Square-Difference Factor Absorbing Primary Hyperideals of Multiplicative Hyperrings

1. Introduction

Generalizing prime and primary ideals has been a central and productive theme in commutative algebra. Among the most influential generalizations are 2-absorbing ideals, introduced by Badawi [6], and 2-absorbing primary ideals [7]. These have since inspired an extensive literature on n-absorbing, $(m,n)$-closed, and related classes of ideals, studied both in rings and in more general algebraic structures.

A particularly elegant recent contribution is the notion of square-difference factor absorbing ideals (sdf-absorbing ideals), introduced by Anderson, Badawi and Coykendall [5]. A proper ideal I of a ring R is sdf-absorbing if, for $0\ne a,b\in R$, $a^2-b^2\in I$ implies $a+b\in I$ or $a-b\in I$. The key structural result of [5] is that every sdf-absorbing ideal is a radical ideal, with the converse holding when $\mathrm{char}\left(R\right)=2$, so that every nonzero proper ideal of R is sdf-absorbing if and only if $R/\sqrt{0}$ is von Neumann regular. Extending this, Khashan, Çelikel and Tekir [11] introduced sdf-absorbing primary ideals, where the condition $a+b\in I$ is relaxed to $a+b\in \sqrt{I}$. They showed that the new class coincides with the primary ideals when $2\in U\left(R\right)$, and provided a complete classification in ZPI-rings and Dedekind domains.

Multiplicative hyperrings, introduced by Rota [14] in 1982, generalize rings by allowing the multiplication to be a hyperoperation, i.e., a map $\circ :H\times H\to {\mathcal{P}}^{*}\left(H\right)$. The theory of hyperrings has grown substantially since Marty’s original work on hypergroups [12], and has been developed through the study of prime and primary hyperideals [8], S-prime hyperideals [10], n-absorbing hyperideals [1,13], and primal hyperideals [2], among others. The multiplicative hyperring of fractions was systematically developed in [3]. Most recently, sdf-absorbing hyperideals in multiplicative hyperrings were studied in [4], establishing hyperring analogues of the main results of [5]. The primary analogue of this notion — in which the condition $x+y\in I$ is relaxed to $x+y\in \sqrt{I}$ — has not yet been investigated in the hyperring setting, leaving the natural completion of this programme open.

In this paper, we introduce and study sdf-absorbing primary hyperideals in commutative multiplicative hyperrings, providing the hyperring analogue of the ring-theoretic work of [11]. Having a single class that subsumes both primary hyperideals and sdf-absorbing hyperideals as special cases clarifies the logical hierarchy of hyperideal theory and enables a unified treatment of results that previously required separate arguments for each class. We show that this class is a proper common generalization of both primary hyperideals and sdf-absorbing hyperideals (Theorem 16 and Example 1), and we prove that in characteristic 2 every proper strong C-hyperideal is sdf-absorbing primary (Proposition 3), while in strongly distributive multiplicative hyperrings with scalar identity the notions of sdf-absorbing primary and primary coincide when $2\in U\left(H\right)$ (Proposition 5 and Theorem 16(ii)). We further establish that the radical of every sdf-absorbing primary strong C-hyperideal is sdf-absorbing when H is strongly distributive with scalar identity, $2\in U\left(H\right)$, all hyperproducts are finite, and $\sqrt{I}$ is itself a strong C-hyperideal (Theorem 7), and we provide a complete characterization via three equivalent conditions (Theorem 16), summarized in a hierarchy table and diagram. We further show that the class is stable under finite intersections sharing a common radical (Proposition 11) and under ascending chains (Proposition 14), and we study the behavior under good homomorphisms (Theorem 9, Corollary 10), localization (Theorem 18, Corollary 20), and Cartesian products (Theorems 22–25). Throughout, parallels with the commutative ring case are made explicit, and the additional hypotheses needed for general multiplicative hyperrings (strong distributivity, scalar identity, finiteness of hyperproducts) are identified in each result.

The paper is organized as follows. Section 2 recalls the basic definitions and fixes notation. Section 3 introduces the new notion, develops its basic properties, and culminates in the characterization theorem together with a hierarchy table and diagram. Section 4 investigates the behavior under localization and Cartesian products. Section 5 provides concluding remarks and identifies open problems.

2. Preliminaries

Throughout this paper, whenever we refer to a ring, we mean a commutative ring with nonzero identity. We briefly recall the basic notions of multiplicative hyperring theory and fix notation; we denote by ${\mathcal{P}}^{*}\left(H\right)$ the set of all non-empty subsets of a set H. A non-empty set H equipped with an operation + and a hyperoperation $\circ :H\times H\to {\mathcal{P}}^{*}\left(H\right)$ is called a commutative multiplicative hyperring if $(H,+)$ is an abelian group; $(H,\circ )$ is a commutative semihypergroup (i.e., an associative commutative hypergroupoid); $x\circ 0=\left\{0\right\}$ for all $x\in H$; and for all $x,y,z\in H$:

$$x\circ (y+z)\subseteq x\circ y+x\circ z,(y+z)\circ x\subseteq y\circ x+z\circ x,x\circ (-y)=-(x\circ y)=(-x)\circ y.$$

When equality holds in the first two conditions above, H is called strongly distributive [14]. Throughout, H denotes a commutative multiplicative hyperring with identity element 1. We write $x^2:=x\circ x$ for any $x\in H$, and more generally $x^n:=x\circ x\circ \dots \circ x$ (n factors) for any $n\in \mathbb{N}$; thus $x^n\in {\mathcal{C}}_H$ for all $n\ge 1$. We also write $U\left(H\right)$ for the group of units of H. For non-empty subsets $A,B\subseteq H$ we write $A+B=\{a+b:a\in A,b\in B\}$ and $A-B=\{a-b:a\in A,b\in B\}$. The identity element 1 is called a scalar identity if $1\circ x=\left\{x\right\}$ for all $x\in H$. When H is strongly distributive with scalar identity, one has $n\circ x=\left\{nx\right\}$ for every $n\in \mathbb{Z}$ and $x\in H$, where $nx$ denotes the n-fold sum in the additive group $(H,+)$.

A fundamental class of examples is given by the hyperrings ${\mathbb{Z}}_{\Omega }$. For any non-empty $\Omega \subseteq \mathbb{Z}\setminus \left\{0\right\}$ with $1\in \Omega$, the triple $({\mathbb{Z}}_{\Omega },+,\circ )$ is a multiplicative hyperring where ${\mathbb{Z}}_{\Omega }=\mathbb{Z}$ and $x\circ y=\{x·\gamma ·y\mid \gamma \in \Omega \}$ for all $x,y\in \mathbb{Z}$ [8]; the condition $1\in \Omega$ ensures that 1 is a multiplicative identity (since $1·1·x=x\in 1\circ x$). When $\Omega =\left\{1\right\}$, this recovers the ordinary ring $\mathbb{Z}$. Note that ${\mathbb{Z}}_{\Omega }$ is not strongly distributive whenever $\left|\Omega \right|\ge 2$; indeed, $(1+1)\circ x=\{2\gamma x:\gamma \in \Omega \}$ while $1\circ x+1\circ x=\{{\gamma }_{1}x+{\gamma }_{2}x:{\gamma }_1,{\gamma }_2\in \Omega \}$, and the latter is strictly larger for any $x\ne 0$ when $\left|\Omega \right|\ge 2$. Example 1 in Section 3 uses the genuinely non-ring case $\Omega =\{1,-1\}$; Example 13 in Section 3 and the localization Example 21 in Section 4 use the ring case $\Omega =\left\{1\right\}$.

A non-empty subset I of H is called a hyperideal if $x-y\in I$ for all $x,y\in I$, and $r\circ x\subseteq I$ for all $r\in H$ and $x\in I$. A hyperideal is called proper if $I\ne H$. For a proper hyperideal I, the set of I-zero-divisors is

$$Z_I\left(H\right)=\{s\in H\mid \mathrm{there}\mathrm{exists}a\in H\setminus I\mathrm{with}s\circ a\subseteq I\}.$$

A proper hyperideal P of H is called a prime hyperideal if $x\circ y\subseteq P$ for $x,y\in H$ implies $x\in P$ or $y\in P$ [8].

Let ${\mathcal{C}}_H=\{c_1\circ c_2\circ \dots \circ c_n\mid c_i\in H,n\in \mathbb{N}\}$ be the collection of all hyperproducts in H. A hyperideal I of H is called a C-hyperideal if, for every $C\in {\mathcal{C}}_H$ with $C\cap I\ne \varnothing$, one has $C\subseteq I$ [8]. Setting ${\tilde{\mathcal{C}}}_H=\{C_1+\dots +C_n\mid n\in \mathbb{N},C_i\in {\mathcal{C}}_H\}$, a hyperideal I is called a strong C-hyperideal if $D\cap I\ne \varnothing$ for $D\in {\tilde{\mathcal{C}}}_H$ implies $D\subseteq I$ [8]. Every strong C-hyperideal is a C-hyperideal, but the converse need not hold.

For a proper hyperideal I of H, the radical of I is defined as

$$\sqrt{I}=\bigcap \{P\subseteq H\mid P\mathrm{is}\mathrm{a}\mathrm{prime}\mathrm{hyperideal},I\subseteq P\}.$$

If no prime hyperideal of H contains I, we set $\sqrt{I}=H$. When I is a C-hyperideal, one has the equivalent description $\sqrt{I}=\{x\in H\mid x^n\subseteq I\mathrm{for}\mathrm{some}n\in \mathbb{N}\}$ [8]. A proper hyperideal I is then called a primary hyperideal if $x\circ y\subseteq I$ for $x,y\in H$ implies $x\in I$ or $y\in \sqrt{I}$ [15].

A proper hyperideal I of H is called an sdf-absorbing hyperideal if, for $0\ne x,y\in H$, $x^2-y^2\subseteq I$ implies $x+y\in I$ or $x-y\in I$ [4]. The characteristic of H, written $\mathrm{char}\left(H\right)$, is the smallest positive integer n with $nx=0$ for all $x\in H$, or 0 if no such integer exists [3]. We write $2:=1+1\in H$ for the element obtained by adding the identity to itself; note that 2 need not be a unit or even non-zero.

3. Basic Properties and Characterizations

We begin with the central definition of this paper.

Definition 1.

A proper hyperideal I of H is called asquare-difference factor absorbing primary hyperideal(orsdf-absorbing primary hyperidealfor short) if, for all $x,y\in H$, $x^2-y^2\subseteq I$ implies

$$x+y\in \sqrt{I}\mathrm{or}x-y\in I.$$

Every sdf-absorbing hyperideal of H is an sdf-absorbing primary hyperideal of H. Indeed, for nonzero x and y the sdf-absorbing condition gives $x+y\in I\subseteq \sqrt{I}$ or $x-y\in I$. When $x=0$ (or symmetrically when $y=0$), we have $0^2-y^2=-y^2\subseteq I$, which forces $y^2\subseteq I$ (since I is closed under negation), and therefore $x+y=y\in \sqrt{I}$. The following example shows that the converse does not hold in general.

Example 1.

Consider the multiplicative hyperring ${\mathbb{Z}}_{\Omega }$ with $\Omega =\{1,-1\}$, and let $I=12\mathbb{Z}$. For any $x,y\in \mathbb{Z}$ the hyperoperation takes the form $x\circ y=\{xy,-xy\}$, so in particular

$$x\circ x=\{x^2,-x^2\},$$

where $x^2$ denotes the ordinary integer square $x·x$. The set-difference $(x\circ x)-(y\circ y)$ therefore consists of all elements of the form ${\epsilon }_1{x}^2-{\epsilon }_2{y}^2$ with ${\epsilon }_1,{\epsilon }_2\in \{1,-1\}$, giving

$$(x\circ x)-(y\circ y)=\left\{x^2-y^2,x^2+y^2,-x^2+y^2,-x^2-y^2\right\}.$$

Claim:$(x\circ x)-(y\circ y)\subseteq 12\mathbb{Z}$ if and only if $6\mid x$ and $6\mid y$.

For the forward direction, the inclusion requires in particular that $12\mid x^2-y^2$ and $12\mid x^2+y^2$. Adding these two conditions gives $12\mid 2x^2$, that is, $6\mid x^2$, while subtracting gives $12\mid 2y^2$, that is, $6\mid y^2$. Since 2 and 3 are prime and both divide $x^2$ (respectively $y^2$), it follows that $6\mid x$ and $6\mid y$. Conversely, if $6\mid x$ and $6\mid y$ then $36\mid x^2$ and $36\mid y^2$, so 12 divides each of $x^2\pm y^2$, and all four elements of $(x\circ x)-(y\circ y)$ belong to $12\mathbb{Z}$.

To compute the radical, observe that $x\circ y=\{xy,-xy\}\subseteq n\mathbb{Z}$ holds if and only if $n\mid xy$. Since every hyperideal of ${\mathbb{Z}}_{\Omega }$ is an additive subgroup of $\mathbb{Z}$ closed under ∘, hence of the form $n\mathbb{Z}$, the nonzero prime hyperideals are exactly $p\mathbb{Z}$ for p prime. The primes containing $12\mathbb{Z}$ are $2\mathbb{Z}$ and $3\mathbb{Z}$ (since $12=2^2·3$), so $\sqrt{12\mathbb{Z}}=2\mathbb{Z}\cap 3\mathbb{Z}=6\mathbb{Z}$.

I is sdf-absorbing primary.If $(x\circ x)-(y\circ y)\subseteq I$, the Claim gives $6\mid x$ and $6\mid y$, so $6\mid (x+y)$, that is, $x+y\in 6\mathbb{Z}=\sqrt{I}$. Thus the sdf-absorbing primary condition holds via the first alternative for every pair $x,y\in \mathbb{Z}$.

I is not sdf-absorbing.Taking $x=6$ and $y=12$, one has $6\mid x$ and $6\mid y$, so $(x\circ x)-(y\circ y)\subseteq I$ by the Claim. However, $x+y=18\notin 12\mathbb{Z}$ and $x-y=-6\notin 12\mathbb{Z}$, so neither alternative of the sdf-absorbing condition holds.

Remark 2.

The role of the hyperring structure in Example 1 is more subtle than it might first appear. In the ordinary ring $\mathbb{Z}$, which corresponds to the special case $\Omega =\left\{1\right\}$, the ideal $12\mathbb{Z}$ isnotsdf-absorbing primary. To see this, take $x=7$ and $y=1$: one has $x^2-y^2=48\in 12\mathbb{Z}$, yet $x+y=8\notin 6\mathbb{Z}=\sqrt{12\mathbb{Z}}$ and $x-y=6\notin 12\mathbb{Z}$, so the sdf-absorbing primary condition fails. This is consistent with the classification of [11]: since $12=2^2·3$ is neither a prime power nor of the form $2q^m$ for an odd prime q, the ideal $12\mathbb{Z}$ is not sdf-absorbing primary in $\mathbb{Z}$.

The contrast with the hyperring case is instructive. In ${\mathbb{Z}}_{\{1,-1\}}$ the containment condition $x^2-y^2\subseteq I$ is considerably stronger than its ring counterpart $x^2-y^2\in I$: it requiresall fourelements $\pm x^2\pm y^2$ to lie in I simultaneously, which forces $6\mid x$ and $6\mid y$ and thereby ensures $x+y\in \sqrt{I}$ automatically. The hyperoperation thus imposes enough additional arithmetic control on the set-valued product to make $12\mathbb{Z}$ sdf-absorbing primary, a property it lacks in the ring. This example therefore illustrates concretely that the passage from commutative rings to multiplicative hyperrings can strictly enlarge the class of sdf-absorbing primary ideals.

The remainder of this section develops the basic properties and characterizations of this class. The first result shows that in characteristic 2, every proper strong C-hyperideal is automatically sdf-absorbing primary.

Proposition 3.

If H is of characteristic 2 and I is a proper strong C-hyperideal of H, then I is an sdf-absorbing primary hyperideal of H.

Proof. 

Let $x,y\in H$ with $x^2-y^2\subseteq I$. Define

$$D=x^2+y^2+x\circ y+x\circ y\in {\tilde{\mathcal{C}}}_H,$$

where D is a sum of four hyperproducts and hence belongs to ${\tilde{\mathcal{C}}}_H$ (not to ${\mathcal{C}}_H$). For any $t\in x^2$, $s\in y^2$, $r\in x\circ y$, since $\mathrm{char}\left(H\right)=2$ we have $r+r=0$, and so

$$t+s+r+r=t+s=t-s\in x^2-y^2\subseteq I.$$

Hence $D\cap I\ne \varnothing$. Since I is a strong C-hyperideal, $D\subseteq I$. By distributivity,

$${(x+y)}^2\subseteq x^2+x\circ y+x\circ y+y^2=D\subseteq I,$$

and therefore $x+y\in \sqrt{I}$. □

Remark 4.

The strong C-hyperideal condition in Proposition 3 is needed to handle the set-valued nature of the multiplication. In the commutative ring case every ideal satisfies this condition automatically, so Proposition 3 recovers ([11] Proposition 1) as a special case.

A natural companion to Proposition 3 asks what happens at the opposite extreme: when 2 is a unit. In this case the sdf-absorbing primary condition turns out to be equivalent to the classical primary condition.

Proposition 5.

Let H be a strongly distributive commutative multiplicative hyperring with scalar identity and $2\in U\left(H\right)$, and let I be a proper sdf-absorbing primary strong C-hyperideal of H. Then I is a primary hyperideal of H.

Proof. 

Let $x,y\in H$ with $x\circ y\subseteq I$; we show $x\in \sqrt{I}$ or $y\in I$.

Define the set $D={(x+y)}^2-{(x-y)}^2$. Since ${(x+y)}^2=(x+y)\circ (x+y)\in {\mathcal{C}}_H$ and $(-{(x-y)}^2)=(y-x)\circ (x-y)\in {\mathcal{C}}_H$, the set D is a sum of two hyperproducts, hence $D\in {\tilde{\mathcal{C}}}_H$.

By strong distributivity and commutativity:

$$\begin{array}{cc}\hfill {(x+y)}^2& =x^2+(x\circ y)+(x\circ y)+y^2,\hfill \\ \hfill {(x-y)}^2& =x^2+((-x)\circ y)+((-x)\circ y)+y^2,\hfill \end{array}$$

where $(-x)\circ y=-(x\circ y)$ by the hyperring axioms; so every element of D has the form $({\alpha }_1-{\alpha }_2)+(c_1+c_2+c_3+c_4)+({\beta }_1-{\beta }_2)$ where ${\alpha }_i\in x^2$, ${\beta }_i\in y^2$, and $c_j\in x\circ y\subseteq I$. Choosing ${\alpha }_1={\alpha }_2$ and ${\beta }_1={\beta }_2$ yields the element $c_1+c_2+c_3+c_4\in I$ (since I, being a hyperideal, is closed under addition). Hence $D\cap I\ne \varnothing$. Since I is a strong C-hyperideal and $D\in {\tilde{\mathcal{C}}}_H$, we conclude $D\subseteq I$.

Setting $a=x+y$ and $b=x-y$, we have $a^2-b^2\subseteq I$. Since I is sdf-absorbing primary: $a+b\in \sqrt{I}$ or $a-b\in I$, that is, $2x\in \sqrt{I}$ or $2y\in I$.

Suppose first that $2y\in I$. Since H has scalar identity, $2\circ y=\left\{2y\right\}$ (see Preliminaries). Then $2^{-1}\circ \left(2y\right)\subseteq I$ (as I is a hyperideal and $2y\in I$). By associativity, $2^{-1}\circ \left(2y\right)=2^{-1}\circ (2\circ y)=(2^{-1}\circ 2)\circ y$; since $1\in 2^{-1}\circ 2$ and $1\circ y=\left\{y\right\}$, it follows that $y\in (2^{-1}\circ 2)\circ y\subseteq 2^{-1}\circ \left(2y\right)\subseteq I$. Hence $y\in I$.

Suppose instead that $2x\in \sqrt{I}$. The same argument, applied with $\sqrt{I}$ in place of I (noting that $\sqrt{I}$ is a hyperideal), gives $x\in \sqrt{I}$. □

Remark 6.

Proposition 5 uses the strong C-hyperideal structure in an essential way: the key step $D\cap I\ne \varnothing$ followed by $D\subseteq I$ replaces the ring-theoretic identity $a^2-b^2=(a-b)·(a+b)$. In a general multiplicative hyperring, the distributivity inclusion gives only $(a-b)\circ (a+b)\subseteq a^2+a\circ b-a\circ b-b^2$, where the set $a\circ b-a\circ b=\{c-c^{\prime }:c,c^{\prime }\in a\circ b\}$ may contain nonzero elements when $a\circ b$ is not a singleton; thus the ring identity $a^2-b^2=(a-b)(a+b)$ does not extend directly. The scalar identity condition ensures that $2\circ y=\left\{2y\right\}$, so that invertibility of 2 can be fully exploited. When H is a commutative ring (i.e. all hyperproducts are singletons), these conditions are automatic and the proposition recovers ([11] Proposition 2) as a special case.

We now examine how sdf-absorbing primary hyperideals interact with the radical operation. The following theorem shows that passing to the radical preserves — and in fact strengthens — the absorbing property.

Theorem 7.

Let H be a strongly distributive commutative multiplicative hyperring with scalar identity and $2\in U\left(H\right)$, and let I be a proper sdf-absorbing primary strong C-hyperideal of H such that $\sqrt{I}$ is also a strong C-hyperideal. If every hyperproduct of two elements in H is a finite set, then $\sqrt{I}$ is an sdf-absorbing hyperideal of H.

Proof. 

We first note that $\sqrt{I}$ is a proper hyperideal. Since I is a C-hyperideal and I is proper ($1\notin I$), the element-wise radical description gives $1\notin \sqrt{I}$: since $1\in x\circ 1$ for every $x\in H$ (weak identity), an induction gives $1\in 1^n$ for every $n\in \mathbb{N}$; hence $1^n\neg \subseteq I$ for all n, and so $1\notin \sqrt{I}$. The proof proceeds in three steps.

Step 1: I is a primary hyperideal. This is immediate from Proposition 5.

Step 2: $\sqrt{I}$ is a prime hyperideal. Let $x,y\in H$ with $x\circ y\subseteq \sqrt{I}$. Since $x\circ y$ is finite by hypothesis, for each $c\in x\circ y$ there exists $n_c\in \mathbb{N}$ with $c^{n_c}\subseteq I$. Setting $N=max\{n_c:c\in x\circ y\}$, we have $c^N\subseteq I$ for every $c\in x\circ y$.

By associativity and commutativity of the semihypergroup $(H,\circ )$, the N-fold set-level hyperproduct ${(x\circ y)}^N$ can be rearranged as $x^N\circ y^N$ (grouping all copies of x together and all copies of y together): since the set-level operation inherits commutativity and associativity from the element-level axioms, one can permute the $2N$ factors $x,y,x,y,\dots$ to $x,x,\dots ,x,y,y,\dots ,y$. For any fixed $c\in x\circ y$, the element-wise power satisfies $c^N\subseteq {(x\circ y)}^N=x^N\circ y^N$ (choosing c in each of the N factors). Since $c^N\subseteq I$, we have $x^N\circ y^N\cap I\ne \varnothing$, and since I is a C-hyperideal and $x^N\circ y^N\in {\mathcal{C}}_H$, we conclude $x^N\circ y^N\subseteq I$.

For each pair $\alpha \in x^N$ and $\beta \in y^N$, we have $\alpha \circ \beta \subseteq x^N\circ y^N\subseteq I$. Since I is primary (Step 1), either $\alpha \in I$ or $\beta \in \sqrt{I}$.

If $x^N\subseteq I$ (i.e., every $\alpha \in x^N$ lies in I), then $x\in \sqrt{I}$ and we are done. Otherwise, there exists ${\alpha }_0\in x^N$ with ${\alpha }_0\notin I$, and then ${\alpha }_0\circ \beta \subseteq I$ forces $\beta \in \sqrt{I}$ for every $\beta \in y^N$, giving $y^N\subseteq \sqrt{I}$. In this case, for each $\beta \in y^N$ there exists $m_{\beta }\in \mathbb{N}$ with ${\beta }^{m_{\beta }}\subseteq I$. Since $y^N$ is finite (as all hyperproducts of two elements are finite, by induction so are all iterated hyperproducts), setting $M=max\{m_{\beta }:\beta \in y^N\}$ yields ${\beta }^M\subseteq I$ for every $\beta \in y^N$. For each $\beta$, ${\beta }^M\subseteq {\left(y^N\right)}^M=y^{NM}$ (choosing $\beta$ in each factor), so $y^{NM}\cap I\ne \varnothing$. Since I is a C-hyperideal and $y^{NM}\in {\mathcal{C}}_H$, we get $y^{NM}\subseteq I$, hence $y\in \sqrt{I}$.

In either case, $x\in \sqrt{I}$ or $y\in \sqrt{I}$, so $\sqrt{I}$ is prime.

Step 3: $\sqrt{I}$ is sdf-absorbing. Let $0\ne a,b\in H$ with $a^2-b^2\subseteq \sqrt{I}$. Define $D=a^2+(a\circ b)+((-a)\circ b)+((-b)\circ b)\in {\tilde{\mathcal{C}}}_H$ (a sum of four hyperproducts, with $(-a)\circ b=-(a\circ b)$ and $(-b)\circ b=-b^2$). By distributivity, $(a-b)\circ (a+b)\subseteq D$. For any $\alpha \in a^2$, $\beta \in b^2$, and $c\in a\circ b$, the element $\alpha +c-c-\beta =\alpha -\beta$ lies in $a^2-b^2\subseteq \sqrt{I}$, so $D\cap \sqrt{I}\ne \varnothing$. Since $\sqrt{I}$ is a strong C-hyperideal, $D\subseteq \sqrt{I}$, and hence $(a-b)\circ (a+b)\subseteq \sqrt{I}$. Since $\sqrt{I}$ is prime (Step 2): $a-b\in \sqrt{I}$ or $a+b\in \sqrt{I}$.

Therefore $\sqrt{I}$ is sdf-absorbing. □

Remark 8.

The proof above avoids the binomial theorem entirely, instead using the strong C-hyperideal property at two critical junctures: first to show $x^N\circ y^N\subseteq I$ in Step 2 (replacing the ring identity ${\left(xy\right)}^N=x^N{y}^N$), and then to show $(a-b)\circ (a+b)\subseteq \sqrt{I}$ in Step 3 (replacing the factorization $a^2-b^2=(a-b)(a+b)$). The hypothesis that $\sqrt{I}$ be a strong C-hyperideal is essential for Step 3 but is automatic in the ring case. When H is a commutative ring with $2\in U\left(H\right)$, all hyperproducts are singletons, and the theorem recovers ([11] Theorem 1) for that case. The general ring case (without $2\in U\left(H\right)$) of ([11] Theorem 1) relies on the binomial theorem, which does not extend to hyperrings.

We now turn to the behavior of sdf-absorbing primary hyperideals under good homomorphisms. Recall that a map $\theta :(H_1,+,\circ )\to (H_2,+,\circ )$ is called a good homomorphism if $\theta (x+y)=\theta \left(x\right)+\theta \left(y\right)$ and $\theta (x\circ y)=\theta \left(x\right)\circ \theta \left(y\right)$ for all $x,y\in H_1$, where the latter means $\left\{\theta \right(t)\mid t\in x\circ y\}=\theta \left(x\right)\circ \theta \left(y\right)$ [9].

Theorem 9.

Let $\theta :(H_1,+,\circ )\to (H_2,+,\circ )$ be a good homomorphism with $\theta \left(1_{H_1}\right)=1_{H_2}$.

(i)

If I is an sdf-absorbing primary C-hyperideal of $H_2$, then ${\theta }^{-1}\left(I\right)$ is an sdf-absorbing primary C-hyperideal of $H_1$.

(ii)

If θ is surjective, $ker\left(\theta \right)\subseteq J$, and J is an sdf-absorbing primary C-hyperideal of $H_1$, then $\theta \left(J\right)$ is an sdf-absorbing primary hyperideal of $H_2$.

Proof.(i). Note first that ${\theta }^{-1}\left(I\right)$ is proper: since $\theta$ is a good homomorphism, $\theta \left(1_{H_1}\right)=1_{H_2}\notin I$ (as I is proper), so $1_{H_1}\notin {\theta }^{-1}\left(I\right)$. Moreover, ${\theta }^{-1}\left(I\right)$ is a C-hyperideal: if $C\in {\mathcal{C}}_{H_1}$ with $C\cap {\theta }^{-1}\left(I\right)\ne \varnothing$, then $\theta \left(C\right)\in {\mathcal{C}}_{H_2}$ (since $\theta$ is a good homomorphism) and $\theta \left(C\right)\cap I\ne \varnothing$, so $\theta \left(C\right)\subseteq I$ (since I is a C-hyperideal), giving $C\subseteq {\theta }^{-1}\left(I\right)$.

Let $x,y\in H_1$ with $x^2-y^2\subseteq {\theta }^{-1}\left(I\right)$. Since $\theta$ is a good homomorphism, applying $\theta$ gives $\theta {\left(x\right)}^2-\theta {\left(y\right)}^2=\theta (x^2-y^2)\subseteq I$. Since I is sdf-absorbing primary, we have $\theta \left(x\right)+\theta \left(y\right)\in \sqrt{I}$ or $\theta \left(x\right)-\theta \left(y\right)\in I$.

In the second case, $\theta (x-y)=\theta \left(x\right)-\theta \left(y\right)\in I$, so $x-y\in {\theta }^{-1}\left(I\right)$. In the first case, $\theta (x+y)\in \sqrt{I}$, so $\theta {(x+y)}^n\subseteq I$ for some n (using the element-wise radical description, which is valid since I is a C-hyperideal [8]). Then $\theta \left({(x+y)}^n\right)=\theta {(x+y)}^n\subseteq I$, giving ${(x+y)}^n\subseteq {\theta }^{-1}\left(I\right)$. For any prime hyperideal Q of $H_1$ with ${\theta }^{-1}\left(I\right)\subseteq Q$: ${(x+y)}^n\subseteq Q$. Applying the primality of Q repeatedly (descending from n to 1): if $x+y\notin Q$, then for each $\alpha \in {(x+y)}^{n-1}$ the inclusion $\alpha \circ (x+y)\subseteq {(x+y)}^n\subseteq Q$ and $x+y\notin Q$ force $\alpha \in Q$, giving ${(x+y)}^{n-1}\subseteq Q$; repeating this argument reaches ${(x+y)}^1=\{x+y\}\subseteq Q$, contradicting $x+y\notin Q$. Hence $x+y\in Q$. Since this holds for every such Q, we conclude $x+y\in \sqrt{{\theta }^{-1}\left(I\right)}$. Hence ${\theta }^{-1}\left(I\right)$ is sdf-absorbing primary.

(ii). Let $x,y\in H_2$ with $x^2-y^2\subseteq \theta \left(J\right)$. Since $\theta$ is surjective, write $x=\theta \left(a\right)$ and $y=\theta \left(b\right)$ for some $a,b\in H_1$. We first note that $\theta \left(J\right)$ is a proper hyperideal of $H_2$: if $1_{H_2}\in \theta \left(J\right)$, then $\theta \left(s\right)=1_{H_2}=\theta \left(1_{H_1}\right)$ for some $s\in J$, giving $s-1_{H_1}\in ker\left(\theta \right)\subseteq J$, hence $1_{H_1}=s-(s-1_{H_1})\in J$ (hyperideal), contradicting J proper. For any $t\in a^2-b^2$, we have $\theta \left(t\right)\in \theta {\left(a\right)}^2-\theta {\left(b\right)}^2\subseteq \theta \left(J\right)$, so $\theta \left(t\right)=\theta \left(s\right)$ for some $s\in J$, giving $t-s\in ker\left(\theta \right)\subseteq J$. Hence $t=(t-s)+s\in J$. Therefore $a^2-b^2\subseteq J$.

Since J is sdf-absorbing primary, we have $a+b\in \sqrt{J}$ or $a-b\in J$. If $a+b\in \sqrt{J}$, then ${(a+b)}^n\subseteq J$ for some n (since J is a C-hyperideal), giving $\theta {(a+b)}^n=\theta \left({(a+b)}^n\right)\subseteq \theta \left(J\right)$. For any prime hyperideal P of $H_2$ with $\theta \left(J\right)\subseteq P$: $\theta {(a+b)}^n\subseteq P$, and the same descending primality argument as in part (i) gives $\theta (a+b)\in P$. Since this holds for every such P, we conclude $x+y=\theta (a+b)\in \sqrt{\theta \left(J\right)}$. If $a-b\in J$, then $x-y=\theta (a-b)\in \theta \left(J\right)$. Hence $\theta \left(J\right)$ is sdf-absorbing primary. □

The following corollary gathers the main structural consequences.

Corollary 10.

Let I be a proper C-hyperideal of H and $Q\subseteq I$ a hyperideal of H.

(i)

If I is sdf-absorbing primary and K is a multiplicative sub-hyperring of H (containing $1_H$), then $I\cap K$ is an sdf-absorbing primary C-hyperideal of K.

(ii)

If I is sdf-absorbing primary, then $I/Q$ is an sdf-absorbing primary hyperideal of $H/Q$.

lcbel=()

I is an sdf-absorbing primary hyperideal of H if and only if $I/Q$ is an sdf-absorbing primary hyperideal of $H/Q$.

Proof. 

For part (i), observe that the inclusion map $\iota :K\hookrightarrow H$ is a good homomorphism with $\iota \left(1_K\right)=1_H$. Theorem 9(i) gives that ${\iota }^{-1}\left(I\right)=I\cap K$ is an sdf-absorbing primary C-hyperideal of K.

For part (ii), consider the canonical surjection $\pi :H\to H/Q$ defined by $\pi \left(x\right)=x+Q$. This is a surjective good homomorphism with $ker\left(\pi \right)=Q\subseteq I$. Applying Theorem 9(ii) to $\pi$ and I yields that $\pi \left(I\right)=I/Q$ is sdf-absorbing primary in $H/Q$.

For part (iii), one direction is part (ii). For the converse, suppose $I/Q$ is sdf-absorbing primary in $H/Q$. Since I is a C-hyperideal of H containing Q, the quotient $I/Q$ is a C-hyperideal of $H/Q$ (for any $C^{\prime }\in {\mathcal{C}}_{H/Q}$ with $C^{\prime }\cap (I/Q)\ne \varnothing$: writing $C^{\prime }=\pi \left(C\right)$ for $C\in {\mathcal{C}}_H$, we have $C\cap I\ne \varnothing$, so $C\subseteq I$, giving $C^{\prime }\subseteq I/Q$). Theorem 9(i) applied to $\pi$ and $I/Q$ gives that ${\pi }^{-1}(I/Q)=I$ is sdf-absorbing primary in H. □

We next investigate how sdf-absorbing primary hyperideals behave under the operations of intersection and union. The key finding is that the intersection preserves the property precisely when all hyperideals share the same radical, while the union is always preserved along chains.

When all hyperideals in a family share the same radical Q, it is convenient to call them Q-sdf-absorbing primary hyperideals. The following proposition shows that this class is closed under finite intersections.

Proposition 11.

Let Q be a strong C-hyperideal of H, and let ${\left\{I_j\right\}}_{j=1}^n$ be a finite family of Q-sdf-absorbing primary strong C-hyperideals of H. Then $I={\bigcap }_{j=1}^n{I}_j$ is also a Q-sdf-absorbing primary strong C-hyperideal of H.

Proof. 

We first verify that I is a strong C-hyperideal. Let $D\in {\tilde{\mathcal{C}}}_H$ with $D\cap I\ne \varnothing$. Since $I\subseteq I_j$ for every j, we have $D\cap I_j\ne \varnothing$ for every j, hence $D\subseteq I_j$ for every j (as each $I_j$ is a strong C-hyperideal), giving $D\subseteq {\bigcap }_j{I}_j=I$.

Let $x,y\in H$ with $x^2-y^2\subseteq I$. Then $x^2-y^2\subseteq I_j$ for every j. Since each $I_j$ is sdf-absorbing primary, we have $x+y\in \sqrt{I_j}=Q$ or $x-y\in I_j$. We first show $\sqrt{I}=Q$. The inclusion $\sqrt{I}\subseteq Q$ follows from $I=\bigcap I_j\subseteq I_j$, which gives $\sqrt{I}\subseteq \sqrt{I_j}=Q$. Conversely, for any $q\in Q$ and each j, since $q\in \sqrt{I_j}$ there exists $n_j\in \mathbb{N}$ with $q^{n_j}\subseteq I_j$. Setting $N={max}_j{n}_j$ (finite family), for each j we have $q^N=q^{n_j}\circ q^{N-n_j}\subseteq I_j\circ H\subseteq I_j$ (since $I_j$ is a hyperideal). Hence $q^N\subseteq {\bigcap }_{j=1}^n{I}_j=I$, giving $q\in \sqrt{I}$. Therefore $\sqrt{I}=Q$.

Now if $x+y\in Q=\sqrt{I}$, we are done. If $x+y\notin Q$, then $x-y\in I_j$ for every j, and therefore $x-y\in {\bigcap }_{j=1}^n{I}_j=I$. Hence I is sdf-absorbing primary. □

The assumption that all $I_j$ share the same radical is essential, as Example 13 below shows.

Remark 12.

The radical Q in Proposition 11 need not be prime. For instance, $18\mathbb{Z}$ is an sdf-absorbing primary ideal of $\mathbb{Z}$ with $\sqrt{18\mathbb{Z}}=6\mathbb{Z}$, which is not a prime ideal. Thus the Q-sdf-absorbing primary family $\left\{18\mathbb{Z}\right\}$ has a non-prime common radical; the proposition requires only that Q itself be a strong C-hyperideal.

Example 13.

In the ring $\mathbb{Z}$ (viewed as the multiplicative hyperring ${\mathbb{Z}}_{\Omega }$ with $\Omega =\left\{1\right\}$), a complete classification of sdf-absorbing primary ideals is known ([11] Example 5): a proper ideal is sdf-absorbing primary if and only if it is primary or of the form $2q^m\mathbb{Z}$ for some odd prime q and $m\ge 1$. In particular, $I_1=6\mathbb{Z}=2·3\mathbb{Z}$ and $I_2=10\mathbb{Z}=2·5\mathbb{Z}$ are both sdf-absorbing primary (each of the form $2q^m\mathbb{Z}$ with $m=1$). However, their intersection $I_1\cap I_2=30\mathbb{Z}$ isnotsdf-absorbing primary: since $30=2·3·5$ is squarefree, $\sqrt{30\mathbb{Z}}=30\mathbb{Z}$, and $30\mathbb{Z}$ is not primary ($6·5=30\in 30\mathbb{Z}$ but $6\notin 30\mathbb{Z}$ and $5\notin \sqrt{30\mathbb{Z}}=30\mathbb{Z}$), nor of the form $2q^m\mathbb{Z}$. Concretely, $a=11,b=1$ gives $a^2-b^2=120\in 30\mathbb{Z}$ while $a+b=12\notin 30\mathbb{Z}=\sqrt{30\mathbb{Z}}$ and $a-b=10\notin 30\mathbb{Z}$. Note that $\sqrt{I_1}=6\mathbb{Z}\ne 10\mathbb{Z}=\sqrt{I_2}$, confirming that the equal-radical hypothesis of Proposition 11 cannot be dropped.

While the intersection result requires a common radical, no such condition is needed for ascending chains: the union of any ascending chain of sdf-absorbing primary hyperideals is again sdf-absorbing primary.

Proposition 14.

Let $I_1\subseteq I_2\subseteq \dots$ be an ascending chain of sdf-absorbing primary strong C-hyperideals of H. If $x^2$ is a finite set for every $x\in H$, then $I={\bigcup }_{n=1}^{\infty }{I}_n$ is an sdf-absorbing primary strong C-hyperideal of H.

Proof. 

Since each $I_n$ is proper, $1\notin I_n$ for any n, hence $1\notin {\bigcup }_{n=1}^{\infty }{I}_n=I$, so I is proper.

We next verify that I is a strong C-hyperideal of H. Let $D\in {\tilde{\mathcal{C}}}_H$ with $D\cap I\ne \varnothing$, and take $d\in D\cap I$. Then $d\in I_{n_0}$ for some $n_0$. Since $I_{n_0}$ is a strong C-hyperideal, $D\subseteq I_{n_0}\subseteq I$.

Now let $x,y\in H$ with $x^2-y^2\subseteq I$. By assumption $x^2-y^2$ is a finite set, say $x^2-y^2=\{t_1,\dots ,t_k\}$. For each i, since $t_i\in I=\bigcup I_n$, there exists $n_i$ with $t_i\in I_{n_i}$. Setting $n_0=max\{n_1,\dots ,n_k\}$, the chain property gives $x^2-y^2\subseteq I_{n_0}$. Since $I_{n_0}$ is sdf-absorbing primary, we have $x+y\in \sqrt{I_{n_0}}$ or $x-y\in I_{n_0}$. As $I_{n_0}\subseteq I$ and $\sqrt{I_{n_0}}\subseteq \sqrt{I}$, we conclude $x+y\in \sqrt{I}$ or $x-y\in I$. Hence I is sdf-absorbing primary. □

Remark 15.

The finiteness condition on $x^2$ in Proposition 14 is satisfied whenever $\left|\Omega \right|<\infty$, as in all the examples considered in this paper. In particular, it holds for every hyperring ${\mathbb{Z}}_{\Omega }$ with Ω finite, and for every classical commutative ring.

The results established so far — the relationship with characteristic, the behavior of the radical, the stability under homomorphisms, and the closure under intersections and chains — now allow us to give a comprehensive characterization of sdf-absorbing primary hyperideals and to place them precisely within the hierarchy of hyperideal classes.

Theorem 16.

Let I be a proper strong C-hyperideal of H. Then the following hold.

(i)

(Equivalent conditions.)The following are equivalent:

(i)

I is an sdf-absorbing primary hyperideal of H.

(ii)

For all $x,y\in H$ with $x+y\notin \sqrt{I}$, the inclusion $x^2-y^2\subseteq I$ implies $x-y\in I$.

lcbel=()

The zero hyperideal of $H/I$ is an sdf-absorbing primary hyperideal of $H/I$.

(ii)

(Relation to primary hyperideals.)Every primary strong C-hyperideal of H is sdf-absorbing primary. If H is strongly distributive with scalar identity and $2\in U\left(H\right)$, the converse holds: a proper sdf-absorbing primary strong C-hyperideal I of H is primary.

lcbel=()

(Relation to sdf-absorbing hyperideals.)Every sdf-absorbing hyperideal of H is sdf-absorbing primary. Conversely, if $I=\sqrt{I}$, then I is sdf-absorbing primary if and only if I is an sdf-absorbing hyperideal of H.

Proof.(i). The equivalence of (a) and (b) follows by separating the disjunction: (b) is the reformulation of (a) obtained by moving the condition $x+y\notin \sqrt{I}$ to the hypothesis side, which is logically equivalent to Definition 1 (since $P\Rightarrow A\vee B$ is equivalent to $P\wedge \neg A\Rightarrow B$). For the equivalence with (c), note that $x^2-y^2\subseteq I$ in H is equivalent to ${\overline{x}}^2-{\overline{y}}^2\subseteq \overline{0}$ in $H/I$, where $\overline{x}=x+I$. Moreover, $x+y\in \sqrt{I}$ if and only if $\overline{x}+\overline{y}\in \sqrt{\overline{0}}$ in $H/I$, and $x-y\in I$ if and only if $\overline{x}-\overline{y}=\overline{0}$. Hence the sdf-absorbing primary condition on I in H is precisely the sdf-absorbing primary condition on $\overline{0}$ in $H/I$.

(ii). Let I be primary and suppose $x^2-y^2\subseteq I$ for $x,y\in H$. Define $D=x^2+(x\circ y)+((-x)\circ y)+((-y)\circ y)\in {\tilde{\mathcal{C}}}_H$ (a sum of four hyperproducts, where $(-x)\circ y=-(x\circ y)$ and $(-y)\circ y=-y^2$ by the hyperring axioms). By distributivity, $(x-y)\circ (x+y)\subseteq x^2+(x\circ y)+((-x)\circ y)+((-y)\circ y)=D$. For any $\alpha \in x^2$, $\beta \in y^2$, and $c\in x\circ y$, the element $\alpha +c-c-\beta =\alpha -\beta$ lies in $x^2-y^2\subseteq I$, so $D\cap I\ne \varnothing$. Since I is a strong C-hyperideal, $D\subseteq I$, and hence $(x-y)\circ (x+y)\subseteq I$. The primary condition now gives $x-y\in I$ or $x+y\in \sqrt{I}$, so I is sdf-absorbing primary. This establishes the direction primary ⇒ sdf-absorbing primary. The reverse direction — sdf-absorbing primary ⇒ primary under strongly distributive, scalar identity, and $2\in U\left(H\right)$ — is Proposition 5.

(iii). If I is sdf-absorbing and $x^2-y^2\subseteq I$: when $x=0$ or $y=0$, we have $0^2=0\circ 0=\left\{0\right\}$ (by the axiom $x\circ 0=\left\{0\right\}$), so $0^2-y^2=-y^2\subseteq I$, giving $y^2\subseteq I$ and hence $x+y=y\in \sqrt{I}$ (and symmetrically when $y=0$); when $0\ne x,y$, the sdf-absorbing condition gives $x+y\in I\subseteq \sqrt{I}$ or $x-y\in I$. Hence I is sdf-absorbing primary. Conversely, suppose $I=\sqrt{I}$ and I is sdf-absorbing primary. For $0\ne x,y\in H$ with $x^2-y^2\subseteq I$, the sdf-absorbing primary condition gives $x+y\in \sqrt{I}=I$ or $x-y\in I$, which is exactly the sdf-absorbing condition. □

Remark 17.

Part(ii)shows that when H is strongly distributive with scalar identity and $2\in U\left(H\right)$, the sdf-absorbing primary condition reduces to the classical primary condition, so the two notions are indistinguishable in that setting. The genuine interest of sdf-absorbing primary hyperideals therefore lies in the case where $2\notin U\left(H\right)$: it is precisely in this regime that the notion furnishes a proper common generalization of both primary and sdf-absorbing hyperideals, strictly containing each class. Part(iii)clarifies that sdf-absorbing hyperideals are precisely those sdf-absorbing primary hyperideals that already equal their own radical.

The relationships established in Theorem 16 are summarized in Table 1 and Figure 1 below.

4. Behavior Under Hyperring Constructions

4.1. Localization

Let H be a commutative multiplicative hyperring and S a multiplicatively closed subset of H, that is, $1\in S$ and $s_1\circ s_2\subseteq S$ for all $s_1,s_2\in S$. The hyperring of fractions$S^{-1}H$ is constructed following [3]: its elements are equivalence classes of pairs $(x,s)\in H\times S$, where $(x,s)\sim (y,t)$ if and only if there exists $u\in S$ such that $u\circ (x\circ t-y\circ s)\subseteq \left\{0\right\}$. We write $x/s$ for the class of $(x,s)$. Addition and hyperoperation are defined by $(x/s)+(y/t)=(x\circ t+y\circ s)/(s\circ t)$ and $(x/s)\circ (y/t)=\{z/u:z\in x\circ y,u\in s\circ t\}$; since S is multiplicatively closed, $s\circ t\subseteq S$. For a proper C-hyperideal I with $S\cap I=\varnothing$, set $S^{-1}I=\{i/s:i\in I,s\in S\}$.

The following lemma identifies the radical of a localized ideal and is used throughout this subsection.

Lemma 1.

Let H be a commutative multiplicative hyperring in which every hyperproduct of two elements is finite, I a proper C-hyperideal of H, and S a multiplicatively closed subset of H with $S\cap I=\varnothing$ and $S\cap Z_I\left(H\right)=\varnothing$. Then $S^{-1}\left(\sqrt{I}\right)=\sqrt{S^{-1}I}$ in $S^{-1}H$.

Proof. 

We first observe that $S^{-1}I$ is a C-hyperideal of $S^{-1}H$: if $C\in {\mathcal{C}}_{S^{-1}H}$ with $C\cap S^{-1}I\ne \varnothing$, write $C=(a_1/s_1)\circ \dots \circ (a_n/s_n)$ and suppose $c_0/u_0\in C\cap S^{-1}I$ with $c_0\in a_1\circ \dots \circ a_n$ and $v\circ c_0\subseteq I$ for some $v\in S$. Since $a_1\circ \dots \circ a_n\in {\mathcal{C}}_H$ and $v\circ c_0\subseteq v\circ (a_1\circ \dots \circ a_n)\in {\mathcal{C}}_H$ with $(v\circ c_0)\cap I\ne \varnothing$, the C-hyperideal property of I gives $v\circ (a_1\circ \dots \circ a_n)\subseteq I$. Hence for any $c\in a_1\circ \dots \circ a_n$: $v\circ c\subseteq I$, so $c/u\in S^{-1}I$ for every $u\in s_1\circ \dots \circ s_n$, giving $C\subseteq S^{-1}I$. The element-wise radical description therefore applies to $S^{-1}I$.

For the inclusion $S^{-1}\left(\sqrt{I}\right)\subseteq \sqrt{S^{-1}I}$, let $x/s\in S^{-1}\left(\sqrt{I}\right)$; so $x\in \sqrt{I}$ and $x^n\subseteq I$ for some $n\in \mathbb{N}$. Then ${(x/s)}^n=x^n/s^n\subseteq S^{-1}I$, giving $x/s\in \sqrt{S^{-1}I}$.

For the reverse inclusion, let $x/s\in \sqrt{S^{-1}I}$; so ${(x/s)}^n=x^n/s^n\subseteq S^{-1}I$ for some n. For each $c\in x^n$ there exists $u_c\in S$ with $u_c\circ c\subseteq I$. Since $x^n=\{c_1,\dots ,c_k\}$ is finite, let U be any element of the iterated hyperproduct $u_{c_1}\circ \dots \circ u_{c_k}\subseteq S$. For each i, letting $V_i$ be the set-level hyperproduct of all $u_{c_j}$ with $j\ne i$, we have $U\in u_{c_i}\circ V_i$, and so

$$U\circ c_i\subseteq (u_{c_i}\circ V_i)\circ c_i=V_i\circ (u_{c_i}\circ c_i)\subseteq V_i\circ I\subseteq I,$$

using commutativity and associativity of $(H,\circ )$, the hypothesis $u_{c_i}\circ c_i\subseteq I$, and the hyperideal property $H\circ I\subseteq I$. Thus $U\circ x^n\subseteq I$. Since $U\in S$ and $S\cap Z_I\left(H\right)=\varnothing$, we conclude $x^n\subseteq I$, hence $x\in \sqrt{I}$ and $x/s\in S^{-1}\left(\sqrt{I}\right)$. □

The main result of this subsection requires only the hyperideal-theoretic hypotheses that $\sqrt{I}$ is a strong C-hyperideal and that all hyperproducts in H are finite; no strong distributivity or scalar identity is needed.

Theorem 18.

Let H be a commutative multiplicative hyperring in which every hyperproduct of two elements is a finite set, S a multiplicatively closed subset of H, and I a proper strong C-hyperideal of H such that $\sqrt{I}$ is also a strong C-hyperideal. Assume $S\cap I=\varnothing$, $S\cap Z_I\left(H\right)=\varnothing$, and $S\cap Z_{\sqrt{I}}\left(H\right)=\varnothing$.

(i)

If I is an sdf-absorbing primary hyperideal of H, then $S^{-1}I$ is an sdf-absorbing primary hyperideal of $S^{-1}H$.

(ii)

If $S^{-1}I$ is an sdf-absorbing primary hyperideal of $S^{-1}H$, then I is an sdf-absorbing primary hyperideal of H.

Proof.(i). Let $p/s,q/t\in S^{-1}H$ with ${(p/s)}^2-{(q/t)}^2\subseteq S^{-1}I$. Recall that in $S^{-1}H$ the hyperproduct is $(p/s)\circ (p/s)=\{z/u:z\in p^2,u\in s^2\}$, so ${(p/s)}^2-{(q/t)}^2$ consists of all fractions of the form $\alpha /\sigma -\beta /\tau =(\alpha \circ \tau -\beta \circ \sigma )/(\sigma \circ \tau )$ with $\alpha \in p^2$, $\beta \in q^2$, $\sigma \in s^2$, $\tau \in t^2$. The condition ${(p/s)}^2-{(q/t)}^2\subseteq S^{-1}I$ therefore means: for every $\alpha \in p^2$, $\beta \in q^2$, $\sigma \in s^2$, $\tau \in t^2$, and every $w\in \alpha \circ \tau -\beta \circ \sigma$, the fraction $w/(\sigma \circ \tau )$ lies in $S^{-1}I$, i.e., there exists $u_w\in S$ with $u_w\circ w\subseteq I$. Since $p^2$, $q^2$, $s^2$, $t^2$ are all finite (by hypothesis), the set of pairs $(\alpha ,\sigma ,\beta ,\tau ,w)$ is finite, and we may choose elements $u_{\alpha ,\beta }\in S$ (one for each $(\alpha ,\beta )$) such that $u_{\alpha ,\beta }\circ w\subseteq I$ for all $w\in \alpha \circ \tau -\beta \circ \sigma$ and all $\sigma \in s^2$, $\tau \in t^2$, by taking $u_{\alpha ,\beta }$ to be any element of a suitable iterated hyperproduct of the individual $u_w$’s (which lies in S by multiplicative closure). Since $p^2$ and $q^2$ are finite (by hypothesis), so the collection $\{u_{\alpha ,\beta }:\alpha \in p^2,\beta \in q^2\}$ is finite; label its elements $u_1,\dots ,u_m$. Let U be any element of the iterated hyperproduct $u_1\circ u_2\circ \dots \circ u_m\subseteq S$, which lies in S since S is multiplicatively closed. We claim that $U\circ w\subseteq I$ for every $({\alpha }_0,{\beta }_0)\in p^2\times q^2$ and every $w\in {\alpha }_0\circ t^2-{\beta }_0\circ s^2$. Indeed, let V be the set-level iterated hyperproduct of all $u_i$ except $u_{{\alpha }_0,{\beta }_0}$, so that $U\in u_{{\alpha }_0,{\beta }_0}\circ V$. Then:

$$U\circ w\subseteq (u_{{\alpha }_0,{\beta }_0}\circ V)\circ w=V\circ (u_{{\alpha }_0,{\beta }_0}\circ w)\subseteq V\circ I\subseteq I,$$

where the set equality uses commutativity and associativity of $(H,\circ )$, and the final inclusion holds because $u_{{\alpha }_0,{\beta }_0}\circ w\subseteq I$ by construction and I is a hyperideal. Hence

Claim: for any $a\in U\circ p\circ t$ and $b\in U\circ q\circ s$, we have $a^2-b^2\subseteq I$. Indeed, $a^2\subseteq {(U\circ p\circ t)}^2=U^2\circ p^2\circ t^2$ and $b^2\subseteq U^2\circ q^2\circ s^2$, so $a^2-b^2\subseteq \widehat{D}$, where $\widehat{D}=(U^2\circ p^2\circ t^2)+((-U)\circ U\circ q^2\circ s^2)\in {\tilde{\mathcal{C}}}_H$ (a sum of two hyperproducts, since $(-U)\circ U\circ q^2\circ s^2\in {\mathcal{C}}_H$ and equals $-(U^2\circ q^2\circ s^2)$ by the hyperring axioms). To show $\widehat{D}\subseteq I$, it suffices (by the strong C-hyperideal property) to exhibit a single element of $\widehat{D}\cap I$. Fix any $u_0\in U^2$, ${\alpha }_0\in p^2$, ${\tau }_0\in t^2$, ${\beta }_0\in q^2$, ${\sigma }_0\in s^2$, and any $v\in {\alpha }_0\circ {\tau }_0-{\beta }_0\circ {\sigma }_0\subseteq {\alpha }_0\circ t^2-{\beta }_0\circ s^2$. By associativity, $u_0\circ v\subseteq U^2\circ v\subseteq U\circ (U\circ v)$; since $U\circ v\subseteq U\circ ({\alpha }_0\circ t^2-{\beta }_0\circ s^2)\subseteq I$ (by hypothesis), and I is a hyperideal, $U\circ (U\circ v)\subseteq I$, so $u_0\circ v\subseteq I$. Moreover, by distributivity $u_0\circ v\subseteq u_0\circ ({\alpha }_0\circ {\tau }_0)-u_0\circ ({\beta }_0\circ {\sigma }_0)\subseteq \widehat{D}$, hence $\widehat{D}\cap I\ne \varnothing$, $\widehat{D}\subseteq I$, and therefore $a^2-b^2\subseteq I$.

Since I is sdf-absorbing primary, $a+b\in \sqrt{I}$ or $a-b\in I$.

Case 1: $a+b\in \sqrt{I}$. Set $D=(U\circ p\circ t)+(U\circ q\circ s)$. Since $U\circ p\circ t$ and $U\circ q\circ s$ are hyperproducts, $D\in {\tilde{\mathcal{C}}}_H$. Moreover $a+b\in D$ (as a and b lie in the two summands), and $a+b\in \sqrt{I}$, so $D\cap \sqrt{I}\ne \varnothing$. Since $\sqrt{I}$ is a strong C-hyperideal, $D\subseteq \sqrt{I}$. Now for any $r={\alpha }^{\prime }+{\beta }^{\prime }\in p\circ t+q\circ s$ with ${\alpha }^{\prime }\in p\circ t$, ${\beta }^{\prime }\in q\circ s$, the distributivity inclusion gives $U\circ r\subseteq U\circ {\alpha }^{\prime }+U\circ {\beta }^{\prime }\subseteq U\circ p\circ t+U\circ q\circ s=D\subseteq \sqrt{I}$. Since $U\in S$ and $S\cap Z_{\sqrt{I}}\left(H\right)=\varnothing$, we get $r\in \sqrt{I}$. Hence $p\circ t+q\circ s\subseteq \sqrt{I}$, so every element of $(p\circ t+q\circ s)/(s\circ t)$ lies in $S^{-1}\sqrt{I}=\sqrt{S^{-1}I}$ (Lemma 1). Hence $p/s+q/t\in \sqrt{S^{-1}I}$.

Case 2: $a-b\in I$. Set $E=(U\circ p\circ t)+((-U)\circ q\circ s)\in {\tilde{\mathcal{C}}}_H$ (a sum of two hyperproducts, with $(-U)\circ q\circ s=-(U\circ q\circ s)$). By the same argument, $E\cap I\ne \varnothing$ gives $E\subseteq I$, and for any $r={\alpha }^{\prime }-{\beta }^{\prime }\in p\circ t-q\circ s$, the inclusion $U\circ r\subseteq U\circ {\alpha }^{\prime }-U\circ {\beta }^{\prime }\subseteq E\subseteq I$ together with $S\cap Z_I\left(H\right)=\varnothing$ gives $r\in I$. Hence $p\circ t-q\circ s\subseteq I$, so $p/s-q/t\in S^{-1}I$.

(ii). Let $x,y\in H$ with $x^2-y^2\subseteq I$. The embedding $\phi \left(a\right)=a/1$ gives ${(x/1)}^2-{(y/1)}^2\subseteq S^{-1}I$. Since $S^{-1}I$ is sdf-absorbing primary, either $x/1+y/1\in \sqrt{S^{-1}I}=S^{-1}\sqrt{I}$ or $x/1-y/1\in S^{-1}I$. In the first case, $v\circ (x+y)\subseteq \sqrt{I}$ for some $v\in S$; since $S\cap Z_{\sqrt{I}}\left(H\right)=\varnothing$, we get $x+y\in \sqrt{I}$. In the second case, $w\circ (x-y)\subseteq I$ for some $w\in S$; since $S\cap Z_I\left(H\right)=\varnothing$, we get $x-y\in I$. □

Remark 19.

The key to extending from rings to general multiplicative hyperrings is that strong distributivity (equality) is not needed: the always-valid inclusion $U\circ ({\alpha }^{\prime }+{\beta }^{\prime })\subseteq U\circ {\alpha }^{\prime }+U\circ {\beta }^{\prime }$ suffices to place $U\circ r$ inside $D=(U\circ p\circ t)+(U\circ q\circ s)$, and the strong C-hyperideal property of $\sqrt{I}$ then forces $D\subseteq \sqrt{I}$. This theorem recovers ([11] Proposition 3) as the special case where all hyperproducts are singletons.

As an immediate consequence of Theorem 18 together with Lemma 1, we characterize when $S^{-1}I$ is a Q-sdf-absorbing primary hyperideal of $S^{-1}H$ in terms of I alone.

Corollary 20.

Let H be a commutative multiplicative hyperring in which every hyperproduct of two elements is finite, P a prime strong C-hyperideal of H, and $S=H\setminus P$. If I is a P-sdf-absorbing primary strong C-hyperideal of H with $S\cap Z_I\left(H\right)=\varnothing$, then $S^{-1}I$ is an $S^{-1}P$-sdf-absorbing primary hyperideal of the local hyperring $H_P=S^{-1}H$, and $S^{-1}P$ is the unique maximal hyperideal of $H_P$.

Proof. 

Since P is proper, $1\notin P$, so $1\in S$. We verify that $S=H\setminus P$ is multiplicatively closed. Let $s_1,s_2\in S$, i.e., $s_1,s_2\notin P$. If $s_1\circ s_2\cap P\ne \varnothing$, then since P is a strong C-hyperideal and $s_1\circ s_2\in {\mathcal{C}}_H$, we would have $s_1\circ s_2\subseteq P$, and by primality $s_1\in P$ or $s_2\in P$ — a contradiction. Hence $s_1\circ s_2\cap P=\varnothing$, i.e., $s_1\circ s_2\subseteq S$.

Since P is prime and $S=H\setminus P$, we have $S\cap P=\varnothing$, so $S\cap I\subseteq S\cap P=\varnothing$ (as $I\subseteq \sqrt{I}=P$). The condition $S\cap Z_{\sqrt{I}}\left(H\right)=S\cap Z_P\left(H\right)=\varnothing$ is automatic: if $s\in S$ and $s\circ a\subseteq P$ for some a, the primality of P gives $s\in P$ or $a\in P$; since $s\notin P$ we get $a\in P$, so there is no $a\notin P$ with $s\circ a\subseteq P$. By Theorem 18(i), $S^{-1}I$ is sdf-absorbing primary in $H_P$. By Lemma 1, $\sqrt{S^{-1}I}=S^{-1}\left(\sqrt{I}\right)=S^{-1}P$. Hence $S^{-1}I$ is $S^{-1}P$-sdf-absorbing primary. The maximality of $S^{-1}P$ in $H_P$ follows since every element outside $S^{-1}P$ in $S^{-1}H$ is of the form $x/s$ with $x\notin P$, hence invertible in $S^{-1}H$ [3]. □

We illustrate Theorem 18 and Lemma 1 with a concrete computation in the ring $\mathbb{Z}$.

Example 21.

Consider the ring $\mathbb{Z}$ (the special case ${\mathbb{Z}}_{\Omega }$ with $\Omega =\left\{1\right\}$), the ideal $I=18\mathbb{Z}$, which is sdf-absorbing primary by ([11] Example 5) (it is of the form $2q^m\mathbb{Z}$ with $q=3$, $m=2$), and the multiplicatively closed set $S=\{n\in \mathbb{Z}:gcd(n,6)=1\}$. Note that S is indeed multiplicatively closed, since if $gcd(s_1,6)=gcd(s_2,6)=1$ then $gcd(s_1{s}_2,6)=1$. Moreover, $S\cap I=\varnothing$ since every element of $18\mathbb{Z}$ is divisible by both 2 and 3, hence not coprime to 6.

We verify that $S\cap Z_I\left(\mathbb{Z}\right)=\varnothing$. Indeed, if $s\in S$ and $sa\in 18\mathbb{Z}$, then $18\mid sa$. Since $gcd(s,18)=1$ (as $gcd(s,6)=1$ implies s is coprime to both 2 and 3, hence to $2·3^2=18$), we conclude $18\mid a$, so $a\in I$. Therefore s is not an I-divisor of zero. The same argument with 6 in place of 18 gives $S\cap Z_{\sqrt{I}}\left(\mathbb{Z}\right)=\varnothing$, since $\sqrt{18\mathbb{Z}}=6\mathbb{Z}$ and $gcd(s,6)=1$ for every $s\in S$.

By Theorem 18(i), $S^{-1}\left(18\mathbb{Z}\right)$ is an sdf-absorbing primary ideal of $S^{-1}\mathbb{Z}$. By Lemma 1, $\sqrt{S^{-1}\left(18\mathbb{Z}\right)}=S^{-1}\sqrt{18\mathbb{Z}}=S^{-1}\left(6\mathbb{Z}\right)$. Concretely, in $S^{-1}\mathbb{Z}$ we have ${\displaystyle \frac{a^2}{s^2}}-{\displaystyle \frac{b^2}{t^2}}={\displaystyle \frac{{\left(at\right)}^2-{\left(bs\right)}^2}{{\left(st\right)}^2}}$. Since $st\in S$ (so ${\left(st\right)}^2\in S$ is a unit in $S^{-1}\mathbb{Z}$), this element lies in $S^{-1}\left(18\mathbb{Z}\right)$ if and only if $18\mid {\left(at\right)}^2-{\left(bs\right)}^2$. In that case, the sdf-absorbing primary condition gives $a/s+b/t\in S^{-1}\left(6\mathbb{Z}\right)=\sqrt{S^{-1}\left(18\mathbb{Z}\right)}$ or $a/s-b/t\in S^{-1}\left(18\mathbb{Z}\right)$, as guaranteed by Theorem 18.

4.2. Cartesian Products

For commutative multiplicative hyperrings $(H_1,{+}_1,{\circ }_1)$ and $(H_2,{+}_2,{\circ }_2)$, the triple $(H_1\times H_2,+,\circ )$ is again a commutative multiplicative hyperring, where addition and multiplication are defined componentwise:

$$(a_1,a_2)+(b_1,b_2)=(a_1{+}_1{b}_1,a_2{+}_2{b}_2),$$

$$(a_1,a_2)\circ (b_1,b_2)=\{(\alpha ,\beta )\mid \alpha \in a_1{\circ }_1{b}_1,\beta \in a_2{\circ }_2{b}_2\}$$

for all $a_i,b_i\in H_i$ [4]. In this subsection we investigate which hyperideals of the form $I_1\times I_2$ are sdf-absorbing primary in $H_1\times H_2$, obtaining necessary conditions, sufficient conditions, and a complete equivalence when one factor is the entire hyperring.

We begin by showing that the sdf-absorbing primary property of a product ideal descends to each component.

Theorem 22.

Let $I_1$ and $I_2$ be proper hyperideals of $H_1$ and $H_2$, respectively. If $I_1\times I_2$ is an sdf-absorbing primary hyperideal of $H_1\times H_2$, then both $I_1$ and $I_2$ are sdf-absorbing primary hyperideals of $H_1$ and $H_2$, respectively.

Proof. 

We show that $I_1$ is sdf-absorbing primary in $H_1$; the argument for $I_2$ is identical. Let $x_1,y_1\in H_1$ with $x_1^2-y_1^2\subseteq I_1$. Since $0\in I_2$, the hyperideal property gives $0\circ 0\subseteq I_2$, so every element of $0^2=0\circ 0$ lies in $I_2$; hence every difference of two such elements lies in $I_2$ (as $I_2$ is closed under subtraction). Moreover, ${(x_1,0)}^2-{(y_1,0)}^2=\{(\alpha -\beta ,\gamma -\delta ):\alpha \in x_1^2,\beta \in y_1^2,\gamma ,\delta \in 0\circ 0\}$. For each such element we have $\alpha -\beta \in x_1^2-y_1^2\subseteq I_1$ and $\gamma -\delta \in I_2$, so ${(x_1,0)}^2-{(y_1,0)}^2\subseteq I_1\times I_2$. Since $I_1\times I_2$ is sdf-absorbing primary, we first establish that $\sqrt{I_1\times I_2}=\sqrt{I_1}\times \sqrt{I_2}$. The prime hyperideals of $H_1\times H_2$ containing $I_1\times I_2$ are exactly those of the form $P_1\times H_2$ (with $P_1\supseteq I_1$ prime in $H_1$) or $H_1\times P_2$ (with $P_2\supseteq I_2$ prime in $H_2$). Indeed, $(1,0)\circ (0,1)=\left\{\right(0,0\left)\right\}$ lies in any proper hyperideal P, so primality forces $(1,0)\in P$ or $(0,1)\in P$; the hyperideal generated by either one yields a factor of the stated form. Taking the intersection over all such primes gives $\sqrt{I_1\times I_2}=\sqrt{I_1}\times \sqrt{I_2}$.

Applying the sdf-absorbing primary condition to $(x_1,0)$ and $(y_1,0)$, we obtain either $(x_1+y_1,0)\in \sqrt{I_1}\times \sqrt{I_2}$, giving $x_1+y_1\in \sqrt{I_1}$, or $(x_1-y_1,0)\in I_1\times I_2$, giving $x_1-y_1\in I_1$. Hence $I_1$ is sdf-absorbing primary. □

When $H_1$ and $H_2$ are commutative rings (i.e., all hyperproducts are singletons), a stronger conclusion holds.

Corollary 23.

Let $R_1$ and $R_2$ be commutative rings and let $I_1$, $I_2$ be proper ideals of $R_1$ and $R_2$, respectively. If $I_1\times I_2$ is an sdf-absorbing primary ideal of $R_1\times R_2$, then $I_1$ and $I_2$ are sdf-absorbing primary, and moreover $2\in \sqrt{I_1}$ or $2\in \sqrt{I_2}$.

Proof. 

That $I_1$ and $I_2$ are sdf-absorbing primary follows from Theorem 22. For the second claim, take $a=(1,1)$ and $b=(1,-1)$ in $R_1\times R_2$. Then $a^2=b^2=(1,1)$ (since ${(-1)}^2=1$ in any ring), so $a^2-b^2=(0,0)\in I_1\times I_2$. The sdf-absorbing primary condition gives $a+b=(2,0)\in \sqrt{I_1}\times \sqrt{I_2}$ or $a-b=(0,2)\in I_1\times I_2$. The first alternative gives $2\in \sqrt{I_1}$; the second gives $2\in I_2\subseteq \sqrt{I_2}$. □

Theorem 22 gives a necessary condition. We now turn to sufficient conditions that guarantee $I_1\times I_2$ is sdf-absorbing primary.

Theorem 24.

Let $I_1$ and $I_2$ be proper hyperideals of $H_1$ and $H_2$, respectively. Each of the following conditions is sufficient for $I_1\times I_2$ to be an sdf-absorbing primary hyperideal of $H_1\times H_2$:

(i)

$I_1$ is an sdf-absorbing strong C-hyperideal of $H_1$ with $2\in I_1$, and $I_2$ is sdf-absorbing primary in $H_2$.

(ii)

$I_2$ is an sdf-absorbing strong C-hyperideal of $H_2$ with $2\in I_2$, and $I_1$ is sdf-absorbing primary in $H_1$.

Proof. 

We prove part (i); part (ii) is entirely symmetric. Let $(x_1,x_2),(y_1,y_2)\in H_1\times H_2$ with ${(x_1,x_2)}^2-{(y_1,y_2)}^2\subseteq I_1\times I_2$. This means $x_1^2-y_1^2\subseteq I_1$ and $x_2^2-y_2^2\subseteq I_2$.

Since $I_1$ is sdf-absorbing and $2\in I_1$, we claim that $x_1+y_1\in I_1$ and $x_1-y_1\in I_1$. If $x_1=0$ or $y_1=0$, say $x_1=0$, then $-y_1^2\subseteq I_1$, giving $y_1^2\subseteq I_1$; since every sdf-absorbing hyperideal is a radical ideal [4], $I_1=\sqrt{I_1}$, hence $y_1\in I_1$ and $x_1\pm y_1=\pm y_1\in I_1$ (and symmetrically for $y_1=0$). When both $x_1$ and $y_1$ are nonzero, by ([4] Theorem 2.9), both $x_1+y_1\in I_1$ and $x_1-y_1\in I_1$. Since $I_2$ is sdf-absorbing primary, we have $x_2+y_2\in \sqrt{I_2}$ or $x_2-y_2\in I_2$. Note that $\sqrt{I_1\times I_2}=\sqrt{I_1}\times \sqrt{I_2}$ (by the same argument as in Theorem 22).

If $x_2+y_2\in \sqrt{I_2}$, then since $x_1+y_1\in I_1\subseteq \sqrt{I_1}$, we have $(x_1+y_1,x_2+y_2)\in \sqrt{I_1}\times \sqrt{I_2}=\sqrt{I_1\times I_2}$, so $(x_1,x_2)+(y_1,y_2)\in \sqrt{I_1\times I_2}$.

If $x_2-y_2\in I_2$, then $(x_1-y_1,x_2-y_2)\in I_1\times I_2$, so $(x_1,x_2)-(y_1,y_2)\in I_1\times I_2$.

In either case, $I_1\times I_2$ is sdf-absorbing primary. □

A particularly clean characterization arises when one factor is the entire hyperring: in this case the sdf-absorbing primary property of $I_1\times H_2$ is completely equivalent to that of $I_1$ alone.

Theorem 25.

Let $I_1$ be a proper hyperideal of $H_1$. Then $I_1\times H_2$ is an sdf-absorbing primary hyperideal of $H_1\times H_2$ if and only if $I_1$ is an sdf-absorbing primary hyperideal of $H_1$.

Proof. 

If $I_1\times H_2$ is sdf-absorbing primary in $H_1\times H_2$, let $x_1,y_1\in H_1$ with $x_1^2-y_1^2\subseteq I_1$. As in the proof of Theorem 22, ${(x_1,0)}^2-{(y_1,0)}^2\subseteq I_1\times H_2$. The sdf-absorbing primary condition gives $(x_1+y_1,0)\in \sqrt{I_1\times H_2}=\sqrt{I_1}\times H_2$ or $(x_1-y_1,0)\in I_1\times H_2$, yielding $x_1+y_1\in \sqrt{I_1}$ or $x_1-y_1\in I_1$. Hence $I_1$ is sdf-absorbing primary.

Conversely, suppose $I_1$ is sdf-absorbing primary in $H_1$. Let $(x_1,x_2),(y_1,y_2)\in H_1\times H_2$ with ${(x_1,x_2)}^2-{(y_1,y_2)}^2\subseteq I_1\times H_2$. Then $x_1^2-y_1^2\subseteq I_1$, so $x_1+y_1\in \sqrt{I_1}$ or $x_1-y_1\in I_1$. In the first case, noting that $\sqrt{I_1\times H_2}=\sqrt{I_1}\times H_2$ (since the primes containing $I_1\times H_2$ are exactly the hyperideals of the form $P_1\times H_2$ with $P_1\supseteq I_1$ prime in $H_1$), we have $(x_1,x_2)+(y_1,y_2)\in \sqrt{I_1}\times H_2=\sqrt{I_1\times H_2}$. In the second case, $(x_1,x_2)-(y_1,y_2)\in I_1\times H_2$. Hence $I_1\times H_2$ is sdf-absorbing primary. □

5. Conclusion

In this paper, we have introduced the notion of sdf-absorbing primary hyperideals in commutative multiplicative hyperrings and established a comprehensive theory for this new class. The central result is a three-way characterization (Theorem 16) showing that sdf-absorbing primary hyperideals provide a proper common generalization of both primary hyperideals and sdf-absorbing hyperideals, coinciding with primary under $2\in U\left(H\right)$ (in the strongly distributive scalar-identity setting) and with sdf-absorbing when $I=\sqrt{I}$.

Several structural results accompany this characterization. We showed that sdf-absorbing primary hyperideals are preserved under good homomorphisms and localization, closed under finite intersections sharing a common radical and under ascending chains, and characterized explicitly in Cartesian products of hyperrings. These results collectively mirror and extend those known for rings [11], while identifying the cases where hyperring-specific phenomena require additional hypotheses or where the extension to general hyperrings remains open.

Two directions stand out as natural continuations of this work. First, Theorem 7 was proved here under the hypotheses that H is strongly distributive with scalar identity, $2\in U\left(H\right)$, I is a strong C-hyperideal, all hyperproducts are finite, and $\sqrt{I}$ is also a strong C-hyperideal; it remains an interesting open problem to determine whether the conclusion that $\sqrt{I}$ is sdf-absorbing can be obtained without these assumptions, i.e., for arbitrary commutative multiplicative hyperrings. Second, one may introduce a reduction function $\phi$ and study the corresponding class of φ-sdf-absorbing primary hyperideals, as has been done for other ideal classes in the hyperring setting [1,4].