#  $n-$absorbing $I-$prime hyperideals in multiplicative hyperrings

Ali A. Mena  Mathematics Department, Faculty of Science
Soran University
44008, Soran, Erbil Kurdistan Region, Iraq<EMAIL_ADDRESS>and  Ismael
Akray  Mathematics Department, Faculty of Science
Soran University
44008, Soran, Erbil Kurdistan Region, Iraq<EMAIL_ADDRESS>
###### Abstract.

In this paper, we define the concept $I-$prime hyperideal in a multiplicative
hyperring $R$. A proper hyperideal $P$ of $R$ is an $I-$prime hyperideal if
for $a,b\in R$ with $ab\subseteq P-IP$ implies $a\in P$ or $b\in P$. We
provide some characterizations of $I-$prime hyperideals. Also we conceptualize
and study the notions $2-$absorbing $I-$prime and $n-$absorbing $I-$prime
hyperideals into multiplicative hyperrings as generalizations of prime ideals.
A proper hyperideal $P$ of a hyperring $R$ is an $n-$absorbing $I-$prime
hyperideal if for $x_{1},\cdots,x_{n+1}\in R$ such that $x_{1}\cdots
x_{n+1}\subseteq P-IP$, then $x_{1}\cdots x_{i-1}x_{i+1}\cdots
x_{n+1}\subseteq P$ for some $i\in\\{1,\cdots,n+1\\}$. We study some
properties of such generalizations. We prove that if $P$ is an $I-$prime
hyperideal of a hyperring $R$, then each of $\frac{P}{J}$, $S^{-1}P$, $f(P)$,
$f^{-1}(P)$, $\sqrt{P}$ and $P[x]$ are $I-$prime hyperideals under suitable
conditions and suitable hyperideal $I$, where $J$ is a hyperideal contains in
$P$. Also, we characterize $I-$prime hyperideals in the decomposite
hyperrings. Moreover, we show that the hyperring with finite number of maximal
hyperideals in which every proper hyperideal is $n-$absorbing $I-$prime is a
finite product of hyperfields.

###### Key words and phrases:

hyperring, multiplicative hyperring, prime hyperideal, I-prime hyperideal

###### 2010 Mathematics Subject Classification:

16Y20, 13A15

##  1\. Introduction

Many concepts in modern algebra was generalized by generalizing their
structures to hyperstructure. The French mathematician F. Marty in 1934
introduced the concept hyperstructure or multioperation by returning a set of
values instead of a single value [11]. The hyperstructures theory was studied
from many points of view and applied to several areas of mathematics
especially in computer science and logic. In [11] the author presented the
concept hypergroup and after that in 1937, the authors H. S. Wall [16] and M.
Kranser [10] also gave their respective definitions of hypergroup as a
generalization of groups.

The hyperrings were introduced by many authors. A type of hyperring where the
multiplication is a hyperoperation while the addition is just an operation
introduced by Rota in 1982 and called a multiplicative hyperring [13]. A well
known example on multiplicative hyperring is that for a ring $(R,+,\cdot)$ and
corresponding to every non-singleton subset $A\in
P^{*}(R)=P(R)\backslash\\{\phi\\}$ where $P(R)$ is the power set of $R$, there
exists a multiplicative hyperring with absorbing zero
$\left(R_{A},+,\circ\right)$ where $R_{A}=R$ and for any $x,y\in R_{A},x\circ
y=\\{x\cdot a\cdot y:a\in A\\}$ (see [12, 15]). Another type of hyperring in
which addition is a hyperoperation while the multiplication is an operation
introduced by M. Krasner in 1983 and called Krasner hyperring [10]. The
hyperrings in which the additions and multiplications are hyperoperations
where introduced by De Salvo [8]. Procesi and Rota in [12] have conceptualized
the notion of primeness of hyperideal in a multiplicative hyperring. A proper
hyperideal $P$ is called prime hyperideal if $ab\subseteq P$, then $a\in P$ or
$b\in P$. The radical of a hyperideal $P$ denoted by $\sqrt{P}$ is the
intersection of all prime hyperideals that contains $P$. Some generalizations
of prime hyperideals can be found in [3, 7, 14] .

In the recent years many generalizations of prime ideals were introduced. Here
state some of them. The authors in [4] and [5] introduced the notions
$2-$absorbing and $n-$absorbing ideals in commutative rings. A proper ideal
$P$ is called $2-$absorbing (or $n-$absorbing) ideal if whenever the product
of three (or $n+1$) elements of $R$ in $P$, the product of two (or $n$) of
these elements is in $P$.

In [1] and [2], the author Akray introduced the notions $I-$prime ideal and
$n-$absorbing $I-$ideal in classical rings as a generalization of prime
ideals. For fixed proper ideal $I$ of a commutative ring $R$ with identity, a
proper ideal $P$ of $R$ is an $I-$prime if for $a,b\in R$ with $a.b\in P-IP$,
then $a\in P$ or $b\in P$. A proper ideal $P$ of $R$ is an $n-$absorbing
$I-$ideal if for $x_{1},\cdots,x_{n+1}\in R$ such that $x_{1}\cdots x_{n+1}\in
P-IP$, then $x_{1}\cdots x_{i-1}x_{i+1}\cdots x_{n+1}\in P$ for some
$i\in\\{1,2,\cdots,n+1\\}$.

In this paper all hyperrings are commutative hyperring with identity. Here we
want to define the $I-$prime hyperideal, $2-$absorbing $I-$hyperideal and
$n-$absorbing $I-$hyperideal in multiplicative hyperrings. For fixed proper
hyperideal $I$ of a multiplicative hyperring $R$, a proper hyperideal $P$ of
$R$ is an $I-$prime if $a,b\in R$ with $a.b\subseteq P-IP$, then $a\in P$ or
$b\in P$. A proper hyperideal $P$ of $R$ is a $2-$absorbing $I-$prime
hyperideal if for $x_{1},x_{2},x_{3}\in R$ such that $x_{1}x_{2}x_{3}\subseteq
P-IP$, then $x_{1}x_{2}\subseteq P$ or $x_{1}x_{3}\subseteq P$ or
$x_{2}x_{3}\subseteq P$. A proper hyperideal $P$ of $R$ is an $n-$absorbing
$I-$prime hyperideal if for $x_{1},\cdots,x_{n+1}\in R$ such that $x_{1}\cdots
x_{n+1}\subseteq P-IP$, then $x_{1}\cdots x_{i-1}x_{i+1}\cdots
x_{n+1}\subseteq P$ for some $i\in\\{1,2,\cdots,n+1\\}$.

In section two, we define $I-$prime hyperideal and we prove some equivalents
of $I-$prime hyperideal (Theorem 2.8). Moreover, we establish $I-$prime
hyperideals in finite product of hyperrings (Theorem 2.10). Section three
devoted for $2-$absorbing $I-$prime and $n-$absorbing $I-$prime hyperideals
and we prove Theorem 3.7 which state (Let $R=\prod_{i=1}^{n+1}R_{i}$ and $P$
be a proper non-zero hyperideal of $R$. If P is an $(n+1)-$absorbing $I-$prime
hyperideal of $R$, then $P=P_{1}\times P_{2}\times\cdots\times P_{n+1}$ for
some proper $n-$absobing $I_{i}-$prime hyperideals $P_{1},\cdots,P_{n+1}$ of
$R_{1},\cdots,R_{n+1}$ respectively, where $I=\prod_{i=1}^{n+1}I_{i}$ and
$I_{i}=R_{i}$, $\forall i=1,2,\cdots,n+1$). Also, we prove Theorem 3.9 that
state (Let $\mid Max(R)\mid\geq n+1\geq 2.$ Then each proper hyperideal of $R$
is an $n-$absorbing $I-$prime hyperideal if and only if each quotient of $R$
is a product of $(n+1)-$fields). Finally, let $P$ be an $n-$absorbing
$I-$hyperideal of a hyperring $R$. Then there are at most $n^{th}$ prime
hyperideals of $R$ that are minimal over $P$ (Theorem 3.10).

##  2\. $I-$prime hyperideals

We start this section by defining the concept of $I-$prime hyperideal and some
example of it. A proper hyperideal $P$ of $R$ is an $I-$prime hyperideal if
for $a,b\in R$ with $ab\subseteq P-IP$ implies $a\in P$ or $b\in P$.

In the following examples we show that the class of $I-$prime hyperideals
contains properly the class of prime hyperideals.

######  Example 2.1.

Consider the hyperring of integers $(\mathbb{Z},+,\circ)$,
$A=\\{0,1\\}\subseteq\mathbb{Z}$ and $n\circ m=\\{nam:a\in A\\}=\\{0,nm\\}$.
So $4\mathbb{Z}$ is not prime hyperideal, since $2\circ 2=\\{0,4\\}\subseteq
4\mathbb{Z}$ and $2\notin 4\mathbb{Z}$. But $4\mathbb{Z}$ is
$2\mathbb{Z}$-prime hyperideal, since $\forall a,b\in\mathbb{Z},a\circ
b=\\{0,ab\\}\nsubseteq$ $4\mathbb{Z}-(2\mathbb{Z}\circ
4\mathbb{Z})=4\mathbb{Z}-8\mathbb{Z}.$

######  Example 2.2.

Let $(\mathbb{Z},+,\circ)$ be the hyperring of integers and
$A=\\{4,8\\}\subseteq\mathbb{Z}$ and $a\circ b=aAb=\\{4ab,8ab\\}$. Then
$1\circ 1=\\{4,8\\}\subseteq 2\mathbb{Z}$ but $1\notin 2\mathbb{Z}$ and hence
$2\mathbb{Z}$ is not prime hyperideal. However $2\mathbb{Z}$ is not
$8\mathbb{Z}-$prime hyperideal, since $2\mathbb{Z}-(8\mathbb{Z}\circ
2\mathbb{Z})=2\mathbb{Z}-(64\mathbb{Z}\cup 128\mathbb{Z})$
$=2\mathbb{Z}-64\mathbb{Z}$ which contains $1\circ 1$. Therefore,
$2\mathbb{Z}$ is neither prime hyperideal nor $8\mathbb{Z}$-prime hyperideal
of $(\mathbb{Z},+,\circ)$.

The intersection of two $I-$prime hyperideals is not $I-$prime hyperideal let
us explain our claim by this example.

######  Example 2.3.

Consider the hyperring of integers $(\mathbb{Z},+,\circ)$, where $a\circ b$ =
$\\{2ab,3ab\\}$. Let $P=2\mathbb{Z}$, $I=3\mathbb{Z}$ and
$P-IP=2\mathbb{Z}-(3\mathbb{Z})\circ(2\mathbb{Z})=2\mathbb{Z}-6A\mathbb{Z}=2\mathbb{Z}-(12\mathbb{Z}\cup
18\mathbb{Z})$. Thus $P$ is $I-$prime hyperideal. Now, for $Q=3\mathbb{Z}$ and
$I=3\mathbb{Z}$ we have
$Q-IQ=3\mathbb{Z}-(3\mathbb{Z})\circ(3\mathbb{Z})=3\mathbb{Z}-9A\mathbb{Z}=3\mathbb{Z}-(18\mathbb{Z}\cup
27\mathbb{Z})$. So $Q$ is $I-$prime hyperideal of $\mathbb{Z}$ while $P\cap
Q=6\mathbb{Z}$ is not $3\mathbb{Z}-$prime hyperideal, since
$6\mathbb{Z}-(3\mathbb{Z})\circ(6\mathbb{Z})=6\mathbb{Z}-(36\mathbb{Z}\cup
54\mathbb{Z})$ $2\circ 3$={12,18}$\subseteq 6\mathbb{Z}-(36\mathbb{Z}\cup
54\mathbb{Z})$, but neither 2 $\in 6\mathbb{Z}$ nor $3\in 6\mathbb{Z}$.

The following lemma is a generalization of Lemma 2.1 in [1].

######  Lemma 2.1.

Let $P$ be a proper hyperideal of a hyperring $(R,+,\circ)$. Then $P$ is an
$I$-prime hyperideal if and only if $P/IP$ is weakly prime hyperideal in
$R/IP$.

###### Proof.

$(\Rightarrow)$ Let $P$ be an $I$-prime hyperideal in $(R,+,\circ)$. Let
$a,b\subseteq R$ with $\\{0\\}\neq(a+$ $IP)(b+IP)=a\circ b+IP\in P/IP$. Then
$a\circ b\subseteq P-IP$ implies $a\subseteq P$ or $b\subseteq P$, hence
$a+IP\subseteq P/IP$ or $b+IP\subseteq P/IP$. So $P/IP$ is weakly prime
hyperideal in $R/IP$. $(\Leftarrow)$ Suppose that $P/IP$ is weakly prime
hyperideal in $R/IP$ and take $r,s\subseteq R$ such that $r\circ s\subseteq
P-IP$. Then $\\{0\\}\neq r\circ s+IP=(r+IP)(s+IP)\subseteq P/IP$ so
$r+IP\subseteq P/IP$ or $s+IP\subseteq P/IP$. Therefore $r\subseteq P$ or
$s\subseteq P$. Thus $P$ is an $I$-prime hyperideal in $R$. ∎

Let $(R,+,\circ)$ be a hyperring and $x$ be an indeterminate. Then
$(R[x],+,\bullet)$ is a polynomial multiplicative hyperring where
$ax^{n}\bullet bx^{m}=(a\circ b)x^{n+m}$ (see [6]).

######  Theorem 2.2.

If $P$ is an $I-$prime hyperideal of $(R,+,\circ)$, then $P[x]$ is
$I[x]-$prime hyperideal of $(R[x],+,\bullet)$ .

###### Proof.

Let $a(x)\bullet b(x)\subseteq P[x]-I[x]\bullet P[x]=P[x]-(IP)[x]$. Without
loss of generality, let $a(x)=cx^{n}$ and $b(x)=dx^{m}$, for $c,d\in R$. Thus
$c\circ dx^{n+m}\subseteq P[x]$ so $c\circ d\subseteq P$ and $c\circ
dx^{n+m}\nsubseteq IP[x]$ implies $c\circ d\nsubseteq IP$. P $I-$prime
hyperideal gives us $c\in P$ or $d\in P$. Hence $a(x)=cx^{n}\in P[x]$ or
$b(x)=dx^{m}\in P[x]$ and so $P[x]$ is an $I[x]-$prime. ∎

######  Corollary 2.2.1.

Let $P$ be an $I-$prime hyperideal of $R$. Then $P[x]$ is an $I-$prime
hyperideal of $R[x]$.

######  Theorem 2.3.

Let $R$ be a hyperring and $f:R\longrightarrow R$ be a good epiomorphism and
let $P$ be an I-prime hyperideal of $R$ with $Kerf\subseteq P$. Then $f(P)$ is
an $I-$prime hyperideal.

###### Proof.

Firstly, we have to show that $f(P)$ is hyperideal of $R$. Let $\bar{r}\in R$
and $y\in f(P)$. Then $x=f^{-1}(y)\in P$ and there exists $r\in R$ such that
$f(r)=\bar{r}$. So $\bar{r}.y=f(r).f(x)=f(r.x)\subseteq f(P)$. Now let us show
that $f(P)$ is an $I-$prime hyperideal. To do this, we have for all $x,y\in R$
there exist $a,b\in R$ such that $x=f(a)$, $y=f(b)$. Then
$x.y=f(a).f(b)=f(a.b)\subseteq f(P)$, so $a.b\subseteq P+Kerf$. As $P$ is an
$I-$prime hyperideal, $a\in P$ or $b\in P$, that is $x=f(a)\in f(P)$ or
$y=f(b)\in f(P)$. So $f(P)$ is an $I-$prime hyperideal of $R$. ∎

######  Theorem 2.4.

Let $(R,+,\circ)$ be a hyperring and $f:R\longrightarrow R$ be a good
homomorphism and let $Q$ be an I-prime hyperideal of $R$. Then $f^{-1}(Q)$ is
an $I-$prime hyperideal.

###### Proof.

Let $a\circ b\subseteq f^{-1}(Q)$. Then $f(a\circ b)=f(a)\circ f(b)\subseteq
Q$ because $f$ is a good homomorphism. As $Q$ is $I-$prime hyperideal,
$f(a)\in Q$ or $f(b)\in Q$. So, $a\in f^{-1}(Q)$ or $b\in f^{-1}(Q)$ and hence
$f^{-1}(Q)$ is an $I-$prime hyperideal of $R$. ∎

The following theorem generalizes Theorem 2.2 of [1].

######  Theorem 2.5.

(1) Let $I\subseteq J$ be two hyperideals of a multiplicative hyperring $R$.
If $P$ is an $I-$prime hyperideal of $R$, then it is a $J-$prime hyperideal.

(2) Let $R$ be a commutative multiplicative hyperring and $P$ an $I$-prime
hyperideal that is not prime hyperideal, then $P^{2}\subseteq IP$. Thus, an
I-prime hyperideal $P$ with $P^{2}\nsubseteq IP$ is a prime hyperideal.

###### Proof.

(1) The proof comes from the fact that if $I\subseteq J$, then $P-JP\subseteq$
$P-IP$.

(2) Suppose that $P^{2}\nsubseteq IP$, we show that $P$ is prime hyperideal.
Let $ab\subseteq P$ for $a,b\in R$. If $ab\nsubseteq IP$, then $P$ $I$-prime
gives $a\in P$ or $b\in P$. So assume that $ab\subseteq IP$. First, suppose
that $aP\nsubseteq IP$; say $ax\nsubseteq IP$ for some $x\in P$. Then
$a(x+b)\subseteq P-IP$. So $a\in P$ or $x+b\in P$ and hence $a\in P$ or $b\in
P$. Hence we can assume that $aP\subseteq IP$ and in a similar way we can
assume that $bP\subseteq IP$. Since $P^{2}\nsubseteq IP$, there exist $y,z\in
P$ with $yz\nsubseteq IP$. Then $(a+y)(b+z)\subseteq P-IP$. So $P$ $I$-prime
gives $a+y\in P$ or $b+z\in P$ and hence $a\in P$ or $b\in P$. Therefore $P$
is a prime hyperideal of $R$ see also [1] . ∎

######  Corollary 2.5.1.

Let $P$ be an I-prime hyperideal of a hyperring $R$ with $IP\subseteq P^{3}$.
Then $P$ is $\cap_{i=1}^{\infty}P^{i}$-prime hyperideal.

###### Proof.

If $P$ is prime hyperideal, then $P$ is $\cap_{i=1}^{\infty}P^{i}$-prime
hyperideal. Assume that $P$ is not prime hyperideal. By Theorem
$2.5,P^{2}\subseteq IP\subseteq P^{3}$. Thus $IP=P^{n}$ for each $n\geq 2$. So
$\cap_{i=1}^{\infty}P^{i}=P\cap P^{2}=P^{2}$ and
$\left(\cap_{i=1}^{\infty}P^{i}\right)P=P^{2}P=P^{3}=IP$. Being $P$ is
$I$-prime hyperideal implies $P$ is $\cap_{i=1}^{\infty}P^{i}$-prime
hyperideal. ∎

######  Remark.

Let $P$ be an $I$-prime hyperideal. Then $P\subseteq\sqrt{IP}$ or
$\sqrt{IP}\subseteq P$. If $P\varsubsetneqq\sqrt{IP}$, then $P$ is not prime
hyperideal since otherwise $IP\subseteq P$ implies
$\sqrt{IP}\subseteq\sqrt{P}=P$. While if $\sqrt{IP}\subsetneq P$, then $P$ is
a prime hyperideal. Now we give a way to construct $I$-prime ideals $P$ when
$\cap_{i=1}^{\infty}P^{i}\subseteq$ $IP\subseteq P^{3}$.

######  Corollary 2.5.2.

Let $P$ be an $I-$prime hyperideal of a hyperring $R$ which is not prime
hyperideal. Then $\sqrt{P}=\sqrt{IP}$.

###### Proof.

By Theorem 2.5, $P^{2}\subseteq IP$ and hence
$\sqrt{P}=\sqrt{P^{2}}\subseteq\sqrt{IP}$. The other containment always holds.
∎

######  Remark.

Assume that $P$ is an $I-$prime hyperideal, but not prime. Then by Theorem
2.5, if $IP\subseteq P^{2}$, then $P^{2}=IP$. In particular, if $P$ is weakly
prime hyperideal (0-prime) but not prime hyperideal, then $P^{2}=\\{0\\}$.
Suppose that $IP\subseteq P^{3}$. Then $P^{2}\subseteq IP\subseteq P^{3}$; So
$P^{2}=P^{3}$ and thus $P^{2}$ is an idempotent.

######  Lemma 2.6.

If $P$ is an $I-$primary hyperideal of a hyperring $R$, then $\sqrt{P}$ is a
$\sqrt{I}-$prime hyperideal of $R$.

###### Proof.

Let $ab\subseteq\sqrt{P}-\sqrt{I}\sqrt{P}=\sqrt{P}-\sqrt{IP}$ for $a,b\in R$.
Then $(ab)^{n}=a^{n}b^{n}\subseteq P$ for some $n\in\mathbb{N}$ and
$(ab)^{m}\nsubseteq IP$ for all $m\in\mathbb{N}$. So $a^{n}b^{n}\subseteq
P-IP$ and as $P$ is an $I-$primary hyperideal of $R$, $a^{n}\subseteq P$ or
$b^{n}\subseteq\sqrt{P}$, that is $a\in\sqrt{P}$ or $b\in\sqrt{P}$ which means
that $\sqrt{P}$ is a $\sqrt{I}-$prime hyperideal of $R$. ∎

The following theorem generalizes the result [1, Theorem 2.8].

######  Theorem 2.7.

(1) Let $R$ and $S$ be two commutative multiplicative hyperrings and $P$ be
$\\{0\\}-$prime hyperideal of $R$. Then $P\times S$ is I-prime hyperideal of
$R\times S$ for each hyperideal $I$ of $R\times S$ with
$\cap_{i=1}^{\infty}(P\times S)^{i}\subseteq I(P\times S)\subseteq P\times S$.
(2) Let $P$ be a finitely generated proper hyperideal of a commutative
hyperring $R$. Assume $P$ is an $I$-prime hyperideal with $IP\subseteq P^{3}$.
Then either $P$ is $\\{0\\}-$prime or $P^{2}\neq\\{0\\}$ is idempotent and $R$
decomposes as $T\times S$ where $S=P^{2}$ and $P=J\times S$ where $J$ is a
$\\{0\\}-$prime. Thus $P$ is I-prime hyperideal for
$\cap_{i=1}^{\infty}P^{i}\subseteq IP\subseteq$ $P$.

###### Proof.

(1) Let $R$ and $S$ be two commutative hyperrings and $P$ be a $\\{0\\}-$prime
hyperideal of $R$. Then $P\times S$ need not be a $\\{0\\}-$prime hyperideal
of $R\times S$; In fact, $P\times S$ is $\\{0\\}-$prime if and only if
$P\times S$ (or equivalently $P)$ is prime hyperideal. However, $P\times S$ is
an $I$-prime hyperideal for each $I$ with $\cap_{i=1}^{\infty}(P\times
S)^{i}\subseteq I(P\times S)$. If $P$ is prime hyperideal, then $P\times S$ is
a prime hyperideal and thus is $I$-prime for all $I$. Assume that $P$ is not a
prime hyperideal. Then $P^{2}=\\{0\\}$ and $(P\times S)^{2}=\\{0\\}\times S$.
Hence $\cap_{i=1}^{\infty}(P\times S)^{i}=\cap_{i=1}^{\infty}P^{i}\times
S=\\{0\\}\times S$. Thus $P\times S-\cap_{i=1}^{\infty}(P\times S)^{i}=P\times
S-\\{0\\}\times S=(P-\\{0\\})\times S$. Since $P$ is $\\{0\\}-$prime
hyperideal, $P\times S$ is $\cap_{i=1}^{\infty}(P\times S)^{i}$-prime
hyperideal and as $\cap_{i=1}^{\infty}(P\times S)^{i}\subseteq I(P\times S)$,
$P\times S$ is $I$-prime hyperideal.

(2) If $P$ is a prime hyperideal, then $P$ is $\\{0\\}-$prime. So we can
assume that $P$ is not prime hyperideal. Then $P^{2}\subseteq IP$ and hence
$P^{2}\subseteq IP\subseteq P^{3}$. So $P^{2}=P^{3}$. Hence $P^{2}$ is
idempotent. Since $P^{2}$ is finitely generated, $P^{2}=<e>$ for some
idempotent $e\in R$. Suppose $P^{2}=\\{0\\}$. Then $IP\subseteq
P^{3}=\\{0\\}$. So $IP=\\{0\\}$ and hence $P$ is $\\{0\\}-$prime. So assume
$P^{2}\neq\\{0\\}$. Put $S=P^{2}=$ ¡e¿ and $T=<1-e>$, so $R$ decomposes as
$T\times S$ where $S=P^{2}$. Let $J=P(1-e)$, so $P=J\times S$ where
$J^{2}=(P(1-e))^{2}=P^{2}(1-e)^{2}=<$ $e><1-e>=\\{0\\}$. We show that $J$ is
$\\{0\\}-$prime hyperideal. Let $a\circ b\subseteq J-\\{0\\}$, so
$(a,1)(b,1)=(a\circ b,1)\subseteq J\times S-(J\times S)^{2}=J\times
S-\\{0\\}\times S\subseteq P-IP$. Since $IP\subseteq P^{3}$ implies
$IP\subseteq P^{3}=(J\times S)^{3}=\\{0\\}\times S$. Hence $(a,1)\in P$ or
$(b,1)\in P$ so $a\in J$ or $b\in J$. Therefore $J$ is a $\\{0\\}-$prime
hyperideal. ∎

######  Corollary 2.7.1.

Let $(R,+,\circ)$ be an indecomposable commutative hyperring and $P$ a
finitely generated $I$-prime hyperideal of $(R,+,\circ)$, where $IP\subseteq
P^{3}$. Then $P$ is a $\\{0\\}-$prime hyperideal.

######  Corollary 2.7.2.

Let $(R,+,\circ)$ be a Noetherian integral hyperdomain. A proper hyperideal
$P$ of $R$ is prime hyperideal if and only if $P$ is $P^{2}$-prime hyperideal.

The next theorem is a generalization of [1, Theorem 2.12].

######  Theorem 2.8.

Let $P$ be a proper hyperideal of a hyperring $R$. Then the following
assertions are equivalent:

  1. (1)

$P$ is I-prime hyperideal.

  2. (2)

For $r\in R-P$, $(P:r)=P\cup(IP:r)$.

  3. (3)

For $r\in R-P$, $(P:r)=P$ or $(P:r)=(IP:r)$.

  4. (4)

For hyperideals $J$ and $K$ of $R$, $JK\subseteq P$ and $JK\nsubseteq IP$
imply $J\subseteq P$ or $K\subseteq P$.

###### Proof.

(1) $\Rightarrow$ (2) Suppose $r\in R-P$. Let $s\in(P:r)$, so $rs\subseteq P$.
If $rs\subseteq P-IP$, then $s\in P$. If $rs\subseteq IP$, then $s\in(IP:r)$.
So $(P:r)\subseteq P\cup(IP:r)$. The other containment always holds.

(2) $\Rightarrow$ (3) Note that if a hyperideal is a union of two hyperideals,
then it is equal to one of them.

$(3)\Rightarrow(4)$ Let $J$ and $K$ be two hyperideals of $R$ with
$JK\subseteq P$. Assume that $J\nsubseteq P$ and $K\nsubseteq P$. We claim
that $JK\subseteq IP$. Suppose $r\in J$. First, let $r\notin P$. Then
$rK\subseteq P$ gives $K\subseteq(P:r)$. Now $K\nsubseteq P$, so
$(P:r)=(IP:r)$. Thus $rK\subseteq IP$. Next, let $r\in J\cap P$. Choose $s\in
J-P$. Then $r+s\in J-P$. By the first case $sK\subseteq IP$ and so
$(r+s)K\subseteq IP.$ Pick $t\in K.$ Then $rt=(r+s)t-st\subseteq IP$ and
$rK\subseteq IP.$ Hence $JK\subseteq IP$.

$(4)\Rightarrow(1)$ Let $rs\in P-IP.$ Then $(r)(s)\subseteq P$. But
$(r)(s)\nsubseteq IP.$ So $(r)\subseteq P$ or $(s)\subseteq P$ which means
$r\in P$ or $s\in P.$ ∎

######  Proposition 2.9.

Let $P$ be an $I$-prime hyperideal of a hyperring $R$ and $J\subseteq P$ be a
hyperideal of $R$. Then $P/J$ is $I-$prime hyperideal of $R/J$.

###### Proof.

Let $x,y\in R$ with $\bar{x}\circ\bar{y}\subseteq P/J-I(P/J)=P/J-(IP+J)/J$
where $\bar{x}$, $\bar{y}$ are the images of $x,y$ in $R/J$. Thus $x\circ
y\subseteq P-IP$. So $x\in P$ or $y\in P$. Therefore $\bar{x}\in P/J$ or
$\bar{y}\in P/J$. So $P/J$ is $I$-prime hyperideal. ∎

Let $R_{1}$ and $R_{2}$ be two hyperrings. It is known that the prime
hyperideals of $R_{1}\times R_{2}$ have the form $P\times R_{2}$ or
$R_{1}\times Q$, where $P$ is a prime hyperideal of $R_{1}$ and $Q$ is a prime
hyperideal of $R_{2}$. We next generalize this result to $I$-prime
hyperideals.

######  Theorem 2.10.

Let $R_{i}$ be a hyperring and $I_{i}$ a hyperideal of $R_{i}$ for $i=1,2$.
Let $I=I_{1}\times I_{2}$. Then the I-prime hyperideals of $R_{1}\times R_{2}$
have exactly one of the following three types:
(1) $P_{1}\times P_{2}$, where $P_{i}$ is a proper hyperideal of $R_{i}$ with
$I_{i}P_{i}=P_{i}$.
(2) $P_{1}\times R_{2}$ where $P_{1}$ is an $I_{1}$-prime hyperideal of
$R_{1}$ and $I_{2}R_{2}=R_{2}$.
(3) $R_{1}\times P_{2}$, where $P_{2}$ is an $I_{2}$-prime hyperideal of
$R_{2}$ and $I_{1}R_{1}=R_{1}$.

###### Proof.

We first prove that a hyperideal of $R_{1}\times R_{2}$ having one of these
three types is $I$-prime hyperideal. The first type is clear since
$P_{1}\times P_{2}-I\left(P_{1}\times P_{2}\right)=$ $P_{1}\times
P_{2}-\left(I_{1}P_{1}\times I_{2}P_{2}\right)=\phi$. Suppose that $P_{1}$ is
$I_{1}$-prime hyperideal and $I_{2}R_{2}=R_{2}$. Let $(a,b)(x,y)\subseteq
P_{1}\times R_{2}-\left(I_{1}P_{1}\times I_{2}R_{2}\right)=P_{1}\times
R_{2}-\left(I_{1}P_{1}\times R_{2}\right)=$
$\left(P_{1}-I_{1}P_{1}\right)\times R_{2}$. Then $ax\subseteq
P_{1}-I_{1}P_{1}$ implies that $a\in P_{1}$ or $x\in P_{1}$, so $(a,b)\in
P_{1}\times R_{2}$ or $(x,y)\in P_{1}\times R_{2}$. Hence $P_{1}\times R_{2}$
is $I$-prime hyperideal. Similarly we can prove the last case. Next, let
$P_{1}\times P_{2}$ be $I$-prime and $ab\subseteq P_{1}-I_{1}P_{1}$. Then
$(a,0)(b,0)=$ $(ab,0)\in P_{1}\times P_{2}-I\left(P_{1}\times P_{2}\right)$,
so $(a,0)\in P_{1}\times P_{2}$ or $(b,0)\in P_{1}\times P_{2}$, that is,
$a\in P_{1}$ or $b\in P_{1}$. Hence $P_{1}$ is $I_{1}$-prime. Likewise,
$P_{2}$ is $I_{2}$-prime.

Assume that $P_{1}\times P_{2}\neq I_{1}P_{1}\times I_{2}P_{2}$, say
$P_{1}\neq I_{1}P_{1}$. Let $x\in P_{1}-I_{1}P_{1}$ and $y\in P_{2}$. Then
$(x,1)(1,y)=(x,y)\in P_{1}\times P_{2}$. So $(x,1)\in P_{1}\times P_{2}$ or
$(1,y)\in P_{1}\times P_{2}$. Thus $P_{2}=R_{2}$ or $P_{1}=R_{1}$. Assume that
$P_{2}=R_{2}$. Then $P_{1}\times R_{2}$ is $I$-prime, where $P_{1}$ is
$I_{1}$-prime.

∎

##  3\. $n-$absorbing $I-$prime hyperideals

We start this section by the definition of $n-$absorbing $I-$prime
hyperideals.

######  Definition 3.1.

A proper hyperideal $P$ of a hyperring $R$ is a $2-$absorbing $I-$prime
hyperideal if for $x_{1},x_{2},x_{3}\in R$ such that $x_{1}x_{2}x_{3}\subseteq
P-IP$, then $x_{1}x_{2}\subseteq P$ or $x_{1}x_{3}\subseteq P$ or
$x_{2}x_{3}\subseteq P$. A proper hyperideal $P$ of $R$ is an $n-$absorbing
$I-$prime hyperideal if for $x_{1},\cdots,x_{n+1}\in R$ such that $x_{1}\cdots
x_{n+1}\subseteq P-IP$, then $x_{1}\cdots x_{i-1}x_{i+1}\cdots
x_{n+1}\subseteq P$ for some $i\in\\{1,2,\cdots,n+1\\}$.

It is clear that the class of $n-$absorbing $I-$prime hyperideals contains
properly the class of $n-$absorbing hyperideals. As we can see this in the
following example.

######  Example 3.1.

Let $K$ be a hyperfield and $R=K[x_{1},\cdots,x_{n+2}]$ be a polynomial
multiplicative hyperring. Consider the hyperideals $P=<x_{1}\cdots
x_{n+1},x_{1}^{2}\cdots x_{n},x_{1}^{2}x_{n+2}>$ and $I=<x_{1}\cdots x_{n}>$.
So $P-IP=<x_{1}\cdots x_{n+1},x_{1}^{2}\cdots x_{n},x_{1}^{2}x_{n+2}>$ $-$
$<x_{1}\cdots x_{n+1},x_{1}^{2}\cdots x_{n},x_{1}^{2}\cdots x_{n}x_{n+2}>$.
Hence $P$ is an $n-$absorbing $I-$prime hyperideal but not $n-$absorbing
hyperideal.

######  Lemma 3.1.

Let $P$ be an $I-$prime hyperideal of $R$ and $K$ be a subset of $R$. For any
$a\in R,aK\subseteq P,\,aK\nsubseteq IP$ and $a\notin P$ implies that
$K\subseteq P$. (or $aK\subseteq P$ and $K\nsubseteq P$ imply that $a\in P$ ).

###### Proof.

Let $aK\subseteq P$ and $a\notin P$ for any $a\in R$. Then we have $aK=\cup
ak_{i}\subseteq P$ for all $k_{i}\in K$. Hence $ak_{i}\subseteq P$ and
$ak_{i}\nsubseteq IP$ for all $k_{i}\in K$. Since $P$ is an $I-$prime
hyperideal and $a\notin P$, $k_{i}\in P,\,\,\forall k_{i}\in K$. Thus
$K\subseteq P$. ∎

######  Lemma 3.2.

Let $P$ be an $I-$prime hyperideal of $R$ and $A,B$ be subsets of $R$. If
$AB\subseteq P$ and $AB\nsubseteq IP$, then $A\subseteq P$ or $B\subseteq P$.

###### Proof.

Assume that $AB\subseteq P,AB\nsubseteq IP$ and $A\nsubseteq P$, $B\nsubseteq
P$. Since $AB=\bigcup a_{i}b_{i}\subseteq P$, $a_{i}b_{i}\subseteq P$, for
$a_{i}\in A,b_{i}\in B$. And as $A\nsubseteq P$ and $B\nsubseteq P$, we have
$x\notin P$ and $y\notin P$ for some $x\in A,y\in B$. Then $xy\subseteq
AB\subseteq P$ and $xy\nsubseteq IP$. From being $P$ an $I-$prime hyperideal,
we have $x\in P$ or $y\in P$ which is a contradiction. Thus $A\subseteq P$ or
$B\subseteq P$. ∎

Every $I-$prime hyperideal is a $2-$absorbing $I-$prime hyperideal. Since for
$(ab)c\subseteq P-IP$, we have $ab\subseteq P$ or $bc\subseteq P$. If
$ab\nsubseteq P$ then by $I-$prime hyperideal of $P$, we have $c\in P$ and so
$ac\in P$ or $bc\in P$. Hence $P$ is a $2-$absorbing $I-$prime hyperideal of
$R$.

######  Lemma 3.3.

Let $P$ be a hyperideal of $R$ and $P_{1},P_{2},\ldots,P_{n}$ be $2-$absorbing
primary hyperideals of $R$ such that $\sqrt{P_{i}}=P$ for all
$\mathrm{i}=1,\ldots,\mathrm{n}$. Then $\bigcap_{i=1}^{n}P_{i}$ is a
2-absorbing $I-$prime hyperideal and $\bigcap_{i=1}^{n}P_{i}=P$.

###### Proof.

Assume $P=\bigcap_{i=1}^{n}P_{i}$ and so
$\sqrt{P}=\sqrt{\cap_{i=1}^{n}P_{i}}=\cap_{i=1}^{n}\sqrt{P_{i}}=P$. Let
$xyz\subseteq P-IP$ with $xy\nsubseteq P$, for $x,y,z\in R$. Thus
$xy\nsubseteq P_{i}$ for some $i=1,2,\cdots,n$. From being $P_{i}$ a
2-absorbing primary hyperideal and $xyz\subseteq P-IP\subseteq P_{i}$, hence
$xz\subseteq\sqrt{P_{i}}=P$ or $yz\subseteq\sqrt{P_{i}}=P$ which means that
$P$ is a 2-absorbing $I-$prime hyperideal of $R$. ∎

######  Theorem 3.4.

Let $h:R\rightarrow L$ be a bijective good homomorphism of hyperrings and $P$
be a 2-absorbing $I-$prime hyperideal of $L$. Then $h^{-1}(P)$ is a
2-absorbing $h^{-1}(I)-$prime hyperideal of $R$.

###### Proof.

Suppose that $abc\subseteq h^{-1}(P),h^{-1}(I)h^{-1}(P)=h^{-1}(P)-h^{-1}(IP)$,
for $a,b,c\in R$. So $h(abc)=h(a)h(b)h(a)\subseteq P$ and $h(abc)\nsubseteq
IP$. From being $P$ a 2-absorbing $I-$prime hyperideal, we have
$h(a)h(b)\subseteq P$ or $h(a)h(c)\subseteq P$ or $h(b)h(c)\subseteq P$, that
is $h(ab)\subseteq P$ or $h(ac)\subseteq P$ or $h(bc)\subseteq P$ which
implies $ab\subseteq$ $h^{-1}(P)$ or $ac\subseteq h^{-1}(P)$ or $bc\subseteq
h^{-1}(P)$. So $h^{-1}(P)$ is a 2-absorbing $h^{-1}(I)-$prime hyperideal of
$R$. ∎

######  Theorem 3.5.

Suppose that $P$ is an $n-$absorbing $I-$prime hyperideal of $R$. Then
$\sqrt{P}$ is an $n-$absorbing $\sqrt{I}-$prime hyperideal of $R$ and
$a^{n}\subseteq P$ for all $a\in\sqrt{P}$.

###### Proof.

Let $a\in\sqrt{P}$. Then $a^{m}\subseteq P$ for some $m\in\mathbb{N}$. If
$m\leq n$, we are done. If $m>n$, by using the $n$-absorbing $I-$prime
property on products $a^{m}$, we conclude that $a^{n}\subseteq P$. Now,
consider $a_{1}\cdots
a_{n+1}\subseteq\sqrt{P}-\sqrt{I}\sqrt{P}=\sqrt{P}-\sqrt{IP}$ for
$a_{1},\cdots,a_{n+1}\in R$. Thus $\left(a_{1}\ldots
a_{n+1}\right)^{n}=a_{1}^{n}\cdots a_{n+1}^{n}\subseteq P$. If
$a_{1}^{n}\cdots a_{n+1}^{n}\subseteq IP$, then $a_{1}\cdots
a_{n+1}\subseteq\sqrt{IP}$ which is a contradiction. Hence $a_{1}^{n}\cdots
a_{n+1}^{n}\subseteq P-IP$ and $P$ $n$-absorbing $I-$prime hyperideal gives us
the desired. ∎

######  Lemma 3.6.

Let $P_{i}$ be an $n_{i}-$absorbing $I-$prime hyperideal of a hyperring $R$
for $i=1,2,\cdots,m$ and $IP_{i}=IP_{j}$, for $i\neq j$, Then
$\cap_{i=1}^{m}P_{i}$ is an $n-$absorbing $I-$prime hyperideal where
$n=\sum_{i=1}^{m}n_{i}$.

###### Proof.

Let $k>n$ and $x_{1}\cdots
x_{k}\subseteq\cap_{i=1}^{m}P_{i}-I\cap_{i=1}^{m}P_{i}$. Then by hypothesis
for each $i=1\cdots m$, there exists a product of $n_{i}$ of these
$k-$elements in $P_{i}$. Let $A_{i}$ be the collection of these elements and
let $A=\cup_{i=1}^{k}A_{i}$. Thus $A$ has at most $n-$elements. Now, as
$P_{i}$ is an $n-$absorbing $I-$prime hyperideal, the product of all elements
of $A$ must be in each $P_{i}$ so $\cap P_{i}$ contains a product of at most
$n-$elements and therefore it is an $n-$absorbing $I-$prime hyperideal of $R$.
∎

######  Theorem 3.7.

Let $R=\prod_{i=1}^{n+1}R_{i}$ and $P$ be a proper non-zero hyperideal of $R$.
If P is an $(n+1)-$absorbing $I-$prime hyperideal of $R$, then $P=P_{1}\times
P_{2}\times\cdots\times P_{n+1}$ for some proper $n-$absobing $I_{i}-$prime
hyperideals $P_{1},\cdots,P_{n+1}$ of $R_{1},\cdots,R_{n+1}$ respectively,
where $I=\prod_{i=1}^{n+1}I_{i}$ and $I_{i}=R_{i}$, $\forall
i=1,2,\cdots,n+1.$

###### Proof.

Let $x_{1},\cdots,x_{n+1}\in R$ with $x_{1}\cdots x_{n+1}\subseteq
P_{1}-I_{1}P_{1}$ and suppose by contrary that $P_{1}$ is not an $n-$absorbing
$I_{1}-$prime hyperideal of $R_{1}$. Set $a_{i}=(x_{i},1,1,\cdots,1)$ for
$i=1,2,\cdots,n+1$ and $a_{n+2}=(1,0,\cdots,0)$. Then we have
$a_{1}a_{2}\cdots a_{n+2}=(x_{1}x_{2}\cdots x_{n+1},0,0,\cdots,0)\subseteq
P-IP$ and $a_{1}\cdots a_{i-1}a_{i+1}\cdots a_{n+2}=(x_{1}x_{2}\cdots
x_{i-1}x_{i+1}\cdots x_{n+1},0,\cdots,0)\nsubseteq P$ for $i=1,\cdots,n+1$,
which contradicts with being $P$ an $(n+1)-$absorbing $I-$prime hyperideal.
Hence $P_{1}$ must be an $n-$absorbing $I_{1}-$prime hyperideal of $R_{1}$. By
similar arguments, we can show that $P_{i}$ is an $n-$absorbing $I_{i}-$prime
hyperideal of $R_{i}$ for $i=1,\cdots,n+1$. ∎

######  Theorem 3.8.

Let $R=\prod_{i=1}^{n+1}R_{i}$, where $R_{i}$ is a hyperring for
$i\in\\{1,\cdots,n+1\\}$. If $P$ is an $n-$absorbing $I-$prime hyperideal of
$R$, then either $P=IP$ or $P=P_{1}\times P_{2}\times\cdots\times
P_{i-1}\times R_{i}\times P_{i+1}\cdots\times P_{n+1}$ for some
$i\in\\{1,\cdots,n+1\\}$ and if $P_{j}\neq R_{i}$ for $j\neq i$, then $P_{j}$
is an $n-$absorbing hyperideal in $R_{i}$.

###### Proof.

Let $P=\prod_{i=1}^{n+1}P_{i}$ be an $n-$absorbing $I-$prime hyperideal of
$R$. Then there exists $(x_{1},\cdots,x_{n+1})\subseteq P-IP$, and so
$(x_{1},1,\cdots,1)(1,x_{2},1$
$\cdots,1)\cdots(1,1,\cdots,1,x_{n+1})=(x_{1},x_{2},\cdots,x_{n+1})\subseteq
P-IP$. As $P$ is an $n-$absorbing $I-$prime hyperideal, we have
$(x_{1},x_{2},\cdots,x_{i-1},1,x_{i+1},\cdots$
$,x_{n+1})\subseteq P$ for some $i\in\\{1,2,\cdots,n+1\\}$. Thus
$(0,0,\cdots,0,1,0,\cdots,0)$
$\in P$ and hence $P=P_{1}\times P_{2}\times\cdots\times P_{i-1}\times
R_{i}\times P_{i+1}\cdots\times P_{n+1}$. If $P_{j}\neq R_{i}$ for $j\neq i$,
then we have to prove $P_{j}$ is an $n-$absorbing hyperideal of $R_{i}$. Let
$i<j$ and take $x_{1}x_{2}\cdots x_{n+1}\subseteq P_{j}$. Then
$(0,0,\cdots,0,1,0,\cdots,0,x_{1}x_{2}\cdots x_{n+1},0\cdots,0)$
$=(0,0,\cdots,1,0,\cdots,0,x_{1}\\\
,0\cdots,0)(0,0,\cdots,1,0,\cdots,0,x_{2},0\cdots,0)\cdots$
$(0,0,\cdots,1,0,\cdots,0,x_{n+1}\\\ ,0\cdots,0)\subseteq P-IP$. Since $P$ is
an $n-$absorbing $I-$prime hyperideal,
$(0,0,\cdots,0,1,0,\cdots,0,x_{1}x_{2}\cdots x_{k-1}x_{k+1}\cdots
x_{n+1},0,\cdots,0)\in P$ for some $k\in\\{1,2,\cdots,n+1\\}$. Thus
$x_{1}x_{2}\cdots x_{k-1}x_{k+1}\cdots x_{n+1}\in P_{j}$ and hence $P_{j}$ is
an $n-$absorbing hyperideal of $R_{i}$. We can do similar arguments for the
case $i>j$. ∎

In the following result, we characterize hyperrings in which every proper
hyperideal of $R$ is an $n-$absorbing $I-$prime hyperideal.

######  Theorem 3.9.

Let $\mid Max(R)\mid\geq n+1\geq 2.$ Then each proper hyperideal of $R$ is an
$n-$absorbing $I-$prime hyperideal if and only if each quotient of $R$ is a
product of $(n+1)-$hyperfields.

###### Proof.

$(\Rightarrow)$ Let $P$ be a proper hyperideal of $R$. Then $\frac{R}{IP}\cong
F_{1}\times\cdots\times F_{n+1}$ and $\frac{P}{IP}\cong
P_{1}\times\cdots\times P_{n+1}$, where $P_{i}$ is a hyperideal of $F_{i}$,
$i=1,\cdots,{n+1}$. If $P=IP$ then there is nothing to prove, otherwise we
have $P_{j}=0,$ for at least one $j\in\\{1,\cdots,{n+1}\\}$ since
$\frac{P}{IP}$ is proper. So, $\frac{P}{IP}$ is an $n-$absorbing $0-$prime
hyperideal of $\frac{R}{IR}$ which means $P$ is an $n-$absorbing $I-$prime
hyperideal of $R$.
$(\Leftarrow)$ Let $m_{1},\cdots,m_{n+1}$ be distinct maximal hyperideals of
$R$. Then $m=\prod_{i=1}^{n+1}m_{i}$ is an $n-$absorbing $I-$hyperideal of
$R$. we claim that $m$ is not an $n-$absorbing hyperideal. First, if
$m_{i}\subseteq\cup_{i\neq j}m_{j}$, then there exist $m_{j}$ with
$m_{i}\subseteq m_{j}$ by Prime Avoidance Lemma and this contradicts the
maximality of $m_{i}$. Hence $m_{i}\nsubseteq\cup_{i\neq j}m_{j}$ and so,
there exists $x_{i}\in m_{i}-\cup_{i\neq j}^{n+1}m_{j}$ so that $x_{1}\cdots
x_{n+1}\subseteq m$. If there exists $j\in\\{1,\cdots,n+1\\}$ with
$a=x_{1}x_{2}\cdots x_{j-1}x_{j+1}\cdots x_{n+1}\subseteq m\subseteq m_{j}$,
then $x_{i}\in m_{j}$ for some $i\neq j$ which is a contradiction. Hence $m$
is not an $n-$absorbing hyperideal and so $m^{n+1}=Im$. Then by Chinese
Remainder Theorem we have
$\frac{R}{Im}\simeq\frac{R}{m_{1}^{n+1}}\times\frac{R}{m_{2}^{n+1}}\times...\times\frac{R}{m_{n+1}^{n+1}}$.
Put $F_{i}=\frac{R}{m_{1}^{n+1}}$. If $F_{i}$ is not a hyperfield, then it has
a nonzero proper hyperideal $H$ and so $0\times 0\times\cdots\times 0\times
H\times 0\times\cdots\times 0$ is an $n-$absorbing $0-$hyperideal of
$\frac{R}{Im}$. Thus, by Lemma 3.8 we have $H=F_{i}$ or $H=0$ which is
impossible. Hence $F_{i}$ is a hyperfield.

∎

######  Corollary 3.9.1.

Suppose $\mid Max(R)\mid\geq n+1\geq 2$. Then each proper hyperideal of $R$ is
an $n-$absorbing $0-$hyperideal if and only if $R\cong F_{1}\times\cdots\times
F_{n+1}$, where $F_{1},\cdots,F_{n+1}$ are hyperfields.

######  Theorem 3.10.

Let $P$ be an $n-$absorbing $I-$prime hyperideal of a hyperring $R$. Then
there are at most $n$ prime hyperideals of $R$ that are minimal over $P$.

###### Proof.

Let $C=\\{q_{i}:q_{i}\textnormal{ is a prime hyperideal minimal over}\,P\\}$
and let $C$ has at least $n$ elements. Assume $q_{1},\cdots,q_{n}\in C$ are
distinct elements and $x_{i}\in q_{i}-\cup_{j\neq i}q_{j}$ for $i=1,\cdots,n$.
By [9, Theorem 2.1], there is a $y_{i}\notin q_{j}$ such that
$y_{i}x_{i}^{t_{i}}\subseteq P$ for $i=1,\cdots,n$ and for some positive
integers $t_{1},\cdots,t_{n}$. Since $x_{i}\notin\cap_{j=1}^{n}q_{j}$ and $P$
an $n-$absorbing $I-$prime hyperideal, we have $y_{i}x_{i}^{n-1}\in P$. As
$x_{i}\notin\cap_{j=1}^{n}q_{j}$ and $y_{i}x_{i}^{n-1}\subseteq
P\subseteq\cap_{j=1}^{n}q_{j}$, we get $y_{i}\in q_{i}-\cup_{j\neq i}q_{j}$,
and so $y_{i}\notin\cap_{j=1}^{n}q_{j}$ for $i=1,\cdots,n$. Since
$y_{i}x_{i}^{n-1}\subseteq P$,
$\sum_{j=1}^{n}y_{j}\prod_{i=1}^{n}x_{i}^{n-1}\subseteq P$ and clearly
$\sum_{j=1}^{n}y_{j}\notin q_{i}$, for $i=1,\cdots,n$, and being $P$ an
$n-$absorbing $I-$prime hyperideal, we have $y_{i}x_{i}^{n-1}\in P$. As
$x_{i}\notin\cap_{j=1}^{n}q_{j}$ and $y_{i}x_{i}^{n-1}\subseteq
P\subseteq\cap_{i=1}^{n}q_{i}$, we get $y_{i}\in q_{i}-\cup_{j\neq i}q_{j}$
and so $y_{i}\notin\cap_{i=1}^{n}q_{i}$ for $i=1,\cdots n$. Since
$y_{i}x_{i}^{n-1}\subseteq P$,
$\varSigma_{j=1}^{n}y_{j}\prod_{i=1}^{n}x_{i}^{n+1}\subseteq P$ and clearly
$\varSigma_{j=1}^{n}y_{j}\notin q_{i}$, for $i=1,\cdots n$ and being $P$ an
$n-$absorbing $I-$prime hyperideal, we have
$\prod_{i=1}^{n}x_{i}^{n-1}\subseteq P$. Now, suppose $q_{n+1}\in C$ such that
$q_{n+1}\neq q_{i}$, for $i=1,\cdots,n$ and consequently $z_{i}\in q_{n+1}$
for $i=1,\cdots,n$ which is a contradiction. Therefore $C$ has at least $n$
elements.

∎

In a multiplicative hyperring $(R,+,\circ)$ a non empty subset $L$ of $R$ is
called a multiplicative set whenever $a,b\in A\Rightarrow a\circ b\cap
A\neq\phi$.

We can contract the localization of a multiplicative hyperring $R$ as follows:
Let $S$ be a multiplicative closed subset of $R$, that is, $S$ is closed under
the hypermultiplication and contains the identity. Let $S^{-1}R$ be the set
$(R\times S/\sim)$ of equivalence classes where

$(r_{1},s_{1})\sim(r_{2},s_{2})\iff\exists s\in S\,\,\text{such that
}ss_{1}r_{2}=ss_{2}r_{1}.$

Let $r/s$ be the equivalence class of $(r,s)\in R\times S$ under the
equivalence relation $\sim$. The operation addition and the hyperoperation
multiplication are defined by

$\frac{r_{1}}{s_{1}}+\frac{r_{2}}{s_{2}}=\frac{s_{1}r_{2}+s_{2}r_{1}}{s_{1}s_{2}}=\\{\frac{a+b}{c}:a\in
s_{1}r_{2},b\in s_{2}r_{1},c\in s_{1}s_{2}\\}$
$\frac{r_{1}}{s_{1}}\cdot{\frac{r_{2}}{s_{2}}}=\frac{r_{1}r_{2}}{s_{1}s_{2}}=\\{\frac{a}{b},a\in
r_{1}r_{2},b\in s_{1}s_{2}\\}.$

Note that the localization map $f:R\to S^{-1}R$, $f(r)=\frac{r}{1}$ is a
homomorphism of hyperrings. It is easy to see that the localization of a
hyperideal is a hyperideal.

######  Proposition 3.11.

Let $P$ be an $I-$prime hyperideal of $R$ with $S\cap P=\emptyset$. Then
$S^{-1}P$ is an $S^{-1}I-$prime hyperideal of $S^{-1}R$.

###### Proof.

$\frac{r_{1}}{s_{1}},\frac{r_{2}}{s_{2}}\in S^{-1}R$ with
$\frac{r_{1}}{s_{1}}\frac{r_{2}}{s_{2}}=\frac{r_{1}r_{2}}{s_{1}s_{2}}\subseteq
S^{-1}P-S^{-1}IS^{-1}P=S^{-1}P-S^{-1}(IP)$. For each $n\in r_{1}r_{2},s\in
s_{1}s_{2}$, we have $\frac{n}{s}\in\frac{r_{1}r_{2}}{s_{1}s_{2}}$ and
$\frac{n}{s}=\frac{a}{t}$, where $a\in P$, $t\in S$. So there exists $q\in S$
such that $qtn=qsa$. Hence $qtn\subseteq P-IP$ and so $qr_{1}r_{2}\subseteq
P-IP$. As $P$ is an $I-$prime hyperideal, we have $qr_{1}\subseteq P$ or
$r_{2}\in P$. Thus $\frac{r_{1}}{s_{1}}=\frac{qr_{1}}{qs_{1}}\in P$ or
$\frac{r_{2}}{s_{2}}\in S^{-1}P$. Therefore $S^{-1}P$ is an $S^{-1}I-$prime
hyperideal of $S^{-1}R$. ∎

Questions Readers can think about the follwoing subjects:

  1. (1)

$2-I-$primal hyperideals

  2. (2)

$2-I-$primal hypersubmodules

  3. (3)

$2-I-$prime hypersubmodules

Conclosion. In this article we transfer notions $I-$prime ideals and
$n-$absorbing $I-$ideals in multiplicative hyperrings and named them $I-$prime
hyperideal and $n-$absorbing $I-$prime hyperideal. We study some properties of
such two concepts and we see that they have analogous properties of prime
ideals. During the study, we found out similar consepts that one can think
about like $2-I-$primal hyperideals, $2-I-$primal hypersubmodules and
$2-I-$prime hypersubmodules.

Acknowleghments. The authors would like to express their sincere thanks to the
anonymous referees for their valuable comments and suggestions on the earlier
version of the manuscript which improved it a lot.

## References

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# Axion-electron decoupling in nucleophobic axion models

Fredrik Björkeroth  INFN, Laboratori Nazionali di Frascati, C.P. 13, 100044
Frascati, Italy Luca Di Luzio  Dipartimento di Fisica, Università di Pisa and
INFN, Sezione di Pisa,
Largo B. Pontecorvo 3, 56127 Pisa, Italy Federico Mescia  Dept. de Física
Quàntica i Astrofísica, Institut de Ciències del Cosmos (ICCUB),
Universitat de Barcelona, Martí Franquès 1, E08028 Barcelona, Spain Enrico
Nardi  INFN, Laboratori Nazionali di Frascati, C.P. 13, 100044 Frascati, Italy
Paolo Panci  Dipartimento di Fisica, Università di Pisa and INFN, Sezione di
Pisa,
Largo B. Pontecorvo 3, 56127 Pisa, Italy Theoretical Physics Department, CERN,
Geneva, Switzerland Laboratori Nazionali del Gran Sasso, Via G. Acitelli, 22,
I-67100 Assergi (AQ), Italy Robert Ziegler  Theoretical Physics Department,
CERN, Geneva, Switzerland

###### Abstract

The strongest upper bounds on the axion mass come from astrophysical
observations like the neutrino burst duration of SN1987A, which depends on the
axion couplings to nucleons, or the white-dwarf cooling rates and red-giant
evolution, which involve the axion-electron coupling. It has been recently
argued that in variants of DFSZ models with generation-dependent Peccei-Quinn
charges an approximate axion-nucleon decoupling can occur, strongly relaxing
the SN1987A bound. However, as in standard DFSZ models, the axion remains in
general coupled to electrons, unless an _ad hoc_ cancellation is engineered.
Here we show that axion-electron decoupling can be implemented without extra
tunings in DFSZ-like models with three Higgs doublets. Remarkably, the
numerical value of the quark mass ratio $m_{u}/m_{d}\sim 1/2$ is crucial to
open up this possibility.

††preprint: CERN-TH-2019-118

##  I Introduction

It has been recently argued Di Luzio  _et al._ (2018) that in variants of
Dine-Fischler-Srednicki-Zhitnitsky (DFSZ) Dine  _et al._ (1981); Zhitnitsky
(1980) models with two Higgs doublets and generation-dependent Peccei Quinn
(PQ) charges, it is possible to strongly suppress axion couplings to nucleons
(axion nucleophobia). This implies that the upper limit on the DFSZ axion mass
from the neutrino burst duration of the supernova (SN) SN1987A, which is
particularly strong and generally considered ineludible, can in fact be
sizeably relaxed.111For Kim-Shifman-Vainshtein-Zakharov (KSVZ) Kim (1979);
Shifman  _et al._ (1980) axions instead, no suppression mechanism for the
axion couplings to nucleons can be enforced, since they are determined in a
model-independent way, and yield the often quoted limit $m_{a}\lesssim
0.02\,$eV Tanabashi  _et al._ (2018). Note, however, that recent analyses of
the axion emissivity from the SN core hint to a weakening of the bound by a
factor of a few Chang  _et al._ (2018); Carenza  _et al._ (2019).  The
parameter space region that opens up for nucleophobic axions is, however, only
marginal. This is because in DFSZ models the axion also couples to electrons,
and then limits from anomalous cooling of white dwarfs and red giants, which
are only moderately less restrictive, apply. On the other hand, generation-
dependent PQ charges imply that the axion couplings to the mass eigenstate
fermions receive corrections from inter-generational mixing. In Ref. Di Luzio
_et al._ (2018) this type of effect was invoked to arrange for a tuned
cancellation between two contributions to the axion-electron coupling: one
proportional to the electron PQ charge, and the other coming from inter-
generational mixing between the leptons. This allows the construction of
models of nucleophobic _and_ electrophobic axions that can evade all the
tightest astrophysical bounds (astrophobic axions). Although the tuning of the
cancellation required to achieve a significant level of electrophobia is at
the level of 10%, astrophobic axion models constructed in this way are
admittedly not particularly elegant. In this short note we put forth a more
natural way to achieve astrophobia, which requires extending the scalar sector
by a third Higgs doublet, but does not involve any _ad hoc_ cancellation
between different contributions to the axion-electron coupling. Remarkably,
this mechanism implies a strong correlation between the couplings whereby the
higher the level of suppression of the axion-nucleon coupling, the more the
axion becomes electrophobic. Intriguingly, the mechanism can be implemented
thanks to the fact that the light quark mass ratio is close to
$m_{u}/m_{d}\approx 1/2$.

##  II The conditions for nucleophobia

Let us first recall the conditions for nucleophobia. We define the axion
couplings to nucleons via

$\frac{C_{\mathcal{N}}}{2f_{a}}\partial_{\mu}a\,\overline{\mathcal{N}}\gamma^{\mu}\gamma_{5}\mathcal{N},$
(1)

with $\mathcal{N}=p,n$, while the fundamental couplings of the axion to quarks
$C_{q}$, with $q=u,d,\dots$ are defined from a similar expression by replacing
$\mathcal{N}\to q$ and $C_{\mathcal{N}}\to C_{q}$. $C_{\mathcal{N}}$ can be
expressed in terms of $C_{q}$ using non-perturbative inputs from nucleon
matrix elements Grilli di Cortona  _et al._ (2016). To understand the
mechanism behind axion-nucleon decoupling, it is convenient to consider the
two linear combinations $C_{p}\pm C_{n}$ and express them in terms of the
$C_{q}$. This yields

$\displaystyle C_{p}+C_{n}$
$\displaystyle=0.50\left(C_{u}+C_{d}-1\right)-2\delta_{s},$ (2) $\displaystyle
C_{p}-C_{n}$ $\displaystyle=1.27\left(C_{u}-C_{d}-f_{ud}\right),$ (3)

where, in the second line, $f_{ud}=f_{u}-f_{d}$, with
$f_{u,d}=m_{d,u}/(m_{d}+m_{u})$ the model-independent contributions induced by
the axion coupling to gluons, chosen in such a way that the axion does not mix
with $\pi^{0}$. In the first line, $1=f_{u}+f_{d}$ is an exact number, while
$\delta_{s}=0.038\,C_{s}+0.012\,C_{c}+0.009\,C_{b}+0.0035\,C_{t}$ is a small
$\mathcal{O}(5\%)$ correction dominated by the $s$-quark contribution.
Nucleophobia requires $C_{p}\pm C_{n}\approx 0$, which is possible in variant
DFSZ models with two Higgs doublets $H_{1,2}$ and non-universal PQ charge
assignment Di Luzio  _et al._ (2018). To see this, let us focus on the first
generation Yukawa terms

$\overline{q}_{1}u_{1}H_{1}+\overline{q}_{1}d_{1}\tilde{H}_{2},$ (4)

where $\tilde{H}_{2}=i\sigma_{2}H_{2}^{*}$. The axion couplings to the light
quark fields (neglecting flavour mixing, which is assumed to be small
throughout this paper222In the presence of flavour mixing, $C_{q}\to
C_{q}+\Delta C_{q}$, where $\Delta C_{q}$ involves quark mass diagonalization
matrices. We refer the reader to Di Luzio  _et al._ (2018) for details. ) are

$\displaystyle C_{u}$
$\displaystyle=\frac{1}{2N}\left(\mathcal{X}_{u_{1}}-\mathcal{X}_{q_{1}}\right)=-\frac{\mathcal{X}_{1}}{2N},$
(5) $\displaystyle C_{d}$
$\displaystyle=\frac{1}{2N}\left(\mathcal{X}_{d_{1}}-\mathcal{X}_{q_{1}}\right)=\frac{\mathcal{X}_{2}}{2N}.$
(6)

Here $\mathcal{X}_{u_{1}}=\mathcal{X}(u_{1})$, etc. denote the PQ charges of
the fermion fields while $\mathcal{X}_{1,2}=\mathcal{X}(H_{1,2})$. The
coefficient of the PQ colour anomaly is then

$2N=\sum_{i=1}^{3}\left(\mathcal{X}_{u_{i}}+\mathcal{X}_{d_{i}}-2\mathcal{X}_{q_{i}}\right).$
(7)

It is also convenient to define the contribution to the color anomaly from
light quarks only:

$2N_{\ell}=\mathcal{X}_{u_{1}}+\mathcal{X}_{d_{1}}-2\mathcal{X}_{q_{1}}=\mathcal{X}_{2}-\mathcal{X}_{1}.$
(8)

The first condition for ensuring approximate nucleophobia then reads (cf. Eq.
(2))

$C_{u}+C_{d}=\frac{N_{\ell}}{N}=1,$ (9)

i.e. only models in which the color anomaly is determined solely by the light
$u,d$ quarks (while the contributions from the two heavier generations cancel
or vanish identically) have a chance to be nucleophobic.333It is worthwhile
mentioning that a certain number of models sharing precisely this property
were found in a recent study of $U(1)$ flavour symmetry for the quark sector
Bjorkeroth  _et al._ (2019).  As emphasized in Di Luzio  _et al._ (2018), this
implies that nucleophobic axion models can be realized only if the PQ charges
are generation-dependent.

Assuming that the first condition Eq. (9) is satisfied, the second condition
(cf. Eq. (3)) reads

$C_{u}-C_{d}=\frac{-\mathcal{X}_{1}-\mathcal{X}_{2}}{2N}=\frac{\mathcal{X}_{1}+\mathcal{X}_{2}}{\mathcal{X}_{1}-\mathcal{X}_{2}}=f_{ud},$
(10)

that is,

$\frac{\mathcal{X}_{1}}{\mathcal{X}_{2}}=-\frac{m_{d}}{m_{u}}.$ (11)

Finally, by imposing the condition which ensures that the physical axion field
is orthogonal to the Goldstone field of hypercharge $U(1)_{Y}$, i.e.
$\mathcal{X}_{1}v^{2}_{1}+\mathcal{X}_{2}v^{2}_{2}=0$, we obtain a relation
between the ratio of the vacuum expectation values (VEVs) $v_{1,2}=\langle
H_{1,2}\rangle$ and the ratio of the quark masses that must be satisfied in
order to ensure nucleophobia:

$\frac{v^{2}_{2}}{v^{2}_{1}}=-\frac{\mathcal{X}_{1}}{\mathcal{X}_{2}}=\frac{m_{d}}{m_{u}}.$
(12)

With only two Higgs doublets responsible for breaking the electroweak symmetry
and for providing masses to all the fermions, the lepton sector is unavoidably
charged under the PQ symmetry and, as mentioned in the introduction,
electrophobia can only be enforced by tuning a cancellation between the
contribution to the axion electron coupling proportional to the electron PQ
charge, and corrections arising from lepton flavour mixing Di Luzio  _et al._
(2018). A possible, and more elegant alternative, is to introduce a third
Higgs doublet $H_{3}$ with PQ charge $\mathcal{X}_{3}$ that couples only to
the leptons, and verify if the condition $\mathcal{X}_{3}\approx 0$ can be
consistently implemented. In this case the whole lepton sector would be
approximately neutral under the PQ symmetry, and in particular the axion would
decouple from the electrons. This possibility is explored in the remainder of
the paper.

##  III The conditions for electrophobia

Let us consider a three-Higgs doublet model (3HDM) wherein $H_{1,2}$ couple to
quarks as above, while $H_{3}$ couples to the leptons. We want to study if
electrophobia can be implemented consistently with nucleophobia. Assuming Eq.
(9) is satisfied, we have four additional conditions: orthogonality between
the physical axion and the hypercharge Goldstone, the second condition for
nucleophobia Eq. (11), and two conditions on the PQ charges that follow from
the requirement that the four $U(1)$ rephasing symmetries of the kinetic term
of the four scalar fields $H_{1,2,3}$ and $\phi$ (the latter being the
Standard Model (SM) singlet, with $\mathcal{X}_{\phi}=1$, responsible for PQ
breaking) are broken down to $U(1)_{Y}\times U(1)_{PQ}$. These last two
conditions can be implemented either by coupling the leptonic Higgs doublet
$H_{3}$ to both hadronic Higgses ($H_{1,2}$), or by coupling one of the two
hadronic Higgses to the other two doublets:

$H_{3}^{\dagger}H_{1}\phi^{m}+H_{3}^{\dagger}H_{2}\phi^{n}~{}~{}\mathrm{or}~{}~{}H_{3}^{\dagger}H_{1,2}\phi^{m}+H_{2}^{\dagger}H_{1}\phi^{n}.$
(13)

For renormalizable operators one has, without loss of generality, $m=1,2$ and
$n=\pm 1,\pm 2$, where negative values of $n$ mean Hermitian conjugation
$\phi^{n}\equiv(\phi^{\dagger})^{|n|}$. All in all, for the first case in Eq.
(13) the four conditions read

$\displaystyle\frac{\mathcal{X}_{1}}{\mathcal{X}_{2}}$
$\displaystyle=-\frac{m_{d}}{m_{u}},$ (14)
$\displaystyle\mathcal{X}_{1}v^{2}_{1}+\mathcal{X}_{2}v_{2}^{2}+\mathcal{X}_{3}v_{3}^{2}$
$\displaystyle=0,$ (15) $\displaystyle-\mathcal{X}_{3}+\mathcal{X}_{1}+m$
$\displaystyle=0,$ (16) $\displaystyle-\mathcal{X}_{3}+\mathcal{X}_{2}+n$
$\displaystyle=0.$ (17)

To see if there is a consistent charge assignment that allows us to decouple
the axion from the leptons, let us set $\mathcal{X}_{3}\to 0$. In this limit,
Eq. (15) reduces to the previous condition, Eq. (12), while Eqs. (16)–(17)
imply $\mathcal{X}_{1}/\mathcal{X}_{2}=m/n$ which, together with Eq. (14),
yields

$\frac{m_{d}}{m_{u}}=-\frac{m}{n}.$ (18)

Hence, with the first choice of operators in Eq. (13), electrophobia can be
consistently implemented for the following values of the light quark mass
ratio: $m_{d}/m_{u}=2,\,1,\,1/2$. It is a fortunate coincidence that the
actual value $m_{u}/m_{d}=0.48(3)$ is perfectly compatible with the first
possibility. This renders it possible to have electrophobia together with
nucleophobia by means of a suitable assignment of PQ charges, rather than by
tuning of some parameters. By contrast, if the breaking $U(1)^{4}\to
U(1)_{Y}\times U(1)_{PQ}$ is enforced via the second set of operators in Eq.
(13), respectively with $H_{1}$ or $H_{2}$ in the first term, electrophobia
would require $m_{d}/m_{u}=1$ or $\infty$ in the first case, and
$m_{d}/m_{u}=1$ or $0$ in the latter, implying that for both these cases
electrophobia would not be compatible with nucleophobia.

##  IV Astrophobic axions in 3HDM models

In the previous sections we have spelled out which conditions need to be
satisfied to enforce axion-nucleon and axion-electron decoupling. Clearly, in
a realistic scenario we expect that these conditions are realized only at some
level of approximation, so that $C_{p,n,e}$, rather than vanish, will just be
suppressed. We will now study more quantitatively the interrelation between
nucleophobia and electrophobia, and the conditions to realize astrophobia with
different levels of accuracy.

According to the results of the previous section, let us assume that the
scalar potential contains the following terms:

$H_{3}^{\dagger}H_{1}\phi^{2}+H_{3}^{\dagger}H_{2}\phi^{{\dagger}}.$ (19)

This corresponds to take $m=2$ and $n=-1$ in Eq. (13) (first case). For the
quarks we assume a $2+1$ structure with the PQ charges for the first
generation equal to the ones of the second generation, as in model $(i_{1})$
in Di Luzio  _et al._ (2018). It is then sufficient to list the Yukawa
operators involving just the second and third generations:

$\displaystyle\overline{q}_{2}u_{2}H_{1},$
$\displaystyle\overline{q}_{3}u_{3}H_{2},$
$\displaystyle\overline{q}_{2}u_{3}H_{1},$
$\displaystyle\overline{q}_{3}u_{2}H_{2},$ (20)
$\displaystyle\overline{q}_{2}d_{2}\tilde{H}_{2},$
$\displaystyle\overline{q}_{3}d_{3}\tilde{H}_{1},$
$\displaystyle\overline{q}_{2}d_{3}\tilde{H}_{2},$
$\displaystyle\overline{q}_{3}d_{2}\tilde{H}_{1}.$

We now assume that the leptons couple to a third Higgs doublet with the same
charges for all generations:

$\overline{\ell}_{i}e_{j}\tilde{H}_{3}.$ (21)

Eqs. (16)–(17) imply $\mathcal{X}_{1}=\mathcal{X}_{3}-2$ and
$\mathcal{X}_{2}=\mathcal{X}_{3}+1$. Neglecting flavour mixings, the diagonal
axion couplings to quarks and leptons read

$\displaystyle C_{u,c}$ $\displaystyle=\frac{2}{3}-\frac{\mathcal{X}_{3}}{3},$
$\displaystyle\quad C_{t}$
$\displaystyle=-\frac{1}{3}-\frac{\mathcal{X}_{3}}{3},$ (22) $\displaystyle
C_{d,s}$ $\displaystyle=\frac{1}{3}+\frac{\mathcal{X}_{3}}{3},$ $\displaystyle
C_{b}$ $\displaystyle=-\frac{2}{3}+\frac{\mathcal{X}_{3}}{3},$ $\displaystyle
C_{e,\mu,\tau}$ $\displaystyle=\frac{\mathcal{X}_{3}}{3}.$

The nucleophobic condition in Eq. (10) then reads

$\mathcal{X}_{3}=\frac{1}{2}-\frac{3}{2}f_{ud}\approx-0.03,$ (23)

which, as already anticipated, automatically yields a suppressed coupling to
electrons: $C_{e}\approx 0.01$ (for comparison, in DFSZ models with
$v_{2}=v_{1}$ one has $C_{e}=1/6$). The SN bound can be expressed as a
constraint on the quantity $C_{N}=\sqrt{C_{p}^{2}+C^{2}_{n}}$ Giannotti  _et
al._ (2017); Tanabashi  _et al._ (2018) that, according to Eqs. (2)–(3), has a
lowest value $C_{N}\approx 0.019$ which is determined by the correction
$\delta_{s}$ in Eq. (2) (for comparisons in KSVZ axion models $C_{N}\approx
0.48$).444An extra tuning with flavour mixings in Eq. (22) can in principle
compensate for $\delta_{s}$ and further reduce the value of $C_{N}$, see Di
Luzio  _et al._ (2018) for details.  To identify the parameter space regions
corresponding to a sizeable suppression of the couplings, and to show the
parametric correlation between nucleophobia and electrophobia, it is
convenient to parameterize the VEVs as

$\displaystyle\begin{split}v_{1}=vc_{1}c_{2},\quad v_{2}=vs_{1}c_{2},\quad
v_{3}=vs_{2},\end{split}$ (24)

where $s_{i}=\sin\beta_{i}$ and $c_{i}=\cos\beta_{i}$, and plot the value of
$C_{N}$ and $C_{e}$ as a function of the angles $\beta_{1,2}$ rather than in
terms of the VEVs ratios $\tan\beta_{1}=v_{2}/v_{1}$ and
$\tan\beta_{2}=\sqrt{v^{2}_{3}/(v^{2}_{2}+v^{2}_{1})}$. Although the latter
are the relevant physical parameters, this has the virtue of zooming in on the
regions in which the values of the VEVs are not strongly hierarchical
($\tan\beta_{1,2}\sim O(1)$), and highlighting the correlation between
electrophobia and nucleophobia. In the parametrization of Eq. (24), the
orthogonality condition in Eq. (15) reads
$\mathcal{X}_{3}=(3c_{1}^{2}-1)c^{2}_{2}$. The requirement that the Yukawa
couplings in the 3HDM remain perturbative restricts the allowed region in the
$(\beta_{1},\beta_{2})$ plane. A conservative limit is obtained by imposing
the tree-level unitarity bound on the $2\to 2$ fermion scattering,
$|\text{Re}\ a_{J=0}|<1/2$, in the 3HDM theory involving Yukawas at
$\sqrt{s}\gg M_{H_{1,2,3}}$. Ignoring running effects, which would make the
bound somewhat stronger, and taking into account group theory factors (see
e.g. Di Luzio  _et al._ (2017a); Di Luzio and Nardecchia (2017)) we get:
$y^{\mathrm{3HDM}}_{t,b}<\sqrt{16\pi/3}$ (from $Q_{L}\overline{u}_{R}\to
Q_{L}\overline{u}_{R}$, with the initial and final states prepared into an
$SU(3)_{c}$ singlet) and $y^{\mathrm{3HDM}}_{\tau}<\sqrt{4\sqrt{2}\pi}$ (from
$Q_{L}\overline{Q}_{L}\to u_{R}\overline{u}_{R}$, with the initial state
prepared into an $SU(2)_{L}$ singlet). The label 3HDM reminds us that these
are not the Yukawa couplings of the SM. The latter are related to the former
via: $y_{t}=y^{\mathrm{3HDM}}_{t}s_{1}c_{2}$,
$y_{b}=y^{\mathrm{3HDM}}_{b}c_{1}c_{2}$,
$y_{\tau}=y^{\mathrm{3HDM}}_{\tau}s_{2}$. The unitarity bounds on
$y^{\mathrm{3HDM}}_{t,b,\tau}$ can be now translated into a perturbativity
bound in the $(\beta_{1},\beta_{2})$ plane, and this results in the hatched
region in Fig. 1. Contour lines for different values of $C_{N}$ and $C_{e}$
are also plotted in Fig. 1, and show how electrophobia and nucleophobia occur
in overlapping regions, so that a single choice of the values of the relevant
parameters simultaneously realizes both properties. It should be also noted
that while for small values of $\beta_{2}$ the region with suppressed
couplings is rather narrow, and astrophobia requires some tuning of the ratio
$v_{2}/v_{1}$ to values sufficiently close to $\sqrt{2}$ ($\beta_{1}\approx
0.95$), at larger values $\beta_{2}\sim\mathcal{O}(1)$ the region opens up and
less tuning is required to simultaneously decouple the axion from nucleons and
electrons.

Figure 1: Contour lines for $C_{N}=\sqrt{C_{p}^{2}+C^{2}_{n}}$ (black) and
$C_{e}$ (yellow) in the $(\beta_{1},\beta_{2})$ plane for the astrophobic
model. For reference, contour lines corresponding to the values of $C_{N}$ and
$C_{e}$ for the KSVZ and the DFSZ axion models are also plotted: $0.48\to
C^{\mathrm{KSVZ}}_{N}$ (dotted line),
$[0.24,0.66]\to[C^{\mathrm{DFSZ}}_{N,\text{min}},\,C^{\mathrm{DFSZ}}_{N,\text{max}}]$
(grey region) and $1/6\to C^{\mathrm{DFSZ}}_{e,(v_{1}=v_{2})}$ (dotted yellow
line) Tanabashi  _et al._ (2018).

We recall here for completeness that since nucleophobia requires generation-
dependent PQ charges, axion couplings to quarks are in general flavour-
violating. As discussed in Di Luzio  _et al._ (2018), limits on FCNCs such as
$K\to\pi a$ yield stringent constraints on nucleophobic models which can be
complementary to astrophysics bounds.

A final remark about the axion coupling to photons is in order. The coupling
is defined by the interaction term

$\frac{\alpha}{8\pi}\frac{C_{\gamma}}{f_{a}}aF_{\mu\nu}\tilde{F}^{\mu\nu},$
(25)

where $C_{\gamma}=E/N-1.92(4)$, with $E$ denoting the coefficient of the
electromagnetic anomaly. In astrophobic models there is no particular reason
for which $C_{\gamma}$ should be suppressed. In the present model the
contributions of the quarks and leptons are respectively
$E_{Q}/N=8/3-2\mathcal{X}_{3}$ and $E_{L}/N=2\mathcal{X}_{3}$, so that their
sum is $E/N=8/3$, a value which is often encountered also in other axion
models Di Luzio  _et al._ (2017b, c).

##  V Conclusions

Astrophobic axion models, wherein the axion couplings to nucleons and
electrons can be simultaneously suppressed well below the values suggested by
well-known benchmark models, can be elegantly implemented in a variant of the
DFSZ model in which the PQ charges of the quarks are generation-dependent, and
the scalar sector contains three Higgs doublets, one of which couples to the
leptons and the other two to the quarks. While consistent astrophobic axion
models were first constructed in Ref. Di Luzio  _et al._ (2018), in the
original scenario axion decoupling from the electrons was achieved by means of
a tuned cancellation between two different contributions to the axion-electron
coupling, one proportional to the electron PQ charge, and the other generated
by the mixing of the electron with the leptons of the heavier generations. The
virtue of the 3HDM construction presented here is that it avoids the need for
this cancellation, and enforces a strong correlation between the suppressions
of the axion couplings to nucleons and electrons, in such a way that
nucleophobia and electrophobia are simultaneously realized in the same region
of parameter space. This renders less contrived the possibility that axions
might exhibit astrophobic properties.

## Acknowledgments

F.B. and E.N. are supported by the INFN “Iniziativa Specifica” Theoretical
Astroparticle Physics (TAsP-LNF). The work of L.D.L. is supported by the ERC
grant NEO-NAT. F.M. is supported by MINECO grant FPA2016-76005-C2-1-P, by
Maria de Maetzu program grant MDM-2014-0367 of ICCUB and 2017 SGR 929. F.M.
acknowledges the INFN Laboratori Nazionali di Frascati for hospitality and
financial support. L.D.L. and E.N. acknowledge hospitality from Chicago O’Hare
International Airport (ORD) and the Aspen Center for Physics (ACP) where this
work was concluded. ACP is supported by National Science Foundation grant
PHY-1607611. The participation of L.D.L. and E.N. at the ACP was supported by
the Simons Foundation.

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