left(I\right)$ is a p-dfa ideal of T.    □

Corollary 2.

Let R be a commutative ring.

(a)

If $R\subseteq T$ is a ring extension and J is a p-dfa ideal of T, then $J\cap R$ is a p-dfa ideal of R.

(b)

If $J\subseteq I$ are ideals of R and I is a p-dfa ideal of R, then $I/J$ is a p-dfa ideal of $R/J$.

(c)

If $J⊊I$ are ideals of R, then $I/J$ is a p-dfa ideal of $R/J$ if and only if I is a p-dfa ideal of R.

Proof. 

Parts (a) and (b) follow immediately from Theorem 2 by applying it to the inclusion map $R\hookrightarrow T$ and the quotient map $\pi :R\twoheadrightarrow R/J$ (with $ker\left(\pi \right)=J\subseteq I$ and $\pi \left(I\right)=I/J$), respectively. For (c), the direction $(\Leftarrow )$ is part (b). For $(\Rightarrow )$, suppose $I/J$ is a p-dfa ideal of $R/J$ and let $a^p-b^p\in I$. Then ${\overline{a}}^p-{\overline{b}}^p\in I/J$, so either $\overline{a}-\overline{b}\in I/J$ or ${\Phi }_p(\overline{a},\overline{b})\in I/J$, giving $a-b\in I$ or ${\Phi }_p(a,b)\in I$. Hence I is a p-dfa ideal of R.    □

The following example shows that the injectivity hypothesis in Theorem 2(a) cannot be dropped.

Example 4.

Let $f:\mathbb{Z}\to {\mathbb{Z}}_9$ be the natural epimorphism and $p=3$. Taking $a=1$ and $b=-2$, we have $a^3-b^3=9\in 9\mathbb{Z}$, but $a-b=3\notin 9\mathbb{Z}$ and ${\Phi }_3(1,-2)=1-2+4=3\notin 9\mathbb{Z}$; so $f^{-1}\left(\left\{0\right\}\right)=9\mathbb{Z}$ is not a 3-dfa ideal of $\mathbb{Z}$. The same pair shows that $\left\{0\right\}$ is not a 3-dfa ideal of ${\mathbb{Z}}_9$ either, so neither part of Theorem 2 applies: f is not injective, and $ker\left(f\right)=9\mathbb{Z}\neg \subseteq \left\{0\right\}$.

We close the section with a collection of examples illustrating how p-dfa ideals behave in concrete rings.

Example 5.

(a)

(Integers)We give a few sample computations in $\mathbb{Z}$; a complete characterization of p-dfa ideals in $\mathbb{Z}$ will be given elsewhere. For $p=2$, the ideal $6\mathbb{Z}$ is a 2-dfa ideal: if $a^2-b^2=(a+b)(a-b)\in 6\mathbb{Z}$, then $a+b$ and $a-b$ have the same parity so 2 divides both; since 3 is prime and $3\mid (a+b)(a-b)$, we get $3\mid (a+b)$ or $3\mid (a-b)$, and combining gives $6\mid (a+b)$ or $6\mid (a-b)$. By contrast, $4\mathbb{Z}$ is not a 2-dfa ideal: taking $a=0$, $b=2$ gives $a^2-b^2=-4\in 4\mathbb{Z}$ while $a-b=-2$ and ${\Phi }_2(0,2)=2$ both miss $4\mathbb{Z}$. For $p=3$, the ideal $4\mathbb{Z}$ is a 3-dfa ideal (a direct case analysis modulo 4 confirms that $a^3-b^3\in 4\mathbb{Z}$ always forces $a-b$ or ${\Phi }_3(a,b)=a^2+ab+b^2$ into $4\mathbb{Z}$), while $9\mathbb{Z}$ is not: $a=1$, $b=-2$ gives $a^3-b^3=9\in 9\mathbb{Z}$ yet $a-b=3$ and ${\Phi }_3(1,-2)=3$ both lie outside $9\mathbb{Z}$.

(b)

(Boolean rings)Let R be a boolean ring, that is, $x^2=x$ for all $x\in R$. A simple induction on $n\ge 1$ shows that $x^n=x$ for all $x\in R$ and all $n\ge 1$: the base cases $n=1,2$ are clear, and if $x^n=x$ then $x^{n+1}=x^n·x=x·x=x^2=x$. In particular, $x^p=x$ for every prime p, so $a^p-b^p=a-b$ for all $a,b\in R$. Hence whenever $a^p-b^p\in I$, we immediately get $a-b\in I$, and thus every proper ideal of a boolean ring is a p-dfa ideal for every prime p.

(c)

(Fields)If R is a field, then its only proper ideal is $\left\{0\right\}$, which is a prime ideal and hence a p-dfa ideal. In particular, $\left\{0\right\}$ is a p-dfa ideal of ${\mathbb{Z}}_p$ for every prime p.

(d)

(Direct products)Consider $R=\mathbb{Z}\times \mathbb{Z}$ and $p=2$. The prime ideal $I=2\mathbb{Z}\times \mathbb{Z}$ is a 2-dfa ideal of R. On the other hand, the ideal $J=3\mathbb{Z}\times 3\mathbb{Z}$ is not a 2-dfa ideal of R: take $a=(-2,-2)$ and $b=(-2,-1)$; then $a^2-b^2=(0,3)\in J$, but $a-b=(0,-1)\notin J$ and ${\Phi }_2(a,b)=a+b=(-4,-3)\notin J$.

3. p-dfa Ideals in Principal Ideal Domains

In a principal ideal domain R, every ideal has the form $\left(\alpha \right)=\alpha R$ for some $\alpha \in R$, unique up to associates. We characterize the p-dfa ideals among principal ideals generated by prime powers and products of two primes, recovering the results of Section 2 for $R=\mathbb{Z}$ as special cases. Throughout, $v_{\pi }$ denotes the $\pi$-adic valuation on R, defined by $v_{\pi }\left(\alpha \right)=sup\{n\ge 0:{\pi }^n\mid \alpha \}$.

Theorem 3.

Let R be a principal ideal domain of characteristic zero, π an irreducible element of R with $p\notin \left(\pi \right)$, and $k\ge 1$ an integer. Then $\left({\pi }^k\right)$ is a p-dfa ideal of R if and only if $k\le p-1$.

Proof. 

Since $\pi$ is irreducible and R is a PID, $\left(\pi \right)$ is a maximal ideal, so $K:=R/\left(\pi \right)$ is a field. The hypothesis $p\notin \left(\pi \right)$ means $v_{\pi }\left(p\right)=0$.

$(\Leftarrow )$ Suppose $k\le p-1$ and $a^p-b^p\in \left({\pi }^k\right)$. Write $i=v_{\pi }(a-b)$ and $j=v_{\pi }\left({\Phi }_p(a,b)\right)$. Since $a^p-b^p=(a-b){\Phi }_p(a,b)$, we have $i+j\ge k$. We consider three cases according to the $\pi$-adic behavior of $a-b$ and b.

Case 1: $\pi \nmid (a-b)$. Then $gcd(a-b,{\pi }^k)=1$ in R (since R is a PID, hence a UFD), so ${\pi }^k\mid (a-b){\Phi }_p(a,b)$ forces ${\pi }^k\mid {\Phi }_p(a,b)$, giving $j\ge k$.

Case 2: $\pi \mid (a-b)$ and $\pi \mid b$. Set $c=a-b$ with $v_{\pi }\left(c\right)=i\ge 1$ and $r:=v_{\pi }\left(b\right)\ge 1$. By the binomial theorem,

$${\Phi }_p(b+c,b)={\displaystyle \sum _{s=0}^{p-1}}{c}^s{\displaystyle \sum _{j=0}^{p-1-s}}\left({\displaystyle \genfrac{}{}{0pt}{}{p-1-j}{s}}\right)b^{p-1-s}.$$

The $s=0$ term equals $p·b^{p-1}$, with $v_{\pi }$-value $(p-1)r$ (since $v_{\pi }\left(p\right)=0$ as $\pi \nmid p$). For $s\ge 1$, the s-th summand has $v_{\pi }$-value at least $(p-1-s)r+si=(p-1)r+s(i-r)\ge (p-1)min(r,i)$ (the inequality holds since if $i\ge r$ then $(p-1)r+s(i-r)\ge (p-1)r$, and if $i<r$ then the minimum over $s\in \{1,\dots ,p-1\}$ is at $s=p-1$, giving $(p-1)i$). Since also $(p-1)r\ge (p-1)min(r,i)$, every term has $v_{\pi }$-value at least $(p-1)min(r,i)$, giving

$$v_{\pi }\left({\Phi }_p(a,b)\right)\ge (p-1)min(r,i)\ge p-1\ge k.$$

Case 3: $\pi \mid (a-b)$ and $\pi \nmid b$. Then $v_{\pi }\left(b\right)=0$, so the constant term $p·b^{p-1}$ satisfies $v_{\pi }(p·b^{p-1})=0$, giving $j=0$. Since $i+j\ge k$, we get $i\ge k$.

$(\Rightarrow )$ Suppose $k\ge p$. When $k=p$, take $a=\pi$ and $b=0$: then $a^p-b^p={\pi }^p\in \left({\pi }^p\right)$, while $a-b=\pi \notin \left({\pi }^p\right)$ and ${\Phi }_p(\pi ,0)={\pi }^{p-1}\notin \left({\pi }^p\right)$. When $k>p$, set $m=k-p+1\ge 2$ and take $a=\pi +{\pi }^m$, $b=\pi$. Then $v_{\pi }(a-b)=m<k$. Writing

$$a^p-b^p={\pi }^p\left[{(1+{\pi }^{m-1})}^p-1\right]={\pi }^p\left[p{\pi }^{m-1}+\left(\genfrac{}{}{0pt}{}{p}{2}\right){\pi }^{2(m-1)}+\dots \right],$$

the dominant term is $p{\pi }^{m-1}$ with $v_{\pi }$-value $m-1$ (using $v_{\pi }\left(p\right)=0$), so $v_{\pi }(a^p-b^p)=p+(m-1)=k$. It follows that $v_{\pi }\left({\Phi }_p(a,b)\right)=k-m=p-1<k$. Thus $a^p-b^p\in \left({\pi }^k\right)$ while $a-b\notin \left({\pi }^k\right)$ and ${\Phi }_p(a,b)\notin \left({\pi }^k\right)$.    □

Corollary 3.

Let $q\ne p$ be primes. Then $q^k\mathbb{Z}$ is a p-dfa ideal of $\mathbb{Z}$ if and only if $k\le p-1$.

Proof. 

Take $R=\mathbb{Z}$ and $\pi =q$ in Theorem 3; the characteristic-zero and $p\notin \left(q\right)$ hypotheses reduce to $q\ne p$.    □

Remark 5.

When $p\in \left(\pi \right)$, i.e. $\pi \mid p$, the situation is more subtle and depends on the π-adic structure of p and the interplay between k and the ramification of p in R. For $R=\mathbb{Z}$ and $\pi =p$, computational evidence shows that the answer is not determined by k alone: for instance, $7^k\mathbb{Z}$ is a 7-dfa ideal for $k=1$ but not for $k=2,3,4,5,6,7,8$. A complete characterization in this case remains open.

We next extend the bijectivity lemma and the two-prime theorem to general PIDs.

Lemma 5.

Let R be a principal ideal domain, π an irreducible element with $p\notin \left(\pi \right)$, and $K=R/\left(\pi \right)$. If K is finite, then the map $x\mapsto x^p$ on K is injective if and only if $p\nmid |K^{*}|$.

Proof. 

Since $p\notin \left(\pi \right)$, we have $\overline{p}\ne 0$ in K, so $\mathrm{char}\left(K\right)\ne p$. Since K is finite, $K\cong {\mathbb{F}}_{q^d}$ for some prime $q\ne p$ and $d\ge 1$, and the multiplicative group $K^{*}:=K\setminus \left\{0\right\}$ is cyclic of order $n:=q^d-1=\left|K^{*}\right|$.

The map $\phi :x\mapsto x^p$ restricted to $K^{*}$ is a group homomorphism, since $\phi \left(xy\right)={\left(xy\right)}^p=x^p{y}^p=\phi \left(x\right)\phi \left(y\right)$ in the commutative ring K. Its kernel is $\{x\in K^{*}:x^p=1\}$, the set of p-th roots of unity in $K^{*}$. Because $K^{*}$ is cyclic of order n, the equation $x^p=1$ has exactly $gcd(p,n)$ solutions in $K^{*}$ (this is a standard fact about cyclic groups: in a cyclic group of order n, the equation $x^m=1$ has exactly $gcd(m,n)$ solutions). Hence $\phi$ is injective on $K^{*}$ if and only if $gcd(p,n)=1$, i.e. $p\nmid n=|K^{*}|$. Since $\phi$ also fixes 0, the map $x\mapsto x^p$ is injective on all of K exactly when it is injective on $K^{*}$, which proves the lemma.    □

Theorem 4.

Let R be a principal ideal domain of characteristic zero whose residue fields $R/\left(\pi \right)$ are finite for all irreducible $\pi \in R$, and let π and τ be non-associate irreducible elements of R. Then $\left(\pi \tau \right)$ is a p-dfa ideal of R if and only if one of the following holds:

(i)

$p\in \left(\pi \right)$ or $p\in \left(\tau \right)$; or

(ii)

$p\notin \left(\pi \right)$, $p\notin \left(\tau \right)$, and the map $x\mapsto x^p$ is injective on both $R/\left(\pi \right)$ and $R/\left(\tau \right)$.

Proof.$(\Leftarrow )$ Suppose first that condition (i) or (ii) holds; we show $\left(\pi \tau \right)$ is a p-dfa ideal of R.

Case (i): $p\in \left(\pi \right)$. Since $\pi \mid p$ and $\pi$ is irreducible, $K_{\pi }=R/\left(\pi \right)$ has characteristic p. Suppose $a^p-b^p\in \left(\pi \tau \right)$; since $\left(\pi \tau \right)\subseteq \left(\pi \right)\cap \left(\tau \right)$, we get both $\pi \mid a^p-b^p$ and $\tau \mid a^p-b^p$. In $K_{\pi }$, the Frobenius identity ${(\overline{a}-\overline{b})}^p={\overline{a}}^p-{\overline{b}}^p$ (valid since $\mathrm{char}\left(K_{\pi }\right)=p$) together with ${\overline{a}}^p={\overline{b}}^p$ gives ${(\overline{a}-\overline{b})}^p=0$; as $K_{\pi }$ is a field (hence has no zero divisors), this forces $\overline{a}=\overline{b}$, i.e. $\pi \mid a-b$.

It remains to show $\tau \mid a-b$ or $\tau \mid {\Phi }_p(a,b)$; we will reach one of these two conclusions via two subcases depending on whether $\tau$ also divides p. Once $\tau \mid a-b$ is established, combining with $\pi \mid a-b$ and $gcd(\pi ,\tau )=1$ gives $\pi \tau \mid a-b$. Alternatively, once $\tau \mid {\Phi }_p(a,b)$ is established, we will use $\pi \mid a-b$ to show $\pi \mid {\Phi }_p(a,b)$ as well, and then combine $\pi \mid {\Phi }_p(a,b)$ with $\tau \mid {\Phi }_p(a,b)$ to conclude $\pi \tau \mid {\Phi }_p(a,b)$.

Subcase $\tau \mid p$. The same Frobenius argument, now applied to $K_{\tau }=R/\left(\tau \right)$ (which likewise has characteristic p), gives $\tau \mid a-b$. Combined with $\pi \mid a-b$ and $gcd(\pi ,\tau )=1$, this yields $\pi \tau \mid a-b$.

Subcase $\tau \nmid p$. Here $p\notin \left(\tau \right)$, so Theorem 3 applies to the ideal $\left(\tau \right)$ with $k=1\le p-1$, giving $\tau \mid a-b$ or $\tau \mid {\Phi }_p(a,b)$. If $\tau \mid a-b$, we again obtain $\pi \tau \mid a-b$ as above. If instead $\tau \mid {\Phi }_p(a,b)$, we use $\pi \mid a-b$ (already established) to compute ${\Phi }_p(a,b)$ modulo $\left(\pi \right)$: since $a\equiv b(mod(\pi \left)\right)$, every term $a^{p-1-j}{b}^j$ in ${\Phi }_p(a,b)={\sum }_{j=0}^{p-1}{a}^{p-1-j}{b}^j$ is congruent to $b^{p-1}$ modulo $\left(\pi \right)$, so

$${\Phi }_p(a,b)\equiv p·b^{p-1}\equiv 0(mod\left(\pi \right))$$

(the last congruence holds because $p\in \left(\pi \right)$ by hypothesis). Thus $\pi \mid {\Phi }_p(a,b)$, and combined with $\tau \mid {\Phi }_p(a,b)$ from this subcase, we get $\pi \tau \mid {\Phi }_p(a,b)$.

Case (ii): $p\notin \left(\pi \right)$, $p\notin \left(\tau \right)$, both maps injective. Suppose $a^p-b^p\in \left(\pi \tau \right)$. Since $\left(\pi \tau \right)\subseteq \left(\pi \right)$, we have $\pi \mid a^p-b^p$, i.e. ${\overline{a}}^p={\overline{b}}^p$ in $K_{\pi }=R/\left(\pi \right)$. As the map $x\mapsto x^p$ is injective on $K_{\pi }$ by hypothesis, this forces $\overline{a}=\overline{b}$, i.e. $\pi \mid a-b$. The identical argument with $\tau$ in place of $\pi$ gives $\tau \mid a-b$. Finally, since $\pi$ and $\tau$ are non-associate irreducibles in the UFD R, they are coprime, so $\pi \mid a-b$ and $\tau \mid a-b$ together give $\pi \tau \mid a-b$.

$(\Rightarrow )$ We prove the contrapositive: if neither (i) nor (ii) holds, then $\left(\pi \tau \right)$ is not a p-dfa ideal.

Case (iii): Neither (i) nor (ii) holds. Then $p\notin \left(\pi \right)$ and $p\notin \left(\tau \right)$, but the map $x\mapsto x^p$ fails to be injective on at least one of the residue fields; say it fails on $K_{\pi }=R/\left(\pi \right)$. We construct explicit elements $a_0,b_0\in R$ witnessing that $\left(\pi \tau \right)$ is not a p-dfa ideal.

Step 1 (finding a p-th root of unity $\ne 1$ in $K_{\pi }$). Since $K_{\pi }$ is finite, Lemma 5 (applied in the contrapositive direction) tells us that non-injectivity of $x\mapsto x^p$ forces $p\mid |K_{\pi }^{*}|$. As $K_{\pi }^{*}$ is a cyclic group of order $|K_{\pi }^{*}|$ divisible by p, it contains an element $\alpha$ of order exactly p (the unique subgroup of order p in a cyclic group is generated by such an element). In particular ${\alpha }^p=1$ but $\alpha \ne 1$.

Step 2 (lifting α via the Chinese Remainder Theorem). Since $\pi$ and $\tau$ are non-associate irreducibles in the PID R, the ideals $\left(\pi \right)$ and $\left(\tau \right)$ are coprime, i.e. $\left(\pi \right)+\left(\tau \right)=R$. The Chinese Remainder Theorem then gives a ring isomorphism $R/\left(\pi \tau \right)\cong R/\left(\pi \right)\times R/\left(\tau \right)$, so we may choose $a_0\in R$ satisfying

$$a_0\equiv \alpha (mod\left(\pi \right)),a_0\equiv 1(mod\left(\tau \right)).$$

Set $b_0=1$.

Step 3 (verifying $a_0^p-b_0^p\in \left(\pi \tau \right)$). Modulo $\left(\pi \right)$: $a_0^p\equiv {\alpha }^p=1=b_0^p$. Modulo $\left(\tau \right)$: $a_0^p\equiv 1^p=1=b_0^p$. So $a_0^p-b_0^p$ is divisible by both $\pi$ and $\tau$, hence by $\pi \tau$ (as $gcd(\pi ,\tau )=1$).

Step 4 (verifying $a_0-b_0\notin \left(\pi \tau \right)$). Modulo $\left(\pi \right)$: $a_0-b_0\equiv \alpha -1$, which is nonzero in the field $K_{\pi }$ since $\alpha \ne 1$. Hence $\pi \nmid a_0-b_0$, and so $\pi \tau \nmid a_0-b_0$.

Step 5 (verifying ${\Phi }_p(a_0,b_0)\notin \left(\pi \tau \right)$). We compute ${\Phi }_p(a_0,b_0)$ modulo each of $\left(\pi \right)$ and $\left(\tau \right)$ separately.

Modulo $\left(\pi \right)$: the telescoping identity $(x-y){\Phi }_p(x,y)=x^p-y^p$, applied in the field $K_{\pi }$ to $x=\alpha$ and $y=1$, gives $(\alpha -1)·\overline{{\Phi }_p(a_0,b_0)}={\alpha }^p-1=0$. Since $K_{\pi }$ is a field and $\alpha -1\ne 0$, this forces $\overline{{\Phi }_p(a_0,b_0)}=0$, i.e. $\pi \mid {\Phi }_p(a_0,b_0)$.

Modulo $\left(\tau \right)$: since $a_0\equiv 1\equiv b_0$, every term $a_0^{p-1-j}{b}_0^j$ in ${\Phi }_p(a_0,b_0)={\sum }_{j=0}^{p-1}{a}_0^{p-1-j}{b}_0^j$ is congruent to 1, so ${\Phi }_p(a_0,b_0)\equiv p(mod\left(\tau \right))$. Since $\tau \nmid p$ (as $p\notin \left(\tau \right)$ by hypothesis), this is nonzero, so $\tau \nmid {\Phi }_p(a_0,b_0)$.

Combining: $\pi \mid {\Phi }_p(a_0,b_0)$ but $\tau \nmid {\Phi }_p(a_0,b_0)$, so $\pi \tau \nmid {\Phi }_p(a_0,b_0)$.

Steps 3–5 together show that $a_0^p-b_0^p\in \left(\pi \tau \right)$ while neither $a_0-b_0$ nor ${\Phi }_p(a_0,b_0)$ lies in $\left(\pi \tau \right)$, so $\left(\pi \tau \right)$ is not a p-dfa ideal.    □

Corollary 4.

Let q and t be distinct primes. Then $qt\mathbb{Z}$ is a p-dfa ideal of $\mathbb{Z}$ if and only if $p\in \{q,t\}$, or both $q\neg \equiv 1(modp)$ and $t\neg \equiv 1(modp)$.

Proof. 

Take $R=\mathbb{Z}$, $\pi =q$, $\tau =t$ in Theorem 4. Since q and p are primes, $p\in \left(q\right)=q\mathbb{Z}$ means $q\mid p$, which forces $q=p$; similarly $p\in \left(t\right)$ forces $t=p$. Hence condition (i) of Theorem 4 becomes $p\in \{q,t\}$.

For condition (ii), suppose $p\notin \{q,t\}$, so that $p\notin \left(q\right)$ and $p\notin \left(t\right)$. Since $\mathbb{Z}/q\mathbb{Z}\cong {\mathbb{F}}_q$ is finite with $|{\mathbb{F}}_q^{*}|=q-1$, Lemma 5 shows that the map $x\mapsto x^p$ is injective on ${\mathbb{F}}_q$ if and only if $p\nmid q-1$, i.e. $q\neg \equiv 1(modp)$; the same argument applied to $\mathbb{Z}/t\mathbb{Z}\cong {\mathbb{F}}_t$ gives the analogous condition $t\neg \equiv 1(modp)$. Thus condition (ii) holds precisely when $p\notin \{q,t\}$ and both $q\neg \equiv 1(modp)$ and $t\neg \equiv 1(modp)$.

Combining via Theorem 4, $qt\mathbb{Z}$ is p-dfa if and only if $p\in \{q,t\}$, or both $q\neg \equiv 1(modp)$ and $t\neg \equiv 1(modp)$; the hypothesis $p\notin \{q,t\}$ accompanying condition (ii) may be omitted from the final statement, since it is automatically satisfied whenever $p\in \{q,t\}$ fails to hold.    □

Remark 6.

A complete characterization of all $\left(\alpha \right)$ for which $\left(\alpha \right)$ is a p-dfa ideal of a PID R remains open in general. Even when $\left({\pi }^k\right)$ and $\left({\tau }^j\right)$ are individually p-dfa ideals, the product $\left({\pi }^k{\tau }^j\right)$ need not be: for example in $\mathbb{Z}$, $4\mathbb{Z}$ and $5\mathbb{Z}$ are both 3-dfa ideals, yet $20\mathbb{Z}$ is not.

4. Further Constructions

We study p-dfa ideals in four standard ring-theoretic constructions, beginning with polynomial rings.

For a commutative ring R and an ideal I of R, we write $I\left[X\right]$ for the ideal $\{f\in R[X]\mid \mathrm{all}\mathrm{coefficients}\mathrm{of}f\mathrm{lie}\mathrm{in}I\}$ of $R\left[X\right]$, and $(I,X)$ for the ideal generated by I and X in $R\left[X\right]$, i.e. $(I,X)=\{a_0+a_{1}X+\dots +a_n{X}^n\mid a_0\in I,a_1,\dots ,a_n\in R\}$.

Proposition 2.

Let I be a nonzero proper ideal of a commutative ring R.

(a)

If $I\left[X\right]$ is a p-dfa ideal of $R\left[X\right]$, then I is a p-dfa ideal of R.

(b)

$(I,X)$ is a p-dfa ideal of $R\left[X\right]$ if and only if I is a p-dfa ideal of R.

Proof. (a) Take $f=a$ and $g=b$ as constant polynomials. Then $a^p-b^p\in I$ implies $a^p-b^p\in I\left[X\right]$, so the p-dfa property of $I\left[X\right]$ gives $a-b\in I\left[X\right]\cap R=I$ or ${\Phi }_p(a,b)\in I\left[X\right]\cap R=I$. Hence I is a p-dfa ideal of R.

(b) The forward direction is proved by the same constant-polynomial argument. For the converse, suppose I is a p-dfa ideal of R and let $f^p-g^p\in (I,X)$ for arbitrary $f,g\in R\left[X\right]$. Writing each polynomial as its constant term plus an X-multiple of the rest, say $f=a+Xh\left(X\right)$ and $g=b+Xl\left(X\right)$ with $a,b\in R$ and $h,l\in R\left[X\right]$, the constant term of $f^p-g^p$ equals $a^p-b^p$, which lies in I since $(I,X)\cap R=I$. The p-dfa property of I then gives $a-b\in I$ or ${\Phi }_p(a,b)\in I$. In the first case, $f-g=(a-b)+X\left(h\right(X)-l(X\left)\right)\in (I,X)$. In the second, the constant term of ${\Phi }_p(f,g)={\sum }_{j=0}^{p-1}{f}^{p-1-j}{g}^j$ equals ${\sum }_{j=0}^{p-1}{a}^{p-1-j}{b}^j={\Phi }_p(a,b)\in I$ and the remaining terms are divisible by X, so ${\Phi }_p(f,g)\in (I,X)$.    □

We turn next to direct products.

Proposition 3.

Let $I_1$ and $I_2$ be nonzero proper ideals of commutative rings $R_1$ and $R_2$, respectively.

(a)

If $I_1\times I_2$ is a p-dfa ideal of $R_1\times R_2$, then $I_1$ and $I_2$ are p-dfa ideals of $R_1$ and $R_2$, respectively.

(b)

If $I_1$ and $I_2$ are p-dfa ideals of $R_1$ and $R_2$ respectively, and $p\in I_1$ or $p\in I_2$, then $I_1\times I_2$ is a p-dfa ideal of $R_1\times R_2$.

Proof. (a) We show $I_1$ is a p-dfa ideal of $R_1$; the argument for $I_2$ is symmetric. Let $a^p-b^p\in I_1$ for $a,b\in R_1$. Then ${(a,0)}^p-{(b,0)}^p=(a^p-b^p,0)\in I_1\times I_2$. Since $I_1\times I_2$ is a p-dfa ideal, either $(a-b,0)\in I_1\times I_2$, giving $a-b\in I_1$, or ${\Phi }_p((a,0),(b,0))=({\Phi }_p(a,b),0)\in I_1\times I_2$, giving ${\Phi }_p(a,b)\in I_1$. Hence $I_1$ is a p-dfa ideal.

(b) Suppose $p\in I_1$; the case $p\in I_2$ is symmetric. Let $(a^p-b^p,c^p-d^p)\in I_1\times I_2$ for $a,b\in R_1$ and $c,d\in R_2$. Since $I_1$ is a p-dfa ideal, either $a-b\in I_1$ or ${\Phi }_p(a,b)\in I_1$. Since $I_2$ is a p-dfa ideal, either $c-d\in I_2$ or ${\Phi }_p(c,d)\in I_2$.

If $a-b\in I_1$ and $c-d\in I_2$, then $(a-b,c-d)\in I_1\times I_2$. If ${\Phi }_p(a,b)\in I_1$ and ${\Phi }_p(c,d)\in I_2$, then $({\Phi }_p(a,b),{\Phi }_p(c,d))\in I_1\times I_2$.

It remains to handle the mixed cases. Suppose $a-b\in I_1$ but ${\Phi }_p(c,d)\in I_2$ (the case ${\Phi }_p(a,b)\in I_1$, $c-d\in I_2$ is analogous). Since $p\in I_1$, $a-b\in I_1$, and $a^p-b^p\in I_1$, Lemma 2(b)⇒(a) gives ${\Phi }_p(a,b)\in I_1$. Hence $({\Phi }_p(a,b),{\Phi }_p(c,d))\in I_1\times I_2$.    □

Remark 7.

The converse of Proposition 3(b) does not hold in general: $p\in I_1$ or $p\in I_2$ is sufficient but not necessary for $I_1\times I_2$ to be a p-dfa ideal. For example, $2{\mathbb{Z}}_4\times 2{\mathbb{Z}}_4$ is a 3-dfa ideal of ${\mathbb{Z}}_4\times {\mathbb{Z}}_4$, yet $3\notin 2{\mathbb{Z}}_4$.

We next consider idealizations. Recall that for a commutative ring R and an R-module M, the idealization of R and M is the commutative ring $R(+)M=R\times M$ with componentwise addition and multiplication $(r,m)(s,n)=(rs,rn+sm)$. Every ideal of $R(+)M$ has the form $I(+)N$ where I is an ideal of R and N is a submodule of M with $IM\subseteq N$; see [2,7].

A key feature of the idealization is that ${(r,m)}^p=(r^p,p·r^{p-1}m)$ for all $r\in R$ and $m\in M$, since ${(0,m)}^2=(0,0)$ makes the binomial expansion collapse. Consequently,

$${(r,m)}^p-{(s,n)}^p=\left(r^p-s^p,p(r^{p-1}m-s^{p-1}n)\right).$$

Moreover, the first component of ${\Phi }_p((r,m),(s,n))$ equals ${\Phi }_p(r,s)$.

Theorem 5.

Let R be a commutative ring, I a nonzero proper ideal of R, M an R-module, and N a submodule of M with $IM\subseteq N$. Then $I(+)N$ is an ideal of $R(+)M$, and:

(a)

If $I(+)N$ is a p-dfa ideal of $R(+)M$, then I is a p-dfa ideal of R.

(b)

If I is a p-dfa ideal of R, then $I(+)M$ is a p-dfa ideal of $R(+)M$.

Proof. (a) Let $a^p-b^p\in I$ for some $a,b\in R$. Consider the elements $(a,0)$ and $(b,0)$ in $R(+)M$. Then

$${(a,0)}^p-{(b,0)}^p=(a^p-b^p,0)\in I(+)N,$$

since $a^p-b^p\in I$ and $0\in N$. Since $I(+)N$ is a p-dfa ideal, either $(a,0)-(b,0)=(a-b,0)\in I(+)N$, giving $a-b\in I$, or ${\Phi }_p((a,0),(b,0))\in I(+)N$. Since the first component of ${\Phi }_p((a,0),(b,0))$ equals ${\Phi }_p(a,b)$, the second case gives ${\Phi }_p(a,b)\in I$. Hence I is a p-dfa ideal of R.

(b) Suppose ${(r,m)}^p-{(s,n)}^p\in I(+)M$ for $(r,m),(s,n)\in R(+)M$. Then $r^p-s^p\in I$. Since I is a p-dfa ideal of R, either $r-s\in I$ or ${\Phi }_p(r,s)\in I$. If $r-s\in I$, then $(r,m)-(s,n)=(r-s,m-n)\in I(+)M$. If ${\Phi }_p(r,s)\in I$, then the first component of ${\Phi }_p((r,m),(s,n))$ equals ${\Phi }_p(r,s)\in I$, so ${\Phi }_p((r,m),(s,n))\in I(+)M$. Hence $I(+)M$ is a p-dfa ideal of $R(+)M$.    □

Remark 8.

The converse of Theorem 5(b) need not hold: it is possible for $I(+)N$ to be a p-dfa ideal of $R(+)M$ with $N⊊M$. For example, $2{\mathbb{Z}}_4(+)2{\mathbb{Z}}_4$ is a 3-dfa ideal of ${\mathbb{Z}}_4(+){\mathbb{Z}}_4$ even though $2{\mathbb{Z}}_4⊊{\mathbb{Z}}_4$. Hence $N=M$ is sufficient but not necessary.

Finally, we study p-dfa ideals in amalgamation rings. Let A and B be commutative rings, $f:A\to B$ a ring homomorphism, and J an ideal of B. The amalgamation of A and B along J with respect to f is the subring

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

of $A\times B$; see [3]. For an ideal I of A, the set $I{\bowtie }^{f}J=\{(i,f\left(i\right)+j)\mid i\in I,j\in J\}$ is an ideal of $A{\bowtie }^{f}J$.

Theorem 6.

Let A, B be commutative rings, $f:A\to B$ a ring homomorphism, J an ideal of B, and I a nonzero proper ideal of A. Then $I{\bowtie }^{f}J$ is a p-dfa ideal of $A{\bowtie }^{f}J$ if and only if I is a p-dfa ideal of A.

Proof.$(\Rightarrow )$ Suppose $I{\bowtie }^{f}J$ is a p-dfa ideal of $A{\bowtie }^{f}J$. Let $a^p-b^p\in I$ for $a,b\in A$. Taking $j_1=j_2=0$, let $x=(a,f(a\left)\right)$ and $y=(b,f(b\left)\right)$ in $A{\bowtie }^{f}J$. Then $x^p-y^p=(a^p-b^p,f{\left(a\right)}^p-f{\left(b\right)}^p)=(a^p-b^p,f(a^p-b^p))\in I{\bowtie }^{f}J$. Since $I{\bowtie }^{f}J$ is a p-dfa ideal, either $x-y=(a-b,f(a-b))\in I{\bowtie }^{f}J$, giving $a-b\in I$, or ${\Phi }_p(x,y)=({\Phi }_p(a,b),f\left({\Phi }_p(a,b)\right))\in I{\bowtie }^{f}J$, giving ${\Phi }_p(a,b)\in I$. Hence I is a p-dfa ideal of A.

$(\Leftarrow )$ Suppose I is a p-dfa ideal of A. Let $x^p-y^p\in I{\bowtie }^{f}J$ for $x=(a,f\left(a\right)+j_1)$ and $y=(b,f\left(b\right)+j_2)$ in $A{\bowtie }^{f}J$. Then $a^p-b^p\in I$. Since I is a p-dfa ideal, either $a-b\in I$ or ${\Phi }_p(a,b)\in I$. If $a-b\in I$, then $x-y=(a-b,f(a-b)+(j_1-j_2))\in I{\bowtie }^{f}J$ since $j_1-j_2\in J$. If ${\Phi }_p(a,b)\in I$, then ${\Phi }_p(x,y)=({\Phi }_p(a,b),f\left({\Phi }_p(a,b)\right)+j^{\prime })$ for some $j^{\prime }\in J$ (since f is a ring homomorphism and ${\Phi }_p(f\left(a\right)+j_1,f\left(b\right)+j_2)-f\left({\Phi }_p(a,b)\right)\in J$), so ${\Phi }_p(x,y)\in I{\bowtie }^{f}J$. Hence $I{\bowtie }^{f}J$ is a p-dfa ideal of $A{\bowtie }^{f}J$.    □

5. Conclusion

In this paper we introduced and studied p-difference factor absorbing ideals of commutative rings, unifying and extending the sdf- and cdf-absorbing ideals of [1] and [4]. The central theme throughout has been the role of the prime p within the ring: whether p is zero, a unit, or neither largely determines the behavior of p-dfa ideals. Among the main results, we characterized p-dfa ideals in principal ideal domains of characteristic zero: for irreducibles $\pi$ with $p\notin \left(\pi \right)$, we determined exactly when $\left({\pi }^k\right)$ is p-dfa (Theorem 3), and for those with finite residue fields we characterized when $\left(\pi \tau \right)$ is p-dfa (Theorem 4). We also established the behavior of p-dfa ideals under the standard ring-theoretic constructions of polynomial rings, direct products, idealizations, and amalgamation rings, and showed that the close relationship between sdf-absorbing ideals and prime ideals breaks down for all primes $p\ge 3$.

Several natural questions remain open. Whether every nonzero p-dfa ideal is necessarily a radical ideal for $p\ge 3$ is perhaps the most immediate; the analogous result for $p=2$ is a cornerstone of the sdf-absorbing theory [1, Theorem 2.2], but the argument does not extend. A second open problem is to identify what additional conditions beyond $p\in U\left(R\right)$ force a p-dfa ideal to be prime. Finally, when $p\in \left(\pi \right)$ in a PID, a complete characterization of which $\left({\pi }^k\right)$ are p-dfa remains elusive, as the answer depends on arithmetic properties that go beyond the valuation-theoretic methods used in Section 3. We hope the results and techniques of this paper will serve as a foundation for further investigation.