#  Krasner $(m,n)$-hyperring of fractions

M. Anbarloei  Department of Mathematics, Faculty of Sciences, Imam Khomeini
International University, Qazvin, Iran<EMAIL_ADDRESS>
###### Abstract.

The formation of rings of fractions and the associated process of localization
are the most important technical tools in commutative algebra. Krasner
$(m,n)$-hyperrings are a generalization of $(m,n)$-rings. Let $R$ be a
commutative Krasner $(m,n)$-hyperring. The aim of this research work is to
introduce the concept of hyperring of fractions generated by $R$ and then
investigate the basic properties such hyperrings.

###### Key words and phrases:

$n$-ary prime hyperideal, $n$-ary multiplicative subset, Krasner
$(m,n)$-hyperring.

###### 2010 Mathematics Subject Classification:

20N20, 19Y99, 20N15

##  1\. Introduction

The notion of Krasner hyperrings was introduced by Krasner for the first time
in [17]. Also, we can see some properties on Krasner hyperrings in [22] and
[26]. In [11], Davvaz and Vougiouklis defined the notion of $n$-ary
hypergroups which is a generalization of hypergroups in the sense of Marty.
The concept of $(m,n)$-ary hyperrings was introduced in [23]. Davvaz and et.
al. introduced Krasner $(m,n)$\- hyperrings as a generalization of
$(m,n)$-rings and studied some results in this context in [24]. We can see
some important hyperideals of the Krasner $(m,n)$-hyperrings in [1] and [14].
Also, Ostadhadi and Davvaz studied the isomorphism theorems of ring theory and
Krasner hyperring theory which are derived in the context of Krasner (m,
n)-hyperrings in [27]. Ameri and Norouzi introduced in [1] the notions of
$n$-ary prime and $n$-ary primary hyperideals in a Krasner $(m,n)$-hyperring
and proved some results in this respect. The notion of $n$-ary 2-absorbing
hyperideals in a Krasner $(m,n)$-hyperring as a generalization of the $n$-ary
prime hyperideals was introduced in [3]. To unify the concepts of the prime
and primary hyperideals under one frame, the notion of $\delta$-primary
hyperideals was defined in Krasner $(m,n)$-hyperrings [4]. The formation of
rings of fractions and the associated process of localization are the most
important technical tools in commutative algebra. They correspond in the
algebra-geometric picture to concentratining attention on an open set or near
a point, and the importance of these notions should be self-evident. Procesi
and Rota in [28] have studied ring of fractions in Krasner hyperrings.

In this paper, we aim to define the notion of a hyperring of fractions of
Krasner $(m,n)$-hyperrings and provide several properties of them. The paper
is organized as follows. In section 2, we have given some basic definitions
and results of n-ary hyperstructures which we need to develop our paper. In
section 3, we have constructed the Krasner $(m,n)$-hyperring of fractions. In
section 4, we have studied the hyperideals of Krasner $(m,n)$-hyperring of
fractions. In section 5, we have investigated construction of qutient Krasner
$(m,n)$-hyperring of fractions.

##  2\. Preliminaries

In this section we recall some definitions and results about $n$-ary
hyperstructures which we need to develop our paper.
A mapping $f:H^{n}\longrightarrow P^{*}(H)$ is called an $n$-ary
hyperoperation, where $P^{*}(H)$ is the set of all the non-empty subsets of
$H$. An algebraic system $(H,f)$, where $f$ is an $n$-ary hyperoperation
defined on $H$, is called an $n$-ary hypergroupoid.
We shall use the following abbreviated notation:
The sequence $x_{i},x_{i+1},...,x_{j}$ will be denoted by $x^{j}_{i}$. For
$j<i$, $x^{j}_{i}$ is the empty symbol. In this convention

$f(x_{1},...,x_{i},y_{i+1},...,y_{j},z_{j+1},...,z_{n})$
will be written as $f(x^{i}_{1},y^{j}_{i+1},z^{n}_{j+1})$. In the case when
$y_{i+1}=...=y_{j}=y$ the last expression will be written in the form
$f(x_{1}^{i},y^{(j-i)},z^{n}_{j+1})$. For non-empty subsets $A_{1},...,A_{n}$
of $H$ we define

$f(A^{n}_{1})=f(A_{1},...,A_{n})=\bigcup\\{f(x^{n}_{1})\ |\ x_{i}\in
A_{i},i=1,...,n\\}$.
An $n$-ary hyperoperation $f$ is called associative if

$f(x^{i-1}_{1},f(x_{i}^{n+i-1}),x^{2n-1}_{n+i})=f(x^{j-1}_{1},f(x_{j}^{n+j-1}),x_{n+j}^{2n-1}),$
hold for every $1\leq i<j\leq n$ and all $x_{1},x_{2},...,x_{2n-1}\in H$. An
$n$-ary hypergroupoid with the associative $n$-ary hyperoperation is called an
$n$-ary semihypergroup.

An $n$-ary hypergroupoid $(H,f)$ in which the equation $b\in
f(a_{1}^{i-1},x_{i},a_{i+1}^{n})$ has a solution $x_{i}\in H$ for every
$a_{1}^{i-1},a_{i+1}^{n},b\in H$ and $1\leq i\leq n$, is called an $n$-ary
quasihypergroup, when $(H,f)$ is an $n$-ary semihypergroup, $(H,f)$ is called
an $n$-ary hypergroup.

An $n$-ary hypergroupoid $(H,f)$ is commutative if for all
$\sigma\in\mathbb{S}_{n}$, the group of all permutations of
$\\{1,2,3,...,n\\}$, and for every $a_{1}^{n}\in H$ we have
$f(a_{1},...,a_{n})=f(a_{\sigma(1)},...,a_{\sigma(n)})$. If an $a_{1}^{n}\in
H$ we denote $a_{\sigma(1)}^{\sigma(n)}$ as the
$(a_{\sigma(1)},...,a_{\sigma(n)})$. We assume throughout this paper that all
Krasner $(m,n)$-hyperrings are commutative.

If $f$ is an $n$-ary hyperoperation and $t=l(n-1)+1$, then $t$-ary
hyperoperation $f_{(l)}$ is given by

$f_{(l)}(x_{1}^{l(n-1)+1})=f(f(...,f(f(x^{n}_{1}),x_{n+1}^{2n-1}),...),x_{(l-1)(n-1)+1}^{l(n-1)+1})$.

######  Definition 2.1.

[24] Let $(H,f)$ be an $n$-ary hypergroup and $B$ be a non-empty subset of
$H$. $B$ is called an $n$-ary subhypergroup of $(H,f)$, if
$f(x^{n}_{1})\subseteq B$ for $x^{n}_{1}\in B$, and the equation $b\in
f(b^{i-1}_{1},x_{i},b^{n}_{i+1})$ has a solution $x_{i}\in B$ for every
$b^{i-1}_{1},b^{n}_{i+1},b\in B$ and $1\leq i\leq n$. An element $e\in H$ is
called a scalar neutral element if $x=f(e^{(i-1)},x,e^{(n-i)})$, for every
$1\leq i\leq n$ and for every $x\in H$.

An element $0$ of an $n$-ary semihypergroup $(H,g)$ is called a zero element
if for every $x^{n}_{2}\in H$ we have
$g(0,x^{n}_{2})=g(x_{2},0,x^{n}_{3})=...=g(x^{n}_{2},0)=0$. If $0$ and
$0^{\prime}$are two zero elements, then $0=g(0^{\prime},0^{(n-1)})=0^{\prime}$
and so the zero element is unique.

######  Definition 2.2.

[18] Let $(H,f)$ be a $n$-ary hypergroup. $(H,f)$ is called a canonical
$n$-ary hypergroup if
(1) there exists a unique $e\in H$, such that $f(x,e^{(n-1)})=x$ for every
$x\in H$;
(2) for all $x\in H$ there exists a unique $x^{-1}\in H$, such that $e\in
f(x,x^{-1},e^{(n-2)})$;
(3) if $x\in f(x^{n}_{1})$, then $x_{i}\in
f(x,x^{-1},...,x^{-1}_{i-1},x^{-1}_{i+1},...,x^{-1}_{n})$ for all $i$ .

We say that e is the scalar identity of $(H,f)$ and $x^{-1}$ is the inverse of
$x$. Notice that $e^{-1}=e$

######  Definition 2.3.

[24] A Krasner $(m,n)$-hyperring is an algebraic hyperstructure $(R,f,g)$
which satisfies the following axioms:
(1) $(R,f$) is a canonical $m$-ary hypergroup;
(2) $(R,g)$ is an $n$-ary semigroup;
(3) the $n$-ary operation $g$ is distributive with respect to the $m$-ary
hyperoperation $f$ , i.e.,

$g(a^{i-1}_{1},f(x^{m}_{1}),a^{n}_{i+1})=f(g(a^{i-1}_{1},x_{1},a^{n}_{i+1}),...,g(a^{i-1}_{1},x_{m},a^{n}_{i+1}))$

for every $a^{i-1}_{1},a^{n}_{i+1},x^{m}_{1}\in R$ and $1\leq i\leq n$;
(4) $0$ is a zero element (absorbing element) of the $n$-ary operation $g$,
i.e.,

$g(0,x^{n}_{2})=g(x_{2},0,x^{n}_{3})=...=g(x^{n}_{2},0)=0$

for every $x^{n}_{2}\in R$.
We denote the Krasner (m,n)-hyperring $(R,f,g)$ simply by $R$. We say that $R$
is with scalar identity if there exists an element $1$ such that
$x=g(x,1^{(n-1)})$ for all $x\in R$. In this paper, we assume that $R$ is with
scalar identity.

A non-empty subset $S$ of $R$ is said to be a subhyperring of $R$ if $(S,f,g)$
is a Krasner $(m,n)$-hyperring. Let $I$ be a non-empty subset of $R$, we say
that $I$ is a hyperideal of $R$ if $(I,f)$ is an $m$-ary subhypergroup of
$(R,f)$ and $g(x^{i-1}_{1},I,x_{i+1}^{n})\subseteq I$, for every $x^{n}_{1}\in
R$ and $1\leq i\leq n$.

######  Definition 2.4.

[1] A proper hyperideal $I$ of a Krasner $(m,n)$-hyperring $R$ is said to be
an $n$-ary prime hyperideal if for hyperideals $I_{1},...,I_{n}$ of $R$,
$g(I_{1}^{n})\subseteq I$ implies that $I_{1}\subseteq I$ or $I_{2}\subseteq
I$ or …or $I_{n}\subseteq I$.

######  Lemma 2.5.

A proper hyperideal $I$ of a Krasner $(m,n)$-hyperring $R$ is an $n$-ary prime
hyperideal if for all $x^{n}_{1}\in R$, $g(x^{n}_{1})\in I$ implies that
$x_{1}\in I\ or\ ...\ or\ x_{n}\in I$. (Lemma 4.5 in [1])

######  Definition 2.6.

[1] Let $R$ be a Krasner $(m,n)$-hyperring. A non-empty subset $S$ of $R$ is
called $n$-ary multiplicative, if $g(s_{1}^{n})\in S$ for $s_{1},...,s_{n}\in
S$.

In this paper, we assume that $1\in S$.

######  Definition 2.7.

[1] A Krasner $(m,n)$-hyperring $R$ is said to be an $n$-ary hyperintegral
domain, if $R$ is a commutative Krasner $(m,n)$-hyperring and $g(x_{1}^{n})=0$
implies that $x_{1}=0$ or $x_{2}=0$ or … or $x_{n}=0$ for all $x_{1}^{n}$.

######  Definition 2.8.

[1] Let $R$ be a Krasner $(m,n)$-hyperring. An element $x\in R$ is said to be
invertible if there exists $y\in R$ with $1=g(x,y,1^{(n-2)})$.

######  Definition 2.9.

[24] Let $(R_{1},f_{1},g_{1})$ and $(R_{2},f_{2},g_{2})$ be two Krasner
$(m,n)$-hyperrings. A mapping $\phi:R_{1}\longrightarrow R_{2}$ is called a
homomorphism if for all $x^{m}_{1}\in R_{1}$ and $y^{n}_{1}\in R_{1}$ we have

$\phi(f_{1}(x_{1},...,x_{m}))=f_{2}(\phi(x_{1}),...,\phi(x_{m}))$

$\phi(g_{1}(y_{1},...,y_{n}))=g_{2}(\phi(y_{1}),...,\phi(y_{n})).$

##  3\.  Krasner $(m,n)$-hypering of fractions

Let $R$ be any Krasner $(m,n)$-hyperring and let $S$ be an $n$-ary
multiplicative subset of $R$ such that $1\in S$. We shall construct the
Krasner $(m,n)$-hyperring of fractions $S^{-1}R$. We define a relation $\sim$
on $R\times S$ by $(r,s)\sim(r^{\prime},s^{\prime})$ if and only if there
exists some $s\in S$ such that

$0\in
g(s,f(g(r,s^{\prime},1^{(n-2)}),g(r^{\prime},s,1^{(n-2)}),0^{(m-2)}),1^{(n-2)})$.

######  Theorem 3.1.

The relation $\sim$ is an equivalence relation on $R\times S$.

###### Proof.

Clearly, $\sim$ is reflexive and symmetric. Suppose that
$(r_{1},s_{1})\sim(r_{2},s_{2})$ and $(r_{2},s_{2})\sim(r_{3},s_{3})$. Then
there exist $s\in S$ such that

$0\in
g(s,f(g(r_{1},s_{2},1^{(n-2)}),-g(r_{2},s_{1},1^{(n-2)}),0^{(m-2)}),1^{(n-2)})$
and

$0\in
g(s^{\prime},f(g(r_{2},s_{3},1^{(n-2)}),-g(r_{3},s_{2},1^{(n-2)}),0^{(m-2)}),1^{(n-2)})$.
Since

$0\in
g(s,f(g(r_{1},s_{2},1^{(n-2)}),-g(r_{2},s_{1},1^{(n-2)}),0^{(m-2)}),1^{(n-2)})$

$\hskip
8.5359pt=f(g(s,r_{1},s_{2},1^{(n-3)}),-g(s,r_{2},s_{1},1^{(n-2)}),0^{(m-2)})$,
we get $g(s,r_{2},s_{1},1^{(n-2)})\in
f(g(s,r_{1},s_{2},1^{(n-3)}),0^{(m-1)})$.
Thus we have
$0=g(g(s,s_{1},1^{(n-2)}),0^{(n-1)})$

$\in
g(g(s,s_{1},1^{(n-2)}),g(s^{\prime},f(g(r_{2},s_{3},1^{(n-2)}),-g(r_{3},s_{2},1^{(n-2)}),0^{(m-2)}),1^{(n-2)}),$

$\hskip 5.69046pt1^{(n-2)})$

$=g(g(s,s_{1},1^{(n-2)}),f(g(s^{\prime},r_{2},s_{3},1^{(n-3)}),-g(s^{\prime},r_{3},s_{2},1^{(n-3)}),0^{(m-2)}),1^{(n-2)})$

$=f(g(s,s_{1},s^{\prime},r_{2},s_{3},1^{(n-5)}),-g(s,s_{1},s^{\prime},r_{3},s_{2},1^{(n-5)}),0^{(m-2)})$

$=f(g(s^{\prime},g(s,r_{2},s_{1},1^{(n-3)}),s_{3}),-g(s,s_{1},s^{\prime},r_{3},s_{2},1^{(n-3)}),0^{(m-3)})$

$\subseteq
f(g(s^{\prime},f(g(s,r_{1},s_{2},1^{(n-3)}),0^{(m-1)}),s_{3},1^{(n-3)}),-g(s,s_{1},s^{\prime},r_{3},s_{2},1^{(n-3)}),$

$\hskip 5.69046pt0^{(m-3)})$

$=f(g(s,s^{\prime},s_{2},r_{1},s_{3}),-g(s,s^{\prime},s_{2},r_{3},s_{1}),0^{(m-2)})$

$=g(g(s,s^{\prime},s_{2},1^{(n-3)}),f(g(r_{1},s_{3},1^{(n-2)}),-g(r_{3},s_{1},1^{(n-2)}),0^{(m-2)}),1^{(n-2)}).$
Since $g(s,s^{\prime},s_{2},1^{(n-3)})\in S$, then
$(r_{1},s_{1})\sim(r_{3},s_{3})$. Consequently, $\sim$ is transitive. ∎

We denote the equivalence class of $(a,s)$ with $\frac{r}{s}$ and let
$S^{-1}R$ denote the set of all equivalence classes. We endow the set
$S^{-1}R$ with a Krasner $(m,n)$-hyperring structure, by defining the $m$-ary
hyperoperation $F$ and the $n$-ary operation $G$ as follows:
$F(\frac{r_{1}}{s_{1}},...,\frac{r_{m}}{s_{m}})=\frac{f(g(r_{1},s_{2}^{m},1^{(n-m)}),g(s_{1},r_{2},s_{3}^{m},1^{(n-m)}),...,g(s_{1}^{m-1},r_{m},1^{(n-m)})}{g(s_{1}^{m},1^{(n-m)})}$
$=\\{\frac{r}{s}\ |\ r\in
f(g(r_{1},s_{2}^{m},1^{(n-m)}),g(s_{1},r_{2},s_{3}^{m},1^{(n-m)}),...,g(s_{1}^{m-1},r_{m},1^{(n-m)}),$

$s=g(s_{1}^{m})\\}$
$G(\frac{r_{1}}{s_{1}},...,\frac{r_{n}}{s_{n}})=\frac{g(r_{1}^{n})}{g(s_{1}^{n})}.$
We need to show that $F$ and $G$ are well defined. If
$\frac{r_{1}}{s_{1}}=\frac{r_{1}^{\prime}}{s_{1}^{\prime}}$,
$\frac{r_{2}}{s_{2}}=\frac{r_{2}^{\prime}}{s_{2}^{\prime}},...,\frac{r_{m}}{s_{m}}=\frac{r_{m}^{\prime}}{s_{m}^{\prime}}$,
then there exist $t_{1},...,t_{m}\in S$ such that

$\displaystyle 0$ $\displaystyle\in
g(t_{1},f(g(r_{1},{s^{\prime}}_{1},1^{(n-2)}),-g({r^{\prime}}_{1},s_{1},1^{(n-2)}),0^{m-2}),1^{(n-2)})$
$\displaystyle(1)$ $\displaystyle 0$ $\displaystyle\in
g(t_{2},f(g(r_{2},{s^{\prime}}_{2},1^{(n-2)}),-g({r^{\prime}}_{2},s_{2},1^{(n-2)}),0^{m-2}),1^{(n-2)})$
$\displaystyle(2)$ $\displaystyle\vdots$ $\displaystyle 0$ $\displaystyle\in
g(t_{m},f(g(r_{m},{s^{\prime}}_{m},1^{(n-2)}),-g({r^{\prime}}_{m},s_{m},1^{(n-2)}),0^{m-2}),1^{(n-2)}).$
$\displaystyle(m)$

$g$-producting (1) by

$g(g(t_{2}^{m},1^{(n-m+1)}),g(1^{(n-m+1)},s_{2}^{m}),g(1^{(n-m+1)},{s^{\prime}}_{2}^{m}),1^{(n-3)})$,
(2) by

$g(g(t_{1},1^{(n-m+1)},t_{3}^{m}),(s_{1},1^{(n-m+1)},s_{3}^{m}),g({s^{\prime}}_{1},1^{(n-m+1)},{s^{\prime}}_{3}^{m}),1^{(n-3)})$

$\vdots$
(m) by

$g(g(t_{1}^{m-1},1^{(n-m+1)}),g(s_{1}^{m-1},1^{(n-m+1)}),g({s^{\prime}}_{1}^{m-1},1^{(n-m+1)}),1^{(n-3)})$.
Thus we get

$0\in
g(g(t_{1}^{m},1^{(n-m)}),f(g(g({s^{\prime}}_{1}^{m},1^{(n-m)}),g(r_{1},s_{2}^{m},1^{(n-m)}),1^{(n-2)}),$

$\hskip
14.22636pt-g((g(s_{1}^{m},1^{(n-m)}),g(r_{1}^{\prime},{s^{\prime}}_{2}^{m},1^{(n-m)}),1^{(n-2)}),0^{(m-2)}),1^{(n-2)})$

$0\in
g(g(t_{1}^{m},1^{(n-m)}),f(g(g({s^{\prime}}_{1}^{m},1^{(n-m)}),g(r_{1},s_{1},s_{3}^{m},1^{(n-m)}),1^{(n-2)}),$

$\hskip
14.22636pt-g(g(s_{1}^{m},1^{(n-m)}),g(r_{2}^{\prime}.{s^{\prime}}_{1},{s^{\prime}}_{3}^{m},1^{(n-m)}),1^{(n-2)}),0^{(m-2)}),1^{(n-2)})$

$\vdots$

$0\in
g(g(t_{1}^{m},1^{(n-m)}),f(g({s^{\prime}}_{1}^{m},1^{(n-m)}),g(r_{m},s_{1}^{m-1},1^{(n-m)}),1^{(n-2)}),$

$\hskip
14.22636pt-g(g(s_{1}^{m},1^{(n-m)}),g(r_{m}^{\prime},{s^{\prime}}_{1}^{m-1},1^{(n-m)}),1^{(n-2)}),0^{(m-2)}),1^{(n-2)})$.
Now, we have

$0\in
f(f(g(g(t_{1}^{m},1^{(n-m)}),g({s^{\prime}}_{1}^{m},1^{(n-m)}),g(r_{1},s_{2}^{m},1^{(n-m)}),1^{(n-3)}),$

$\hskip
8.5359ptg(g(t_{1}^{m},1^{(n-m)}),g({s^{\prime}}_{1}^{m},1^{(n-m)}),g(r_{1},s_{1},s_{3}^{m},1^{(n-m)}),1^{(n-3)}),$

$\hskip 8.5359pt\cdots,$

$\hskip
8.5359ptg(g(t_{1}^{m},1^{(n-m)}),g({s^{\prime}}_{1}^{m},1^{(n-m)}),g(r_{m},s_{1}^{m-1},1^{(n-m)}),1^{(n-3)})),$

$-f(g(g(t_{1}^{m},1^{(n-m)}),g(s_{1}^{m},1^{(n-m)}),g(r_{1}^{\prime},{s^{\prime}}_{2}^{m},1^{(n-m+1)}),1^{(n-3)}),$

$\hskip
8.5359ptg(g(t_{1}^{m},1^{(n-m)}),g(s_{1}^{m},1^{(n-m)}),g(r_{2}^{\prime}.{s^{\prime}}_{1},{s^{\prime}}_{3}^{m},1^{(n-m)}),1^{(n-3)}),$

$\hskip 8.5359pt\cdots,$

$\hskip
8.5359ptg(g(t_{1}^{m},1^{(n-m)}),g(s_{1}^{m},1^{(n-m)}),g(r_{m}^{\prime},{s^{\prime}}_{1}^{m-1},1^{(n-m)}),1^{(n-3)}),0^{(m-2)})$.

We put $t=g(t_{1}^{m},1^{(n-m)}),s=g(s_{1}^{m},1^{(n-m)}))$ and
$s^{\prime}=g({s^{\prime}}_{1}^{m},1^{(n-m)})$.
Therefore we have

$0\in
f(g(t,g(s^{\prime},f(g(r_{1},s_{2}^{m},1^{(n-m)}),g(r_{2}^{\prime}.{s^{\prime}}_{1},{s^{\prime}}_{3}^{m},1^{(n-m)}),\cdots,g(r_{m},s_{1}^{m-1},1^{(n-m)}),$

$\hskip
14.22636pt1^{(n-2)}),-g(t,g(s,f(g(r_{1}^{\prime},{s^{\prime}}_{2}^{m},1^{(n-m)}),g(r_{2}^{\prime}.{s^{\prime}}_{1},{s^{\prime}}_{3}^{m},1^{(n-m)}),\cdots,g(r_{m}^{\prime},{s^{\prime}}_{1}^{m-1},$

$\hskip 14.22636pt1^{(n-m)}),1^{(n-2)},0^{(m-2)})$.
Thus
$F(\frac{r_{1}}{s_{1}},...,\frac{r_{m}}{s_{m}})=F(\frac{r^{\prime}_{1}}{s^{\prime}_{1}},...,\frac{r^{\prime}_{m}}{s^{\prime}_{m}})$,
i. e., $F$ is well defined.

Now, suppose that $\frac{r_{1}}{s_{1}}=\frac{r_{1}^{\prime}}{s_{1}^{\prime}}$,
$\frac{r_{2}}{s_{2}}=\frac{r_{2}^{\prime}}{s_{2}^{\prime}},...,\frac{r_{n}}{s_{n}}=\frac{r_{n}^{\prime}}{s_{n}^{\prime}}$,
then there exist $t_{1},...,t_{n}\in S$ such that

$0\in
g(t_{1},f(g(r_{1},s_{1}^{\prime},1^{(n-2)}),-g(r_{1}^{\prime},s_{1},1^{(n-2)}),0^{(m-2)}),1^{(n-2)})$

$0\in
g(t_{2},f(g(r_{2},s_{2}^{\prime},1^{(n-2)}),-g(r_{2}^{\prime},s_{2},1^{(n-2)}),0^{(m-2)}),1^{(n-2)})$

$\vdots$

$0\in
g(t_{n},f(g(r_{n},s_{n}^{\prime},1^{(n-2)}),-g(r_{n}^{\prime},s_{n}^{\prime},1^{(n-2)}),0^{(m-2)}),1^{(n-2)})$.
Then we conclude that

$0\in
f(g(g(t_{1},r_{1},s_{1}^{\prime},1^{(n-3)}),g(t_{2},r_{2},s_{2}^{\prime},1^{(n-3)}),...,g(t_{n},r_{n},s_{n}^{\prime},1^{(n-3)}),1^{(n-m)}),$

$\hskip
8.5359pt-g(g(t_{1},r_{1}^{\prime},s_{1},1^{(n-3)}),g(t_{2},r_{2}^{\prime},s_{2},1^{(n-3)}),...,g(t_{n},r_{n}^{\prime},s_{n}^{\prime},1^{(n-3)}),1^{(n-m)}),0^{(m-2)}).$
It means

$0\in
f(g(g(t_{1}^{n}),g(r_{1}^{n}),g({s^{\prime}}_{1}^{n}),1^{(n-3)}),-g(g(t_{1}^{n}),g({r^{\prime}}_{1}^{n}),g(s_{1}^{n}),1^{(n-3)}),0^{(m-2)}).$
Put $t=g(t_{1}^{n})$. We have

$0\in
f(g(t,g(r_{1}^{n}),g({s^{\prime}}_{1}^{n}),1^{(n-3)}),-g(t,g({r^{\prime}}_{1}^{n}),g(s_{1}^{n}),1^{(n-3)}),0^{(m-2)})$
and so

$0\in
g(t,f(g(g(r_{1}^{n}),g({s^{\prime}}_{1}^{n}),1^{(n-2)}),-g(g(t,g({r^{\prime}}_{1}^{n}),g(s_{1}^{n}),1^{(n-2)}),0^{(m-2)}),1^{(n-2)})$.
It implies that
$\frac{g(r_{1}^{n})}{g(s_{1}^{n})}=\frac{g({r^{\prime}}_{1}^{n})}{g({s^{\prime}}_{1}^{n})}$
and so
$G(\frac{r_{1}}{s_{1}},...,\frac{r_{n}}{s_{n}})=G(\frac{r^{\prime}_{1}}{s^{\prime}_{1}},...,\frac{r^{\prime}_{n}}{s^{\prime}_{n}})$,
i. e., $G$ is well defined.

######  Lemma 3.2.

Let $R$ be a Krasner $(m,n)$-hyperring and $S$ be an $n$-ary multiplicative
subset of $R$ with $1\in S$. Then:

$1)$ For all $s\in S$, $\frac{0}{1}=\frac{0}{s}=0_{S^{-1}R}$.

$2)$ $\frac{r}{s}=0_{S^{-1}R}$, for $r\in R,s\in S$ if and only if there
exists $t\in S$ such that

$\hskip 14.22636ptg(t,r,1^{n-2})=0$.

$3)$ For all $s\in S$, $\frac{s}{s}=\frac{1}{1}=1_{S^{-1}R}$.

$4)$
$\frac{g(r,s^{(m-1)},1^{(n-m)})}{g(s^{\prime},s^{(m-1)},1^{(n-m)})}=\frac{g(r,1^{(n-1)})}{g(s^{\prime},1^{(n-1)})}$,
for $r\in R$ and $s,s^{\prime}\in S$.

###### Proof.

$(1)$ Let $t\in S$. Then for all $s\in S$ we have

$0=g(t,s,0,1^{(n-3)})$

$\hskip 8.5359pt=g(t,g(0,s,1^{(n-2)}),1^{(n-2)})$

$\hskip 8.5359pt=g(t,f(g(0,s,1^{(n-2)}),0^{(m-1)}),1^{(n-2)})$

$\hskip
8.5359pt=g(t,f(g(0,s,1^{(n-2)}),-g(1,0,1^{(n-2)}),0^{(m-2)}),1^{(n-2)}).$
Then we conclude that $\frac{0}{1}=\frac{0}{s}=0_{S^{-1}R}$. Now, we show that
$\frac{0}{1}=0_{S^{-1}R}$. Let $r\in R$ and $s\in S$. Then

$F(\frac{r}{s},\frac{0}{1}^{(m-1)})=\\{\frac{u}{v}\ |\ u\in
f(g(r,1^{(n-1)}),g(0,s,1^{(n-2)})^{(m-1)}),s=g(s,1^{n-1})\\}$

$\hskip 59.75095pt=\\{\frac{u}{v}\ |\ u\in
f(g(r,1^{(n-1)}),0^{(m-1)}),v=g(s,1^{n-1})\\}$

$\hskip 59.75095pt=\\{\frac{u}{v}\ |\ u\in f(r,0^{(m-1)}),s=g(s,1^{n-1})\\}$

$\hskip 59.75095pt=\\{\frac{u}{v}\ |\ u=r,s=g(s,1^{n-1})\\}.$
Thus $F(\frac{r}{s},\frac{0}{1}^{(m-1)})=\frac{r}{s}$. Consequently
$\frac{0}{1}=0_{S^{-1}R}$.

$(2)$ $(\Longrightarrow):$ Let $\frac{r}{s}=0_{S^{-1}R}$ for $r\in R,s\in S$.
By $(1)$, we have $\frac{r}{s}=\frac{0}{1}$. Hence there exists $t\in S$ such
that

$\hskip 17.07182pt0\in
g(t,f(g(r,1^{n-1)}),-g(0,s,1^{(n-2)}),0^{(m-2)}),1^{(n-2)})$.
Therefore $0\in g(t,f(r,0^{(m-1)}),1^{(n-2)})$. It means $g(t,r,1^{(n-2)})=0$.
$(\Longleftarrow):$ Let $g(t,r,1^{(n-2)})=0$ for some $t\in S$. Then
$0=g(t,f(r,0^{(m-1)}),1^{(n-2)})$. Since $r=g(r,1^{n-1)})$ and
$0=g(0,s,1^{(n-2)})$, we get

$0=g(t,f(g(r,1^{n-1)}),g(0,s,1^{(n-2)}),0^{(m-2)}),1^{(n-2)})$.
Then $\frac{r}{s}=\frac{0}{1}$ and so $\frac{r}{s}=0_{S^{-1}R}$, by $(1)$.

$(3)$ Let $s\in S$. It is clear that $0=g(0,1^{(n-1)})$. Then we get

$\hskip
14.22636pt0=g(1,f(g(s,1,0^{(n-2)}),-g(1,s,1^{(n-2)}),0^{(m-2)}),1^{(n-2)}).$
It means $\frac{s}{s}=\frac{1}{1}$. Now, we show that
$\frac{1}{1}=1_{S^{-1}R}$. Let $r\in R$ and $s\in S$. Then we have

$\hskip
17.07182ptG(\frac{r}{s},\frac{1}{1}^{(n-1)})=\frac{g(r,1^{(n-1)}}{g(s,1^{(n-1)})}=\frac{r}{s}.$
This implies that $\frac{1}{1}=1_{S^{-1}R}$.

$(4)$ Let $r\in R$ and $s,s^{\prime}\in S$. Clearly,

$F(\frac{r}{s^{\prime}},{\frac{0}{s}}^{(m-1)})=\frac{f(g(r,s^{(m-1)},1^{(n-m)}),g(s^{\prime},0,s^{(m-2)},1^{(n-m)})^{(m-1)})}{g(s^{\prime},s,1^{(n-2)})}$

$\hskip
62.59596pt=\frac{f(g(r,s^{(m-1)},1^{(n-m)}),0^{(m-1)})}{g(s^{\prime},s^{(m-1)},1^{(n-m)})}$

$\hskip
62.59596pt=\frac{g(r,s^{(m-1)},1^{(n-m)})}{g(s^{\prime},s^{(m-1)},1^{(n-m)})}$.
On the other hand,

$F(\frac{r}{s^{\prime}},{\frac{0}{1}}^{(m-1)})=\frac{f(g(r,1^{(n-1)}),g(s^{\prime},0,1^{(n-2)})^{(m-1)})}{g(s^{\prime},1^{(n-1)})}$

$\hskip
62.59596pt=\frac{f(g(r,1^{(n-1)}),0^{(m-1)})}{g(s^{\prime},1^{(n-1)})}$

$\hskip 62.59596pt=\frac{g(r,1^{(n-1)})}{g(s^{\prime},1^{(n-1)})}.$ ∎

######  Definition 3.3.

Let $R$ be a Krasner $(m,n)$-hyperring and $S$ be an $n$-ary multiplicative
subset of $R$ with $1\in S$. The mapping $\phi:R\longrightarrow S^{-1}R$,
defined by $r\longrightarrow\frac{r}{1}$, is called natural map.

######  Theorem 3.4.

The natural map $\phi$ is a homomorphism of Krasner $(m,n)$-hyperring.

###### Proof.

Let $R$ be a Krasner $(m,n)$-hyperring and $S$ be an $n$-ary multiplicative
subset of $R$ with $1\in S$. For all $r_{1}^{m}\in R$, we get

$\hskip 28.45274pt\phi(f(r_{1}^{m}))=\frac{f(r_{1}^{m})}{1}$

$\hskip
71.13188pt=\frac{f(g(r_{1},1^{(n-1)}),g(r_{2},1^{(n-1)}),...,g(r_{m},1^{(n-1)}))}{g(1^{(m)},1^{(n-m)})}$

$\hskip 71.13188pt=\\{\frac{r}{1}\ |\ r\in
f(g(r_{1},1^{(n-1)}),g(r_{2},1^{(n-1)}),...,g(r_{m},1^{(n-1)})\\}$

$\hskip 71.13188pt=F(\frac{r_{1}}{1},...,\frac{r_{m}}{1})$

$\hskip 71.13188pt=F(\phi(r_{1}),...,\phi(r_{m}))$.
Also, for all $r_{1}^{n}\in R$, we have

$\hskip 28.45274pt\phi(g(r_{1}^{n}))=\frac{g(r_{1}^{n})}{1}$

$\hskip 71.13188pt=\frac{g(r_{1}^{n})}{g(1^{(n)})}$

$\hskip 71.13188pt=G(\frac{r_{1}}{1},...,\frac{r_{1}}{1})$

$\hskip 71.13188pt=G(\phi(r_{1}),...,\phi(r_{n}))$. ∎

######  Theorem 3.5.

Let $\frac{r}{s}$ be an nonzero element of $S^{-1}R$. Then

$1)$ For all $s\in S$, $\phi(s)$ is an invertible element of $S^{-1}R$.

$2)$ If $\phi(r)=0$, then there exists $t\in S$ such that
$g(t,r,1^{(n-2)})=0$.

$3)$ $\frac{r}{s}=G(\phi(r),\phi(s)^{-1},{\frac{1}{1}}^{(n-2)})$, for all
$\frac{r}{s}\in S^{-1}R$.

###### Proof.

$(1)$ Let $s\in S$. Then we have

$\hskip
28.45274ptG(\frac{s}{1},\frac{1}{s},{\frac{1}{1}}^{(n-2)})=\frac{g(s,1^{(n-1)})}{g(1,s,1^{(n-2)})}$

$\hskip 96.73918pt=\frac{g(s,1^{(n-1)})}{g(s,1^{(n-1)})}$

$\hskip 96.73918pt=\frac{1}{1}\hskip 85.35826pt\text{by Lemma \ref{11} (3) }$

$\hskip 96.73918pt=1_{S^{-1}R}.$

$(2)$ It is clear by 3.2 (2).

$(3)$ Let $\frac{r}{s}\in S^{-1}R$. Then

$\hskip 28.45274pt\frac{r}{s}=\frac{g(r,1^{(n-1)})}{g(s,1^{(n-1)})}$

$\hskip 36.98866pt=G(\frac{r}{1},\frac{1}{s},{\frac{1}{1}}^{(n-2)})$

$\hskip 36.98866pt=G(\phi(r),\phi(s)^{-1},{\frac{1}{1}}^{(n-2)})$. ∎

######  Theorem 3.6.

Let $(R_{1},f_{1},h_{1})$ and $(R_{2},f_{2},g_{2})$ be two Krasner
$(m,n)$-hyperrings and $S$ be an $n$-ary multiplicative subset of $R_{1}$ with
$1\in S$. Let $k:R_{1}\longrightarrow R_{2}$ be a homomorphism such that for
each $s\in S$, $k(s)$ is an invertible element of $R_{2}$. Then there exists
an unique homomorphism $h:S^{-1}R_{1}\longrightarrow R_{2}$ such that
$ho\phi=k$.

###### Proof.

Let $(R_{1},f_{1},h_{1})$, $(R_{2},f_{2},g_{2})$ and $(S^{-1}R_{1},G,F)$ be
Krasner $(m,n)$-hyperrings such that $S$ is an $n$-ary multiplicative subset
of $R_{1}$ and $1\in S$. Define mapping $h$ from $S^{-1}R_{1}$ to $R_{2}$ as
follows:

$h(\frac{r}{s})=g_{2}(k(a),k(s)^{-1},1^{(n-2)})$.
We need to show that $h$ is well defined. Let
$\frac{r_{1}}{s_{1}}=\frac{r^{\prime}}{s^{\prime}}$. Then there exists $t\in
S$ such that

$0\in
g_{1}(t,f_{1}(g_{1}(r,s^{\prime},1^{(n-2)}),-g_{1}(r^{\prime},s,1^{(n-2)}),0^{(m-2)}),1^{(n-2)})$.

$\hskip
8.5359pt=f_{1}(g_{1}(t,r,s^{\prime},1^{(n-2)}),-g_{1}(t,r^{\prime},s,1^{(n-2)}),0^{(m-2)})$.
Hence

$0\in
k(f_{1}(g_{1}(t,r,s^{\prime},1^{(n-2)}),-g_{1}(t,r^{\prime},s,1^{(n-2)}),0^{(m-2)})$

$\hskip
8.5359pt=f_{2}(k(g_{1}(t,r,s^{\prime},1^{(n-2)}),k(-g_{1}(t,r^{\prime},s,1^{(n-2)}),k(0)^{(m-2)})$

$\hskip
8.5359pt=f_{2}(k(g_{1}(g_{1}(t,1^{(n-1)}),g_{1}(r,1^{(n-1)}),g_{1}(s^{\prime},1^{(n-1)}),1^{(n-3)})),$

$\hskip
19.91684ptk(-g_{1}(g_{1}(t,1^{(n-1)}),g_{1}(r^{\prime},1^{(n-1)}),g_{1}(s,1^{(n-1)}),1^{(n-3)})),k(0)^{(m-2)})$

$\hskip
8.5359pt=f_{2}(g_{2}(k(g_{1}(t,1^{(n-1)})),k(g_{1}(r,1^{(n-1)})),k(g_{1}(s^{\prime},1^{(n-1)})),1^{(n-3)}),$

$\hskip 19.91684pt-
g_{2}(k(g_{1}(t,1^{(n-1)})),k(g_{1}(r^{\prime},1^{(n-1)})),k(g_{1}(s,1^{(n-1)})),1^{(n-3)})),k(0)^{(m-2)})$

$\hskip 8.5359pt=f_{2}(g_{2}(k(t),k(r)),k(s^{\prime}),1^{(n-3)}),$

$\hskip 19.91684pt-g_{2}(k(t),k(r^{\prime}),k(s),1^{(n-3)})),0^{(m-2)})$

$\hskip
8.5359pt=f_{2}(g_{2}(k(t),g_{2}(k(r)),k(s^{\prime}),1^{(n-2)}),1^{(n-2)}),$

$\hskip 19.91684pt-
g_{2}(k(t),g_{2}(k(r^{\prime}),k(s),1^{(n-2)}),1^{(n-2)}),0^{(m-2)})$

$\hskip 8.5359pt=g_{2}(k(t),f_{2}(g_{2}(k(r)),k(s^{\prime}),1^{(n-2)}),$

$\hskip 19.91684pt-
g_{2}(k(r^{\prime}),k(s),1^{(n-2)})),0^{(m-2)}),1^{(n-2)})$.
Since $k(t)$, $k(s)$ and $k(s^{\prime})$ are invertible elements in $R_{2}$,
we get

$0=g_{2}(k(t)^{-1},k(s)^{-1},k(s^{\prime})^{-1},1^{(n-4)},0)$

$\hskip 8.5359pt\in
g_{2}(g_{2}(k(t)^{-1},k(s)^{-1},k(s^{\prime})^{-1},1^{(n-3)}),k(t),f_{2}(g_{2}(k(r),k(s^{\prime}),1^{(n-2)}),$

$\hskip 19.91684pt-g_{2}(k(r^{\prime}),k(s),1^{(n-2)})),0^{(m-2)}),1^{(n-3)})$

$\hskip
8.5359pt=g_{2}(g_{2}(k(t)^{-1},k(t),1^{(n-2)}),g_{2}(k(s)^{-1},k(s^{\prime})^{-1},1^{(n-2)}),f_{2}(g_{2}(k(r),k(s^{\prime}),1^{(n-2)}),$

$\hskip 19.91684pt-g_{2}(k(r^{\prime}),k(s),1^{(n-2)})),0^{(m-2)}),1^{(n-3)})$

$\hskip
8.5359pt=g_{2}(1,g_{2}(k(s)^{-1},k(s^{\prime})^{-1},1^{(n-2)}),f_{2}(g_{2}(k(r),k(s^{\prime}),1^{(n-2)}),$

$\hskip 19.91684pt-g_{2}(k(r^{\prime}),k(s),1^{(n-2)})),0^{(m-2)}),1^{(n-3)})$

$\hskip
8.5359pt=f_{2}(g_{2}(g_{2}(k(s)^{-1},k(s^{\prime})^{-1},1^{(n-2)}),g_{2}(k(r),k(s^{\prime}),1^{(n-2)}),1^{(n-2)})$

$\hskip 19.91684pt-
g_{2}(g_{2}(k(s)^{-1},k(s^{\prime})^{-1},1^{(n-2)}),g_{2}(k(r^{\prime}),k(s),1^{(n-2)}),1^{(n-2)}),0^{(m-2)})$

$\hskip
8.5359pt=f_{2}(g_{2}(g_{2}(k(s^{\prime})^{-1},k(s^{\prime}),1^{(n-2)}),g_{2}(k(r),k(s)^{-1},1^{(n-2)}),1^{(n-2)})$

$\hskip 19.91684pt-
g_{2}(g_{2}(k(s),k(s)^{-1},1^{(n-2)}),g_{2}(k(r^{\prime}),k(s^{\prime})^{-1},1^{(n-2)}),1^{(n-2)}),0^{(m-2)})$

$\hskip
8.5359pt=f_{2}(g_{2}(k(r),k(s)^{-1},1^{(n-2)})-g_{2}(k(r^{\prime}),k(s^{\prime})^{-1},1^{(n-2)}),0^{(m-2)})$

$\hskip
8.5359pt=f_{2}(h(\frac{r}{s}),h(\frac{r^{\prime}}{s^{\prime}}),0^{(m-2)}).$
Then we coclude that $h(\frac{r}{s})=h(\frac{r^{\prime}}{s^{\prime}})$.
We must show that the mapping $h$ is an homomorphism. Let $r_{1}^{m}\in R_{1}$
and $s_{1}^{m}\in S$. Then we get

$h(F(\frac{r_{1}}{s_{1}},...,\frac{r_{m}}{s_{m}})$

$=h(\frac{f_{1}(g_{1}(r_{1},s_{2}^{m},1^{(n-m)}),g_{1}(s_{1},r_{2},s_{3}^{m},1^{(n-m)}),...,g_{1}(s_{1}^{m-1},r_{m},1^{(n-m)})}{g_{1}(s_{1}^{m},1^{(n-m)})})$

$=g_{2}(k(f_{1}(g_{1}(r_{1},s_{2}^{m},1^{(n-m)}),...,g_{1}(s_{1}^{m-1},r_{m},1^{(n-m)})),k(g_{1}(s_{1}^{m},1^{(n-m)}))^{-1},1^{(n-2)})$

$=g_{2}(f_{2}(k(g_{1}(r_{1},s_{2}^{m},1^{(n-m)}),...,k(g_{1}(s_{1}^{m-1},r_{m},1^{(n-m)})),k(g_{1}(s_{1}^{m},1^{(n-m)}))^{-1},1^{(n-2)})$

$=g_{2}(f_{2}(g_{2}((k(r_{1}),k(s_{2}),...,k(s_{m}),k(1)^{(n-m)}),...,g_{2}(k(s_{1}),...,k(s_{m-1}),k(r_{m}),k(1)^{(n-m)})),$

$\hskip
11.38092ptg_{2}(k(s_{1})^{-1},...,k(s_{m})^{-1},k(1)^{(n-m)})),1^{(n-2)})$

$=f_{2}(g_{2}(g_{2}(k(s_{1})^{-1},...,k(s_{m})^{-1},k(1)^{(n-m)}),g_{2}(k(r_{1}),k(s_{2}),...,k(s_{m}),k(1)^{(n-m}),1^{(n-2)})),$

$\hskip 11.38092pt...,$

$\hskip
11.38092ptg_{2}(g_{2}(k(s_{1})^{-1},...,k(s_{m})^{-1},k(1)^{(n-m)}),g_{2}(k(s_{1}),...,k(s_{m-1}),k(r_{m}),k(1)^{(n-m)})),1^{(n-2)}))$

$=f_{2}(g_{2}(k(r_{1}),k(s_{1})^{-1},1^{(n-2)}),...,g_{2}(k(r_{m}),k(s_{m})^{-1},1^{(n-2)}))$

$=f_{2}(h(\frac{r_{1}}{s_{1}}),...,h(\frac{r_{m}}{s_{m}}))$.
Also, we have

$h(G(\frac{r_{1}}{s_{1}},...,\frac{r_{n}}{s_{n}})=h(\frac{g_{1}(r_{1}^{n})}{g_{1}(s_{1}^{n})})$

$\hskip
65.44142pt=g_{2}(k(g_{1}(r_{1}^{n})),k(g_{1}(s_{1}^{n}))^{-1},1^{(n-2)})$

$\hskip
65.44142pt=g_{2}(g_{2}(k(r_{1}),...,k(r_{n})),g_{2}(k(s_{1})^{-1},...,k(s_{n})^{-1}),1^{(n-2)})$

$\hskip
65.44142pt=g_{2}(g_{2}(k(r_{1}),k(s_{1})^{-1}),1^{(n-2)}),...,g_{2}(k(r_{n}),k(s_{n})^{-1}),1^{(n-2)}))$

$\hskip 65.44142pt=g_{2}(h(\frac{r_{1}}{s_{1}}),...,h(\frac{r_{n}}{s_{n}}))$
for $r_{1}^{n}\in R_{1}$ and $s_{1}^{n}\in S$. Consequently, $h$ is a
homomorphism. Now, suppose that $h^{\prime}$ is another homomorphism frome
$S^{-1}R_{1}$ to $R_{2}$ with $h^{\prime}o\phi=k$. Then we obtain

$h(\frac{r}{s})=h(G(\frac{r}{1},\frac{1}{s},\frac{1}{1}^{(n-2)}))$

$\hskip
22.76228pt=g_{2}(h(\frac{r}{1}),h(\frac{1}{s}),h(\frac{1}{1})^{(n-2)})$

$\hskip 22.76228pt=g_{2}(h(\phi(r)),h(\phi(s)^{-1}),1^{(n-2)})$

$\hskip 22.76228pt=g_{2}(h(\phi(r)),(h(\phi(s))^{-1},1^{(n-2)})$

$\hskip 22.76228pt=g_{2}(k(r),k(s)^{-1},1^{(n-2)})$

$\hskip
22.76228pt=g_{2}(h^{\prime}(\phi(r)),(h^{\prime}(\phi(s))^{-1},1^{(n-2)})$

$\hskip
22.76228pt=g_{2}(h^{\prime}(\frac{r}{1})),(h^{\prime}(\frac{s}{1}))^{-1},1^{(n-2)})$

$\hskip
22.76228pt=g_{2}(h^{\prime}(\frac{r}{1})),h^{\prime}(\frac{1}{s}),1^{(n-2)})$

$\hskip 22.76228pt=h^{\prime}(G(\frac{r}{1},\frac{1}{s},\frac{1}{1}^{(n-2)}))$

$\hskip 22.76228pt=h^{\prime}(\frac{r}{s}).$
for every $\frac{r}{s}\in S^{-1}R$. It implies that the homomorphism $h$ is
unique. Thus the proof is completed. ∎

######  Corollary 3.7.

Let $(R_{1},f_{1},h_{1})$ and $(R_{2},f_{2},g_{2})$ be two Krasner
$(m,n)$-hyperrings and $S$ be an $n$-ary multiplicative subset of $R_{1}$ with
$1\in S$. Let $k:R_{1}\longrightarrow R_{2}$ be a homomorphism such that

$i)$ $k(s)$ is an invertible element of $R_{2}$ for each $s\in S$.

$ii)$ $k(r_{1})=0$ for $r_{1}\in R_{1}$ implies that
$g_{1}(t,r_{1},1^{(n-2)})=0$, for some $t\in S$.

$iii)$ for each $r_{2}\in R_{2}$, $r_{2}=g_{2}(k(r_{1}),k(s)^{-1},1^{(n-2)})$
where $r_{1}\in R_{1}$ and $s\in S$.
Then there exists an unique isomomorphism $h:S^{-1}R_{1}\longrightarrow R_{2}$
such that $ho\phi=k$.

###### Proof.

By using an argument similar to that in the proof of Theorem 3.6, one can
easily complete the proof. ∎

######  Theorem 3.8.

If $R$ is an $n$-ary hyperintegral domain, then $S^{-1}R$ is an $n$-ary
hyperintegral domain.

###### Proof.

Let $G(\frac{r_{1}}{s_{1}},...,\frac{r_{1}}{s_{1}})=0_{S^{-1}R}$ for
$r_{1}^{n}\in R$ and $s_{1}^{n}\in S$. Thus
$\frac{g(a_{1}^{n})}{g(s_{1}^{n})}=0_{S^{-1}R}$. By Lemma 3.2 (2), we have
$g(t,g(a_{1}^{n}),1^{(n-2)})=0$ for some $t\in S$. Since $R$ is an $n$-ary
hyperintegral domain and $t\neq 0$, we have $g(a_{1}^{n})=0$ which implies
$a_{1}=0$ or $a_{2}=0$ or … or $a_{n}=0$. Hence we get
$\frac{a_{1}}{s_{1}}=0_{S^{-1}R}$ or $\frac{a_{2}}{s_{2}}=0_{S^{-1}R}$ or … or
$\frac{a_{n}}{s_{n}}=0_{S^{-1}R}$. Thus $S^{-1}R$ is an $n$-ary hyperintegral
domain. ∎

######  Theorem 3.9.

Let $R$ be an $n$-ary hyperintegral domain and $S=R-\\{0\\}$. Then each
nonzero element of $S^{-1}R$ is invertible.

###### Proof.

Let $\frac{r}{s}$ be an nonzero element of $S^{-1}R$. Since $r\neq 0$, then
$r\in S$ and so $\frac{s}{r}\in S^{-1}R$. Thus
$G(\frac{r}{s},\frac{s}{r},{\frac{1}{1}}^{(n-2)})=\frac{g(r,s,1^{(n-2)})}{g(s,r,1^{(n-2)})}=\frac{1}{1}=1_{S^{-1}R}$,
by Lemma 3.2 (3). ∎

##  4\. hyperideals in Krasner $(m,n)$-hyperring of fractions

Let $I$ be a hyperideal of Krasner $(m,n)$-hyperring $R$ and $S$ be an $n$-ary
multiplicative subset of $R$ with $1\in S$, then we can define that
$S^{-1}I=\\{\frac{a}{s}\ |\ a\in I,s\in S\\},$ which is a hyperideal of
$S^{-1}R$.

######  Theorem 4.1.

Let $R$ be a Krasner $(m,n)$-hyperring and $S$ be an $n$-ary multiplicative
subset of $R$ with $1\in S$. Let $I$ be a hyperideal of $R$. Then $I\cap
S\neq\varnothing$ if and only if $S^{-1}I=S^{-1}R$.

###### Proof.

$(\Longrightarrow):$ Let $a\in I\cap S$. Then $\frac{1}{1}=\frac{a}{a}\in
S^{-1}I$. Since $I$ is a hyperideal of $R$, we have
$G(\frac{1}{1},\frac{r}{s},\frac{1}{1}^{(n-2)})\in S^{-1}I$ for all
$\frac{r}{s}\in S^{-1}R$. Since
$G(\frac{1}{1},\frac{r}{s},\frac{1}{1}^{(n-2)})=\frac{g(1,r,1^{(n-2)})}{g(1,s,1^{(n-2)})}=\frac{r}{s}$,
then $\frac{r}{s}\in S^{-1}I.$ Thus $S^{-1}I=S^{-1}R$.
$(\Longleftarrow):$ By the homomorphism $\phi:R\longrightarrow S^{-1}R$, it
implies that $\phi(1)=\frac{1}{1}$. Since $S^{-1}I=S^{-1}R$ and $\phi(1)\in
S^{-1}R$, then $\phi(1)\in S^{-1}I$. Hence, there exist $a\in I$, $s\in S$
such that $\frac{1}{1}=\phi(1)=\frac{a}{s}$. So, there exists $t\in S$ such
that

$0\in g(t,f(g(a,1,1^{(n-2)}),-g(1,s,1^{(n-2)}),0^{(m-2)}),1^{(n-2)})$

$\hskip 8.5359pt=g(t,f(g(a,1^{(n-1)}),-g(s,1^{(n-1)}),0^{(m-2)}),1^{(n-2)})$

$\hskip
8.5359pt=f(g(t,g(a,1^{(n-1)}),1^{(n-2)}),g(t,-g(s,1^{(n-1)}),1^{(n-2)}),0^{(m-2)})$

$\hskip 8.5359pt=f(g(t,a,1^{(n-2)}),-g(t,s,1^{(n-2)}),0^{(m-2)})$.
Since $g(t,a,1^{(n-2)})\in I$, then $g(t,s,1^{(n-2)})\in I$. Also, since $S$
is an $n$-ary multiplicative subset of $R$, then $g(t,s,1^{(n-2)})\in S$.
Consequently, $I\cap S\neq\varnothing$. ∎

If $(a,s)\in S^{-1}I$ we don,t get necessarily $a\in I$, maybe
$(a,s)=(a^{\prime},s)$ such that $a^{\prime}\in I$ but $a\notin I$.

######  Theorem 4.2.

Let $R$ be a Krasner $(m,n)$-hyperring and $S$ be an $n$-ary multiplicative
subset of $R$ with $1\in S$. Then every hyperideal of $S^{-1}R$ is an extended
hyperideal.

###### Proof.

Suppose that $J$ is a hyperideal of $S^{-1}R$. Put $B=\\{r\in R\ |\ \exists
s\in S;\frac{r}{s}\in J\\}$. Easily, it is proved that $B$ is a hyperideal of
$R$. We show that $B^{e}=S^{-1}B=J$. Let $\frac{r}{s}\in J$. Then $r\in B$ and
so $\frac{r}{s}\in S^{-1}B$ which means $J\subseteq S^{-1}B$. Now, assume that
$\frac{b}{s}\in S^{-1}B$. Then there exist $b^{\prime}\in B$ and
$s^{\prime}\in S$ such that $\frac{b}{s}=\frac{b^{\prime}}{s^{\prime}}$. It
means there exists $t\in S$ such that

$0\in
g(t,f(g(b,s^{\prime},1^{(n-2)}),-g(b^{\prime},s,1^{(n-2)}),0^{(m-2)}),1^{(n-2)})$

$\hskip
8.5359pt=f(g(t,b,s^{\prime},1^{(n-3)}),-g(t,b^{\prime},s,1^{(n-3)}),0^{(m-2)}),1^{(n-2)})$.
Since $g(t,b^{\prime},s,1^{(n-3)})\in B$, then
$g(g(t,s^{\prime},1^{(n-2}),b,1^{(n-2)})=g(t,b,s^{\prime},1^{(n-3)})\in B$.
Put $t^{\prime}=g(t,s^{\prime},1^{(n-2})$. Therefore we have
$g({t^{\prime}}^{(m-1)},b,1^{(n-m)})\in B$. Hence there exists
$t^{\prime\prime}\in S$ such that
$\frac{g({t^{\prime}}^{(m-1)},b,1^{(n-m)})}{t^{\prime\prime}}\in J$ and so
$\frac{g({t^{\prime}}^{(m-1)},b,1^{(n-m)})}{g({t^{\prime\prime}}^{(m-1)},1^{(n-m+1)})}\in
J$. Then we have

$G(\frac{g({t^{\prime\prime}}^{(m-1)},1^{(n-m+1)})}{g(t^{\prime},s,1^{(n-2)})},\frac{g({t^{\prime}}^{(m-1)},b,1^{(n-m)})}{g({t^{\prime\prime}}^{(m-1)},1^{(n-m+1)})})=\frac{g(g(t^{\prime\prime},t^{\prime},1^{(n-2)})^{(m-1)},b,1^{(n-m)})}{g(g(t^{\prime\prime},t^{\prime},1^{(n-2)})^{(m-1)},s,1^{(n-m)})}=\frac{b}{s}\in
J$.
This means $S^{-1}B\subseteq J$. Consequently, $S^{-1}B=J$. ∎

Let $R$ be a Krasner (m, n)-hyperring. Then the hyperideal $M$ of $R$ is said
to be maximal if for every hyperideal $I$ of $R$, $M\subseteq I\subseteq R$
implies that $I=M$ or $I=R$ [1].

######  Lemma 4.3.

Let $R$ be a Krasner $(m,n)$-hyperring such that $M$ is a hyperideal of $R$.
If each $x\in R-M$ is invertible, then $M$ is a maximal hyperideal of $R$.

###### Proof.

The proof is similar to ordinary algebra. ∎

######  Theorem 4.4.

Let $R$ be a Krasner $(m,n)$-hyperring and $P$ be an $n$-ary prime hyperideal
of $R$. If $S=R-P$, then $M=\\{\frac{a}{s}\ |\ a\in P,s\in S\\}$ is the only
maximal hyperideal of $S^{-1}R$.

###### Proof.

Clearly, $S=R-P$ is an $n$-ary multiplicative subset of $R$. Let
$\frac{a_{1}}{s_{1}},...,\frac{a_{m}}{s_{m}}\in M$ such that $a_{1}^{m}\in P$
and $s_{1}^{m}\in S$. Then

$F(\frac{a_{1}}{s_{1}},...,\frac{a_{m}}{s_{m}})=\frac{f(g(a_{1},s_{2}^{m},1^{(n-m)}),g(s_{1},a_{2},s_{3}^{m},1^{(n-m)}),...,g(s_{1}^{m-1},a_{m},1^{(n-m)}))}{g(s_{1}^{m},1^{(n-m)})}.$
Since $a_{1}^{m}\in P$, then

$g(a_{1},s_{2}^{m},1^{(n-m)}),g(s_{1},a_{2},s_{3}^{m},1^{(n-m)}),...,g(s_{1}^{m-1},a_{m},1^{(n-m)})\in
P$
and so

$f(g(a_{1},s_{2}^{m},1^{(n-m)}),g(s_{1},a_{2},s_{3}^{m},1^{(n-m)}),...,g(s_{1}^{m-1},a_{m},1^{(n-m)}))\subseteq
P$.
Thus we conclude that $F(\frac{a_{1}}{s_{1}},...,\frac{a_{m}}{s_{m}})\subseteq
M$.
Clearly, if $\frac{a}{r}\in M$, then $-\frac{a}{r}=\frac{-a}{r}\in M$. Also,
since $0\in P$,then $0_{R_{P}}=\frac{0}{s}\in M$ for all $s\in S$. Hence
$(M,F)$ is a canonical $n$-ary hypergroup.
Now, let $r_{1}^{n}\in R$, $s_{1}^{n}\in S$ and $k\in\\{1,...,n\\}$. Then

$G(\frac{r_{1}}{s_{1}},...,\frac{r_{k-1}}{s_{k-1}},M,\frac{r_{k+1}}{s_{k+1}},...,\frac{r_{n}}{s_{n}})=\bigcup\\{G(\frac{r_{1}}{s_{1}},...,\frac{r_{k-1}}{s_{k-1}},\frac{a}{s},\frac{r_{k+1}}{s_{k+1}},...,\frac{r_{n}}{s_{n}})\
|\ \frac{a}{s}\in M\\}$

$\hskip
133.72786pt=\bigcup\\{\frac{g(r_{1}^{k-1},a,r_{k-1}^{n})}{g(s_{1}^{k-1},s,s_{k-1}^{n})}\
|\ a\in P,s\in S\\}$.
Since $a\in P$, then $g(r_{1}^{k-1},a,r_{k-1}^{n})\in P$ and so
$G(\frac{r_{1}}{s_{1}},...,\frac{r_{k-1}}{s_{k-1}},M,\frac{r_{k+1}}{s_{k+1}},...,\frac{r_{n}}{s_{n}})\subseteq
M$. Thus, $(M,F,G)$ is a hyperideal of $R_{P}$.
Suppose that $1_{R_{P}}=\frac{1}{1}\in M$. Then there exist $a\in P$ and $s\in
S$ such that $\frac{1}{1}=\frac{a}{s}$. It implies that there exists $t\in S$
such that

$0\in g(t,f(g(a,1,1^{(n-2)}),-g(1,s,1^{(n-2)}),0^{(m-2)}),1^{(n-2)})$

$\hskip 8.5359pt=f(g(t,a,1^{(n-2)}),-g(t,s,1^{(n-2)}),0^{(m-2)})$.
Since $g(t,a,1^{(n-2)})\in P$, then $g(t,s,1^{(n-2)})\in P$. Since $P$ is an
$n$-ary prime hyperideal of $R$, then we obtain $t\in P$ or $s\in P$ which is
a contradiction. Then $M$ is a proper hyperideal of $R$.
Now, suppose that $\gamma\in R_{P}-M$. It means $\gamma=\frac{r}{s}$ such that
$r\in R-P$ and $s\in S$. Then $r\in S$ and so $\frac{s}{r}\in M$. Hence
$\frac{1}{1}=G(\frac{r}{s},\frac{s}{r},\frac{1}{1}^{(n-2)})\in M$ which is a
contradiction. Consequently, $M$ is the only maximal hyperideal of $R_{P}$. ∎

######  Theorem 4.5.

Let $R$ be a Krasner $(m,n)$-hyperring and $S$ be an $n$-ary multiplicative
subset of $R$ with $1\in S$. If $P$ is an $n$-ary prime hyperideal of $R$ with
$P\cap S=\varnothing$, then $S^{-1}P$ is an $n$-ary prime hyperideal of
$S^{-1}R$.

###### Proof.

Let $G(\frac{a_{1}}{s_{1}},...,\frac{a_{n}}{s_{n}})\in S^{-1}P$ for
$\frac{a_{1}}{s_{1}},...,\frac{a_{n}}{s_{n}}\in S^{-1}R$. Then we have
$\frac{g(a_{1}^{n})}{g(s_{1}^{n})}\in S^{-1}P$. It implies that there exists
$t\in S$ such that $g(t,g(a_{1}^{n}),1^{(n-2)})\in P$. Since $P$ is an $n$-ary
prime hyperideal of $R$ and $P\cap S=\varnothing$, then $g(a_{1}^{n})\in P$
which means there exists $1\leq i\leq n$ with $a_{i}\in P$. Hence we conclude
that $\frac{a_{i}}{s_{i}}\in S^{-1}P$ for some $1\leq i\leq n$. Thus $S^{-1}P$
is an $n$-ary prime hyperideal of $S^{-1}R$. ∎

######  Example 4.6.

The set $R=\\{0,1,2\\}$ with the following 3-ary hyperoeration $f$ and 3-ary
operation $g$ is a Krasner $(3,3)$-hyperring such that $f$ and $g$ are
commutative.

$f(0,0,0)=0,\ \ \ f(0,0,1)=1,\ \ \ f(0,1,1)=1,\ \ \ f(1,1,1)=1,\ \ \
f(1,1,2)=R,$ $f(0,1,2)=R,\ \ \ f(0,0,2)=2,\ \ \ f(0,2,2)=2,\ \ \ f(1,2,2)=R,\
\ \ f(2,2,2)=2,$

$\ \ \ g(1,1,1)=1,\ \ \ \ g(1,1,2)=g(1,2,2)=g(2,2,2)=2,$
and for $x_{1},x_{2}\in R,\ g(0,x_{1},x_{2})=0$.

$S=\\{1,2\\}$ is a 3-ary multiplicative subset of Krasner $(3,3)$-hyperring
$(R,f,g)$ and hyperideal $P=\\{0\\}$ is a 3-ary prime hyperideal of $R$ (see
example 4.10 in [1]). Thus $S^{-1}P=\\{\frac{0}{1}\\}$ is a $3$-ary prime
hyperideal of $S^{-1}R$.

Let $I$ be a hyperideal in a Krasner $(m,n)$-hyperring $R$ with scalar
identity. The radical (or nilradical) of $I$, denoted by ${\sqrt{I}}^{(m,n)}$
is the hyperideal $\bigcap P$, where the intersection is taken over all
$n$-ary prime hyperideals $P$ which contain $I$. If the set of all $n$-ary
hyperideals containing $I$ is empty, then ${\sqrt{I}}^{(m,n)}$ is defined to
be $R$. Ameri and Norouzi showed that if $x\in{\sqrt{I}}^{(m,n)}$, then there
exists $t\in\mathbb{N}$ such that $g(x^{(t)},1_{R}^{(n-t)})\in I$ for $t\leq
n$, or $g_{(l)}(x^{(t)})\in I$ for $t=l(n-1)+1$ [1].

######  Lemma 4.7.

Let $R$ be a Krasner $(m,n)$-hyperring and $S$ be an $n$-ary multiplicative
subset of $R$ with $1\in S$. If $I$ is an $n$-ary hyperideal of $R$, then
$\sqrt{S^{-1}I}^{(m,n)}=S^{-1}\sqrt{I}^{(m,n)}.$

###### Proof.

Let $\frac{a}{s}\in\sqrt{S^{-1}I}^{(m,n)}$. Then there exists $k\in\mathbb{N}$
with $G(\frac{a}{s}^{(k)},\frac{1}{1}^{(n-k)})\in S^{-1}I$ for $k\leq n$, or
$G_{(l)}(\frac{a}{s}^{(k)})\in S^{-1}I$ for $k=l(n-1)+1$. If
$G(\frac{a}{s}^{(k)},\frac{1}{1}^{(n-k)})\in S^{-1}I$, then
$\frac{g(a^{(k)},1^{(n-k)})}{g(a^{(k)},1^{(n-k)})}\in S^{-1}I$. Therefore
$g(t,g(a^{(k)},1^{(n-k)}),1^{(n-2)})\in I$ for some $t\in S$ and so
$g(g(t,a,1^{(n-2)})^{(k)},1^{(n-k)})=g(t^{(k)},g(a^{(k)},1^{(n-k)}),1^{(n-k-1)})\in
I$. It means $g(t,a,1^{(n-2)})\in\sqrt{I}^{(m,n)}$ and so
$g(t^{(m-1)},a,1^{(n-m)})\in\sqrt{I}^{(m,n)}$. Hence we get
$\frac{g(t^{(m-1)},a,1^{(n-m)})}{g(t^{(m-1)},s,1^{(n-m)})}=\frac{a}{s}\in
S^{-1}\sqrt{I}^{(m,n)}$, by Lemma 3.2 (4). Similarly for the other case. Thus
$\sqrt{S^{-1}I}^{(m,n)}\subseteq S^{-1}\sqrt{I}^{(m,n)}.$
Now, let $\frac{a}{s}\in S^{-1}\sqrt{I}^{(m,n)}$. Then we conclude
$g(t,a,1^{(n-2)})\in\sqrt{I}^{(m,n)}$ for some $t\in S$ and so
$g(t^{(m-1)},a,1^{(n-m)})\in\sqrt{I}^{(m,n)}$. It means that there exists
$k\in\mathbb{N}$ with $g(g(t^{(m-1)},a,1^{(n-m)})^{(k)},1^{(n-k)})\in I$ for
$k\leq n$, or $g_{(l)}(g(t^{(m-1)},a,1^{(n-2)})^{(k)})\in I$ for $k=l(n-1)+1$.
If $g(g(t^{(m-1)},a,1^{(n-2)})^{(k)},1^{(n-k)})\in I$, then we have

$G(\frac{a}{s}^{(k)},\frac{1}{1}^{(n-k)})=G(\frac{g(t^{(m-1)},a,1^{(n-2)})}{g(t^{(m-1)},s,1^{(n-2)})}^{(k)},\frac{1}{1}^{(n-k)})$

$\hskip
68.28644pt=\frac{g(g(t^{(m-1)},a,1^{(n-2)})^{(k)},1^{(n-k)})}{g(g(t^{(m-1)},s,1^{(n-2)})^{(k)},1^{(n-k)})}\in
S^{-1}I$.
Therefore we get $\frac{a}{s}\in\sqrt{S^{-1}I}^{(m,n)}$. Similarly for the
other case. Thus $S^{-1}\sqrt{I}^{(m,n)}\subseteq\sqrt{S^{-1}I}^{(m,n)}$.
Consequently, $\sqrt{S^{-1}I}^{(m,n)}=S^{-1}\sqrt{I}^{(m,n)}$. ∎

A hyperideal $Q\neq R$ in a Krasner $(m,n)$-hyperring $(R,f,g)$ with the
scalar identity $1_{R}$ is said to be $n$-ary primary if $g(x^{n}_{1})\in Q$
and $x_{i}\notin Q$ implies that
$g(x_{1}^{i-1},1_{R},x_{i+1}^{n})\in{\sqrt{Q}}^{(m,n)}$ [1].

######  Theorem 4.8.

Let $R$ be a Krasner $(m,n)$-hyperring and $S$ be an $n$-ary multiplicative
subset of $R$ with $1\in S$. If $P$ is an $n$-ary primary hyperideal of $R$
with $P\cap S=\varnothing$, then $S^{-1}P$ is an $n$-ary primary hyperideal of
$S^{-1}R$.

###### Proof.

Let $\frac{a_{1}}{s_{1}},...,\frac{a_{n}}{s_{n}}\in S^{-1}R$ such that
$G(\frac{a_{1}}{s_{1}},...,\frac{a_{n}}{s_{n}})\in S^{-1}P$ . Then we have
$\frac{g(a_{1}^{n})}{g(s_{1}^{n})}\in S^{-1}P$. It implies that there exists
$t\in S$ such that $g(t,g(a_{1}^{n}),1^{(n-2)})\in P$. Since $P$ is an $n$-ary
primary hyperideal of $R$, then there exist $1\leq i\leq n$ such that at least
one of the cases hold: $a_{i}\in P$,
$g(a_{1}^{i-1},1,a_{i+1}^{n})\in\sqrt{P}^{(m,n)}$, $t\in\sqrt{P}^{(m,n)}$ or
$g(a_{1}^{n})\in\sqrt{P}^{(m,n)}$. If $a_{i}\in P$, then
$\frac{a_{i}}{s_{i}}\in S^{-1}P$ and we are done. If
$g(a_{1}^{i-1},1,a_{i+1}^{n})\in\sqrt{P}^{(m,n)}$, then
$G(\frac{a_{1}}{s_{1}},...,\frac{a_{i-1}}{s_{i-1}},\frac{1}{1},\frac{a_{i+1}}{s_{i+1}},...,\frac{a_{n}}{s_{n}})=\frac{g(a_{1}^{i-1},1,a_{i+1}^{n})}{g(s_{1}^{i-1},1,s_{i+1}^{n})}\in
S^{-1}\sqrt{P}^{(m,n)}=\sqrt{S^{-1}P}^{(m,n)}$, by Lemma 4.7. If
$t\in\sqrt{P}^{(m,n)}$, then
$g(t^{(m-1)},a_{k},1^{(m-n)})\in\sqrt{P}^{(m,n)}$, for all $1\leq k\leq n$.
Therefore
$\frac{g(t^{(m-1)},a_{k},1^{(m-n)})}{g(t^{(m-1)},s_{k},1^{(m-n)})}\in
S^{-1}\sqrt{P}^{(m,n)}=\sqrt{S^{-1}P}^{(m,n)}$ and so
$\frac{a_{k}}{s_{k}}\in\sqrt{S^{-1}P}^{(m,n)}$. Therefore for each $i\neq k$,
$G(\frac{a_{1}}{s_{1}},...,\frac{a_{i-1}}{s_{i-1}},\frac{1}{1},\frac{a_{i+1}}{s_{i+1}},...,\frac{a_{n}}{s_{n}})\in\sqrt{S^{-1}P}^{(m,n)}$.
Let $g(a_{1}^{n})\in\sqrt{P}^{(m,n)}$. Theorem 4.28. in [1] shows that
$\sqrt{P}^{(m,n)}$ is an $n$-ary prime hyperideal of $R$. Hence there exists
$1\leq k\leq n$ such that $a_{i}\in\sqrt{P}^{(m,n)}$. It implies that
$\frac{a_{k}}{s_{k}}\in S^{-1}\sqrt{P}^{(m,n)}=\sqrt{S^{-1}P}^{(m,n)}$.
Therefore for each $i\neq k$,
$G(\frac{a_{1}}{s_{1}},...,\frac{a_{i-1}}{s_{i-1}},\frac{1}{1},...,\frac{a_{i+1}}{s_{i+1}},\frac{a_{n}}{s_{n}})\in\sqrt{S^{-1}P}^{(m,n)}$.
Thus $S^{-1}P$ is an $n$-ary primary hyperideal of $S^{-1}R$. ∎

A proper hyperideal $I$ of a Krasner $(m,n)$-hyperring $(R,f,g)$ with the
scalar identity $1_{R}$ is said to be $n$-ary 2-absorbing if for $x_{1}^{n}\in
R$, $g(x_{1}^{n})\in I$ implies that $g(x_{i},x_{j},1_{R}^{(n-2)})\in I$ for
some $1\leq i<j\leq n$ [3].

######  Theorem 4.9.

Let $R$ be a Krasner $(m,n)$-hyperring and $S$ be an $n$-ary multiplicative
subset of $R$ with $1\in S$. If $P$ is an $n$-ary 2-absorbing hyperideal of
$R$ with $P\cap S=\varnothing$, then $S^{-1}P$ is an $n$-ary 2-absorbing
hyperideal of $S^{-1}R$.

###### Proof.

Let $G(\frac{a_{1}}{s_{1}},...,\frac{a_{n}}{s_{n}})\in S^{-1}P$, for
$\frac{a_{1}}{s_{1}},...,\frac{a_{n}}{s_{n}}\in S^{-1}R$. Then we have
$\frac{g(a_{1}^{n})}{g(s_{1}^{n})}\in S^{-1}P$. It implies that there exists
$t\in S$ such that $g(t,g(a_{1}^{n}),1^{(n-2)})\in P$. Since $P$ is an $n$-ary
2-absorbing hyperideal of $R$, then there exist $1\leq i<j\leq n$ such that
$g(t,a_{i},1^{(n-2)})\in P$ or $g(a_{i},a_{j},1^{(n-2)})\in P$. Hence we
conclude that $\frac{a_{i}}{s_{i}}\in S^{-1}P$ for some $1\leq i\leq n$. Thus
$S^{-1}P$ is an $n$-ary prime hyperideal of $S^{-1}R$. If for some $1\leq
i\leq n$, $g(t,a_{i},1^{(n-2)})\in P$, then $g(t^{(m-1)},a_{i},1^{(n-m)})\in
P$ and so
$\frac{g(t^{(m-1)},a_{i},1^{(n-m)})}{g(t^{(m-1)},s_{i},1^{(n-m)})}\in
S^{-1}P$. Hence $\frac{a_{i}}{s_{i}}\in S^{-1}P$, by Lemma 3.2 (4). Therefore
for every $1\leq j\leq n$,
$G(\frac{a_{i}}{s_{i}},\frac{a_{j}}{s_{j}},\frac{1}{1}^{(n-2)})\in S^{-1}P$
and we are done. If $g(a_{i},a_{j},1^{(n-2)})\in P$, for some $1\leq i<j\leq
n$, then $\frac{g(a_{i},a_{j},1^{(n-2)})}{g(s_{i},s_{j},1^{(n-2)})}\in
S^{-1}P$ which means
$G(\frac{a_{i}}{s_{i}},\frac{a_{j}}{s_{j}},\frac{1}{1}^{(n-1)})\in S^{-1}P$.
Consequently, $S^{-1}P$ is an $n$-ary 2-absorbing hyperideal of $S^{-1}R$. ∎

##  5\. qutient Krasner $(m,n)$-hyperring of fractions

Let $R$ be a Krasner $(m,n)$-hyperring and $I$ be a hyperideal of $R$. Then we
consider the set $R/I$ as follows:

$R/I=\\{f(r,I,0^{(m-2)})\ |\ r\in R\\}$.

######  Lemma 5.1.

Let $R$ be a Krasner $(m,n)$-hyperring and $S$ be an $n$-ary multiplicative
subset of $R$ with $1\in S$. Let $I$ be a hyperideal of $R$ such that $S\cap
I=\varnothing$. Then $\bar{S}=\\{f(s,I,0^{(m-2)})\ |\ s\in S\\}$ is an $n$-ary
multiplicative subset of $R/I$.

###### Proof.

Let $f(s_{1},I,0^{(m-2)}),...,f(s_{n},I,0^{(m-2)})\in\bar{S}$, for
$s_{1}^{n}\in S$. Then we have

$g(f(s_{1},I,0^{(m-2)}),...,f(s_{n},I,0^{(m-2)}))=f(g(s_{1}^{n}),I,0^{m-2)})$.
Since $S$ is an $n$-ary multiplicative subset of $R$, then $g(s_{1}^{n})\in
S$. It implies that
$g(f(s_{1},I,0^{(m-2)}),...,f(s_{n},I,0^{(m-2)}))\in\bar{S}$. ∎

######  Theorem 5.2.

Let $R$ be a Krasner $(m,n)$-hyperring and $S$ be an $n$-ary multiplicative
subset of $R$ with $1\in S$. Let $I$ be a hyperideal of $R$ such that $S\cap
I=\varnothing$. If $\bar{S}=\\{f(s,I,0^{(m-2)})\ |\ s\in S\\}$, then
$\bar{S}^{-1}(R/I)\cong S^{-1}R/S^{-1}I$.

###### Proof.

Define mapping $k:R/I\longrightarrow S^{-1}R/S^{-1}I$ as following:

$k(f(r,I,0^{(m-2)})=F(\frac{r}{1},S^{-1}I,0_{S^{-1}R}^{(m-2)})$.
It is easy to see the mapping is a homomorphism. Let
$f(s,I,0^{(m-2)})\in\bar{S}$. Then
$k(f(s,I,0^{(m-2)}))=F(\frac{s}{1},S^{-1}I,0_{S^{-1}R}^{(m-2)}).$ Since
$F(\frac{1}{s},S^{-1}I,0_{S^{-1}R}^{(m-2)})\in S^{-1}R/S^{-1}I$, then we
obtain

$G(F(\frac{s}{1},S^{-1}I,0_{S^{-1}R}^{(m-2)}),F(\frac{1}{s},S^{-1}I,0_{S^{-1}R}^{(m-2)}),F(\frac{1}{1},S^{-1}I,0_{S^{-1}R}^{(m-2)})^{(n-2)}))$

$\hskip
28.45274pt=F(G(\frac{s}{1},\frac{1}{s},\frac{1}{1}^{(n-2)}),S^{-1}I,0_{S^{-1}R}^{(m-2)})$

$\hskip 28.45274pt=F(\frac{1}{1},S^{-1}I,0_{S^{-1}R}^{(m-2)})).$
Assume that $k(f(r,I,0^{(m-2)})=S^{-1}I.$ Then we have
$F(\frac{r}{1},S^{-1}I,0_{S^{-1}R}^{(m-2)})=S^{-1}I$. It means $\frac{r}{1}\in
S^{-1}I$. Then there exists $t\in S$ such that $g(t,r,1^{(n-2)})\in I$.
Clearly, $f(t,I,0^{(m-2)})\in\bar{S}$ and we have

$g(f(t,I,0^{(m-2)},f(r,I,0^{(m-2)}),f(1,I,0^{(m-2)})^{(n-2)})$

$\hskip 28.45274pt=f(g(t,r,1^{(n-2)}),I,0^{(m-2)})=I$.
Now, suppose that $F(\frac{r}{s},S^{-1}I,0_{S^{-1}R})\in S^{-1}R/S^{-1}I$.
Thus we have

$F(\frac{r}{s},S^{-1}I,0_{S^{-1}R})=G(F(\frac{r}{1},S^{-1}I,0_{S^{-1}R}),F(\frac{1}{s},S^{-1}I,0_{S^{-1}R}),F(\frac{1}{1},S^{-1}I,0_{S^{-1}R})^{(n-2)})$

$\hskip
28.45274pt=G(k(f(r,I,0^{(m-2)}),k(f(r,I,0^{(m-2)}),F(\frac{1}{1},S^{-1}I,0_{S^{-1}R})^{(n-2)})$.
Hence, there exists an isomorphism from $\bar{S}^{-1}(R/I)$ to
$S^{-1}R/S^{-1}I$, by Corollary 3.7. It means $\bar{S}^{-1}(R/I)\cong
S^{-1}R/S^{-1}I$. ∎

Let $P$ be an $n$-ary prime hyperideal of Krasner $(m,n)$-hyperring $R$. Put
$S=R-P$. Then $S$ is an $n$-ary multiplicative subset of $R$ such that $1\in
S$ and $0\notin S$. In this case, we denote $S^{-1}R=R_{P}$. Moreover, If
$S^{-1}I$ is a hyperideal of $R_{P}$, then it is denoted by $IR_{P}$.

######  Example 5.3.

Let $R$ be a Krasner $(m,n)$-hyperring such that $P$ is an $n$-ary prime
hyperideal of $R$. Put $S=R-P$. Then

$\bar{S}=\\{f(s,P,0^{(n-2)})\ |s\in S\\}=R/P-\\{f(P,0^{(n-1)})\\}$
is an $n$-ary multiplicative subset of $R/P$. By Theorem 4.6 in [1], $R/P$ is
an n-ary hyperintegral domain. Theorem 3.8 and 3.9 show that
$\bar{S}^{-1}(R/P)$ is an n-ary hyperintegral domain and each nonzero element
of $\bar{S}^{-1}(R/P)$ is invertible. Moreover, we have

$\bar{S}^{-1}(R/P)\cong\frac{S^{-1}R}{S^{-1}P}=\frac{R_{p}}{PR_{p}}$,
by Theorem 5.2.

######  Example 5.4.

Let $R$ be a Krasner $(m,n)$-hyperring such that $P$ and $Q$ are two $n$-ary
prime hyperideals of $R$ such that $Q\subseteq P$. Put $S=R-P$. Then

$\bar{S}=\\{f(s,Q,0^{(n-2)})\ |s\in S\\}=R/Q-R/P.$
It is clear that $P/Q$ is an $n$-ary prime hyperideal of $R/Q$. Therefore
$\bar{S}^{-1}(R/Q)=(R/Q)_{P/Q}$. By Theorem 5.2, we get
$(R/Q)_{P/Q}\cong\frac{R_{P}}{QR_{P}}$.

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11institutetext: Institut d’Astrophysique Spatiale, CNRS-Université Paris-Sud
11, 91405 Orsay Cedex, France
22institutetext: National Astronomical Observatory, Chinese Academy of
Sciences, Beijing 100012, China

# Hanle signatures of the coronal magnetic field in the linear polarization of
the hydrogen L$\alpha$ line

M. Derouich Present address: Colorado Research Associates Division, NorthWest
Research Associates, Inc., 3380 Mitchell Ln., Boulder, CO 80301, U.S.A.11 F.
Auchère 11 J. C. Vial 11 and M. Zhang 22

(Received 06 July 2009 / Accepted 21 October 2009 )

###### Abstract

Aims.  This paper is dedicated to the assessment of the validity of future
coronal spectro-polarimetric observations and to prepare their interpretation
in terms of the magnetic field vector.

Methods.  We assume that the polarization of the hydrogen coronal L$\alpha$
line is due to anisotropic scattering of an incident chromospheric radiation
field. The anisotropy is due to geometrical effects but also to the
inhomogeneities of the chromospheric regions which we model by using
Carrington maps of the L$\alpha$. Because the corona is optically thin, we
fully consider the effects of the integration over the line-of-sight (LOS). As
a modeling case, we include a dipolar magnetic topology perturbed by a non-
dipolar magnetic structure arising from a prominence current sheet in the
corona. The spatial variation of the hydrogen density and the temperature is
taken into account. We determine the incident radiation field developed on the
tensorial basis at each point along the LOS. Then, we calculate the local
emissivity vector to obtain integrated Stokes parameters with and without
coronal magnetic field.

Results.  We show that the Hanle effect is an interesting technique for
interpreting the scattering polarization of the L$\alpha$ $\lambda$1216 line
in order to diagnose the coronal magnetic field. The difference between the
calculated polarization and the zero magnetic field polarization gives us an
estimation of the needed polarimetric sensitivity in future polarization
observations. We also obtain useful indications about the optimal
observational strategy.

Conclusions. Quantitative interpretation of the Hanle effect on the scattering
linear polarization of L$\alpha$ line can be a crucial source of information
about the coronal magnetic field at a height over the limb $h$
$<0.7\;R_{\sun}$. Therefore, one needs the development of spatial
instrumentation to observe this line.

###### Key Words.:

Line: polarization – Sun: corona – Sun: UV radiation – Scattering

††offprints: M. Derouich
e-mail<EMAIL_ADDRESS>
##  1 Introduction

One of the most powerful tools for the diagnostics of magnetic fields in the
Sun is the interpretation of polarimetric observations (e.g. the monograph by
Landi Degl’Innocenti & Landolfi 2004 and the recent review by Trujillo Bueno
2009). However, these diagnostics are mostly concerned with the fields at the
photospheric and chromospheric levels. The coronal magnetic field presents
more intrinsic difficulties to measure and interpret. This is especially true
for the case of the UV coronal lines. Only rather recently, Raouafi et al.
(2002) performed the first measurement and interpretation of the linear
polarization of a UV line (O vi $\lambda$1032 line) polarized under
anisotropic scattering by the underlying solar radiation field. In addition,
Manso Sainz & Trujillo Bueno (2009) proposed a polarizing mechanism showing
the adequate sensitivity of other coronal UV lines to the direction of the
magnetic field. These successful works suggest that new UV polarimeters with
high sensitivity associated with theoretical and numerical modeling obtained
with a high degree of realism are a fundamental step to be performed in order
to extract information on the coronal plasmas. In this context, the Hanle
effect on the L$\alpha$ polarization constitutes an excellent opportunity
which merits to be exploited.

The scattering polarization of the coronal L$\alpha$ line of neutral hydrogen,
which we are revisiting in this paper, has been computed by Bommier & Sahal-
Bréchot (1982) and by Trujillo Bueno et al. (2005). These authors, however,
neglected the effects of the integration over the line-of-sight (LOS) by
considering a local position of the scattering hydrogen atom. Since the corona
is optically thin, the LOS integration problem has to be solved. Fineschi et
al. (1992) treated the case of the L$\alpha$ line polarization and took into
account the LOS integration. However, Fineschi et al. considered the effect of
a deterministic magnetic field vector having a direction and strength
independent of the position of the scattering volume. They also treated the
case of a random magnetic field.

To improve upon these previous works, we take into account the variation of
the direction and the strength of the magnetic field for each scattering event
along the LOS. The calculation of the polarization generated by scattering
depends strongly on the level of anisotropy of the incident radiation, which
in turn depends strongly on the geometry of the scattering process and the
brightness variation of the chromospheric regions. In order to accurately
compute the degree of the anisotropy at each scattering position, we use
Carrington maps of the chromospheric incident radiation of the L$\alpha$ line
obtained by Auchère (2005). In addition, the coronal density of the scattering
atoms and the local temperature are included according to a quiet coronal
model (Cranmer et al. 1999). We perform a comparison of the L$\alpha$ linear
polarization in the zero-field reference case with the amplitude corresponding
to the polarization in the presence of a magnetic field. In our forward
modeling, we adopt a dipolar magnetic distribution as a first step and then we
add a magnetic field associated to an equatorial current sheet.

The paper is organized as follows. We describe the theoretical background and
formulate the problem in Sect. 2. Section 3 deals with the calculations of the
Hanle effect without integration over the LOS in order to compare with known
results. The generalization of these calculations to integrate over the LOS
and the discussion of the possibility of obtaining a coronal magnetic field
through polarization measurements are presented in Sect. 4. The technique that
could be used to measure the scattering polarization of the L$\alpha$ D2 line
is given in Sect. 5; in particular we show how the linear scattering
polarization could be measured using a L$\alpha$ disk imager and coronagraph
called LYOT (LYman Orbiting Telescope). In Sect. 6 we summarize our
conclusions.

##  2 Formulation of the problem

###  2.1 Hanle effect

The term Hanle effect represents the ways in which the scattering polarization
can be modified by weak magnetic fields. The well-known Zeeman effect and the
Hanle effect are complementary because they respond to magnetic fields in very
different parameter regimes. The Zeeman effect depends on the ratio between
the Zeeman splitting and the Doppler line width. The Hanle effect though
depends on the ratio between the Zeeman splitting and the inverse life time of
the atomic levels involved in the process of the formation of the polarized
line. For the permitted UV lines, the Zeeman effect is of limited interest for
the determination of the magnetic fields in the quiet corona. This is because
the ratio between the Zeeman splitting and the Doppler width is small due to
the weakness of the magnetic field and the high Doppler width in such hot
coronal plasmas. On the contrary, the measurement and physical interpretation
of the scattering polarization of the UV lines are a very efficient diagnostic
tool for determining the coronal magnetic field through its Hanle effect.

###  2.2 Atomic linear polarization

The possibility of the creation of a linear polarization by anisotropic
scattering can be only explained correctly in the framework of the quantum-
mechanical scattering theory. In fact, the intrinsic capacity of a line to be
polarized is intimately linked to subtle quantum behaviors pertaining to the
atomic levels involved in the transition. Let us denote by $m_{J}$ the
projection of the orbital angular momentum $J$ of the hydrogen atom; $m_{J}$
takes the values $-J$, $J+1$,…, $J$. The term “atomic linear polarization” in
a $J$-level consists in (e.g. Cohen-Tannoudji & Kastler 1966, Omont 1977,
Sahal-Bréchot 1977, Blum 1981):

–

an unbalance of the populations of the Zeeman sub-levels having different
absolute values $|m_{J}|$

–

a presence of interferences between these Zeeman sub-levels.

This means that by definition, only levels having $J>1/2$ can be linearly
polarized.

###  2.3 Linear scattering polarization in the L$\alpha$ line

The so-called scattering polarization is simply the observational
manifestation of the atomic polarization. The Hanle effect is nothing but a
perturbation of the atomic polarization by a magnetic field. The Hanle
signatures in the spectrum of the linear polarization are a variation of the
polarization degree and a rotation of the polarization plane. These Hanle
signatures can be used to retrieve information on coronal magnetic fields. The
two components D1 and D2 of the L$\alpha$ connect the hydrogen ground state
${}^{2}S_{J=1/2}$ to the electronic excited states ${}^{2}P_{J=1/2}$ and
${}^{2}P_{J=3/2}$, respectively. The upper level ${}^{2}P_{3/2}$ of the D2
line can be polarized due to the difference of the populations between the
Zeeman sub-levels with $|m_{J}|=1/2$ and $|m_{J}|=3/2$. However, the states
${}^{2}S_{1/2}$ and ${}^{2}P_{1/2}$ cannot be polarized since $|m_{J}|$ is
necessarily 1/2 implying that no difference of population inside these states
can be generated by anisotropic scattering. Consequently, the D1 line is not
linearly polarizable.

It is useful to keep in mind that in the description of the emitting hydrogen
atom, we neglect the contribution of the hyperfine structure (HFS). For
instance if the HFS is not neglected, the level $J=1/2$ of the ground state
${}^{2}S_{1/2}$ is split into hyperfine levels $F=0$ and $F=1$ due to coupling
with the nuclear spin of the hydrogen $I=1/2$. The hyperfine level $F=1$ can
be linearly polarized111In other words, population imbalances and quantum
interferences between the sub-levels having $|m_{F}|=1$ and $|m_{F}|=0$ can be
created due to the scattering of anisotropic light. The same is true for the
hyperfine levels of the upper states ${}^{2}P_{3/2}$ and ${}^{2}P_{1/2}$.,
which means that the D1 line can be polarized and the polarization of the D2
line can be affected. As previously suggested by Bommier & Sahal-Bréchot
(1982), we neglect the effect of the HFS in the process of formation of
L$\alpha$ line.

Figure 1:  Geometry of the scattering of chromospheric L$\alpha$ photons by
residual coronal neutral hydrogen. The anisotropy of the incident light is due
to geometrical effects but also to the inhomogeneities of the chromospheric
regions.

###  2.4 Expression of the Stokes parameters

The emission of the L$\alpha$ $\lambda$1216 line in the solar corona has been
discovered by Gabriel et al. (1971). They concluded that in most coronal
structures the process responsible for the formation of the L$\alpha$ line is
the photo-excitation by underlying radiation. The creation of population
imbalances and the quantum interferences in the ${}^{2}P_{3/2}$ and thus the
existence of the scattering polarization in the D2 L$\alpha$ line are caused
by the photo-excitation of coronal neutral hydrogen by anisotropic
chromospheric radiation (see Fig 1).

The components of the incident radiation field at a frequency $\nu_{0}$ are
usually denoted by $\bar{J}_{q}^{k}(\nu_{0})$ where $k$ is the tensorial order
and $q$ represents the coherences in the tensorial basis ($-k\leq q\leq k$);
the order $k$ can be equal to 0 (with $q=0$) or 2 (with $q=0$, $\pm$1,
$\pm$2). This radiation field with six components constitutes a generalization
of the unpolarized light field where only the quantity
$\bar{J}_{0}^{0}(\nu_{0})$ is considered. In fact, $\bar{J}_{0}^{0}(\nu_{0})$
is proportional to the intensity of the radiation.

If the incident radiation is no longer anisotropic, the components
$\bar{J}_{q}^{k=2}(\nu_{0})$ become zero, which means that no linear
polarization can be created as a result of scattering processes. Regardless of
the anisotropy of the incident radiation, the radiation component associated
with the circular polarization usually denoted by $\bar{J}_{q}^{k=1}$ is
negligible. This means that no odd order $k$ can be created inside the
scattering hydrogen atom. As a result, the Stokes $V$ of the scattered
radiation is zero.

We denote by $\zeta$ the angle between the direction of the incident light MP
and the local vertical through the scattering center OP. The incident
radiation comes from a chromospheric spherical cap limited by an angle
$\zeta_{\textrm{\scriptsize{max}}}$ corresponding to the tangent to the solar
limb (see Fig. 1). $\chi$ is the azimuth angle around the normal with respect
to an arbitrary reference. Note that $0\leq$ $\zeta$ $\leq$
$\zeta_{\textrm{\scriptsize{max}}}$ and 0 $\leq$ $\chi$ $\leq$ 2 $\pi$. When
the distance from the solar surface increases, the anisotropy of the light
becomes larger and the polarization degree increases. The maximum of
polarization is reached when the radiation is purely directive, i.e. the
spherical cap is seen by the scattering hydrogen atom as a point. It is useful
to notice that if the chromosphere is assumed to be uniform the radiation has
a cylindrical symmetry around its preferred direction, implying that the
coherence components with $q\neq 0$ are zero. In fact, $\bar{J}_{q=\pm
1}^{k=2}(\nu_{0})$ and $\bar{J}_{q=\pm 2}^{k=2}(\nu_{0})$ components quantify
the breaking of the cylindrical symmetry around the axis of quantification
which is here the local vertical.

In the framework of the two level approximation, where only the upper level is
polarized, the statistical equilibrium equations are solved analytically. The
upper level density matrix elements are simply proportional to the incident
radiation elements $\bar{J}_{q}^{k}$. The emissivity vector is then expressed
as a function of the incident radiation field. Consequently, we do not
explicitly calculate the density matrix elements, but instead we determine the
incident radiation tensor at each scattering position along the line of sight.
For an unmagnetized atmosphere, in an arbitrary reference, the emissivity
vector can be written as (e.g. Landi Degl’Innocenti & Landolfi 2004):

$\displaystyle\epsilon_{j}(\Omega)=n_{\textrm{\scriptsize{H}}}\frac{h\nu_{0}B_{J_{l}J_{u}}}{4\pi}\sum_{k,q}W_{k}(J_{l},J_{u})\mathcal{T}_{q}^{k}(j,\Omega)(-1)^{q}\bar{J}_{-q}^{k}(\nu_{0})$
(1)

where $\Omega$ is the solid angle giving the direction of the LOS,
$n_{\textrm{\scriptsize{H}}}$ is the local number density of scattering
hydrogen atoms, $h$ is the Planck constant, and $B_{J_{l},J_{u}}$ is the
Einstein coefficient for absorption. We recall that
$\mathcal{T}_{q}^{k}(j,\Omega)$ is the spherical tensor for polarimetry which
contains the angular distribution of the emitted radiation, and $j$ is the
index of the Stokes parameter ($j$ = 0, 1, 2, and 3 for the Stokes $I,Q,U,$
and $V$, respectively).

In order to determine the magnetic field one has to include its Hanle effect
on the polarization of the L$\alpha$ light, then, for a given magnetic field
vector ${\bf B}$, $\epsilon_{j}(\Omega)$ becomes (e.g. Landi Degl’Innocenti &
Landolfi 2004):

$\displaystyle\epsilon_{j}(\Omega,{\bf
B})=n_{\textrm{\scriptsize{H}}}\frac{h\nu_{0}B_{J_{l}J_{u}}}{4\pi}\times$ (2)
$\sum_{k,q}W_{k}(J_{l},J_{u})\mathcal{T}_{q}^{k}(j,\Omega)(-1)^{q}\bar{J}_{-q}^{k}(\nu_{0})\frac{1}{1+\textrm{i}qH_{u}}$

This expression of $\epsilon_{j}(\Omega,{\bf B})$ is correct only in a
reference system having the quantization $z$-axis in the magnetic field
direction. $H_{u}$ is the so-called reduced magnetic field strength,
associated to the level ${}^{2}P_{3/2}$, given by:

$\displaystyle H_{u}=\frac{0.879\;g_{u}\;\textrm{B}}{A_{J_{u}J_{l}}}$ (3)

where the Einstein coefficient for spontaneous emission $A_{J_{u}J_{l}}$ is
given in [$10^{7}$ s-1], $g_{u}=4/3$ is the Landé factor of the level
${}^{2}P_{3/2}$ and the magnetic field strength B is given in Gauss. $H_{u}=1$
corresponds to the magnetic field strength $B=53$ Gauss around which one may
expect a noticeable change in the scattering polarization of L$\alpha$ with
respect to the unmagnetized reference case. The quantity $W_{k}(J_{l},J_{u})$
was first introduced by Landi Degl’Innocenti (1984) and depends only on the
quantum numbers of the lower and upper levels ($J_{l}$ and $J_{u}$) involved
in the transition. For $k$=2, $W_{2}(J_{l},J_{u})$ can be seen as the
efficiency of creation of the linear polarization in the scattering processes.
That is why the $W_{2}(J_{l}=1/2,J_{u}=1/2)$ =0 for the D1 line, which is not
polarizable, but $W_{2}(J_{l}=1/2,J_{u}=3/2)$=1/2 for the polarizable D2 line.

##  3 Hanle effect without integration over the LOS

We developed a numerical code allowing for the calculation of the theoretical
polarization taking into account the effects of the LOS. In order to validate
the code, we considered typical cases of a horizontal magnetic field having
different azimuth angles $\theta$ (angles between the magnetic field vector
and the LOS). LOS integrations are avoided in order to be able to compare our
results with well known Hanle effect results. We retrieve the Hanle behaviors
typically encountered in the literature, for instance:

–

when the magnetic field is zero or very small or oriented along the symmetry
axis of the radiation field, the polarization is not affected

–

when the field increases until reaching the critical value corresponding to
$H_{u}=1$, the polarization decreases rapidly. Moreover, for a very large
$H_{u}$ (i.e. very large magnetic field strength) we obtain an asymptotical
curve of polarization $p[B\to\infty]$ which depends only on the value of
$\theta$ but not on the magnetic field strength. The asymptotic value of
$p[B\to\infty]$ divided by $p[B=0]$ equals 1/5 when the distribution of the
magnetic field is isotropic222The case of isotropic field distribution is
encountered in the photosphere of the Sun (second solar spectrum) where the
magnetic geometries are unresolved within the spatiotemporal resolution of the
current observational capabilities. and 1/4 when the field has a cylindrical
symmetry (i.e. horizontal magnetic field with random azimuth)

–

no rotation of the plane of the polarization in the case of a highly symmetric
distribution (e.g. isotropic or cylindrical) because the contributions of
opposite magnetic polarities tend to cancel out

–

we find that a meridian magnetic vector (i.e. horizontal with
$\theta=\frac{\pi}{2}$) presents a depolarizing effect without rotation of the
polarization direction.

##  4 Hanle effect integrated over the LOS

The corona being an optically thin medium for the L$\alpha$ line, it is then
necessary to consider the effects of the integration over the LOS. We adopt
the analytical magnetic field model proposed by Fong et al. (2002) and Low et
al. (2003). It is a sum of two terms: a purely dipolar term and a term
corresponding to the magnetic field of a current sheet structure. The model is
axisymmetric and the prominence is treated as a cold plasma sheet forming a
flat ring around the Sun. We take into consideration that the current is in
the equator and that it represents a prominence sheet extending from
$r=R_{\sun}$ to $r=\sqrt{4/5}\;R_{\sun}$. In the analytical expression of the
magnetic field the contribution of the current sheet, relative to the dipolar
background field is controlled by a constant ratio $\gamma$ (see Eq. 12 of Low
et al. 2003).

Figure 2:  A purely dipolar magnetic field structure presented in the plane
$(x,z)$ in units of the solar radius $R_{\sun}$. We use a system of orthogonal
Cartesian coordinates $(x,y,z)$ with the origin at the Sun center and the
$z$-axis pointing toward the north solar pole.

###  4.1 Purely dipolar magnetic field: $\gamma$=0

As a first step, we avoid the effect of the term associated to the current
sheet by taking $\gamma$=0. Figure 2 represents the dipolar term of the
magnetic field. This configuration represents a typical coronal magnetic field
of 15 to 20 Gauss close to the base of the corona. The Hanle effect of the
dipolar magnetic field depends on the angle $\phi$ between the axis of
symmetry of the incident light333Rigorously speaking, this is only the axis of
symmetry of the spherical cap where the chromospheric radiation is uniform.
The fact that the incident radiation is inhomogeneous implies that this
symmetry around the preferred axis of radiation is broken.  and the axis of
symmetry of the magnetic structure. It also depends on the height above the
solar surface $h$ mainly because the magnetic field strength decreases and the
anisotropy of the incident light increases. The parameters $\phi$ and $h$ are
represented in Fig. 3.

Figure 3:  The polarization at each position P of the scattering event depends
on the height over the limb $h$ and the angle $\phi$ between the preferred
axis of the radiation and the symmetry axis of the magnetic structure. The LOS
is perpendicular to the plane of the figure.

In theory, the expression of the emissivity vector is valid regardless of the
location of the scattering atom. However, the integration over the LOS must
take into account the inhomogeneities of the solar conditions like the
variation of the hydrogen density, the temperature, the magnetic field and the
variation of the incident radiation field. The density of neutral hydrogen and
the temperature are assumed to be a function of the radial distance $r$ and
the latitude (see Cranmer et al. 1999 for details). In order to model the
inhomogeneities of the chromospheric intensity, we use the Carrington maps of
the L$\alpha$ chromospheric line built by Auchère (2005). In the optically
thin limit, the integrated Stokes parameters of the scattered radiation reduce
to a volume integration over the LOS:

$\displaystyle\mathcal{E}_{j}(\Omega)$ $\displaystyle=$
$\displaystyle\int_{\textrm{\scriptsize{LOS}}}\epsilon_{j}(\Omega)dl$ (4)

then the polarization degree is

$\displaystyle p$ $\displaystyle=$
$\displaystyle\frac{\sqrt{\mathcal{E}_{1}^{2}+\mathcal{E}_{2}^{2}}}{\mathcal{E}_{0}}$
(5)

and the rotation of the direction of the polarization $\alpha_{0}$ is given
by:

$\displaystyle tg(2\alpha_{0})$ $\displaystyle=$
$\displaystyle\frac{\mathcal{E}_{2}}{\mathcal{E}_{1}}$ (6)

Figure 4 shows the variation of the linear polarization with the inclination
$\phi$ for two different magnetic structures corresponding to two heights
above the solar surface. Note that the ratio $p[B]/p[B=0]$ obtained for the
height $h$=0.3 R☉ is smaller than the one corresponding to $h$=0.5 R☉ since
the magnetic field decreases with $h$. Furthermore, as shown in Fig. 5, a
notable Hanle rotation of about $10^{o}$ is obtained for $h$=0.3 R☉ and for
$h$=0.5 R☉. Both Hanle signatures on the L$\alpha$ line, i.e. depolarization
and rotation (see Figs. 4 and 5), are clearly sizable for $\phi>$ $40^{o}$.
This important result suggests that in order to measure a dipolar magnetic
field by its Hanle effect one should observe regions rather far away from the
pole.

Figure 4:  Linear polarization degree obtained for a dipolar magnetic field
divided by zero-field polarization versus the angle $\phi$. Full lines
represent the polarization at $h$=0.5 R☉ and dashed lines represent the
polarization at $h$=0.3 R☉.  Figure 5:  Rotation angle obtained after
integration over the LOS versus the inclination $\phi$ at $h$=0.3 R☉ and
$h$=0.5 R☉. Full lines represent the rotation at $h$=0.5 R☉ and dashed lines
represent the rotation at $h$=0.3 R☉.  Figure 6:  Perturbation of the lines of
the dipolar magnetic field due to an equatorial current sheet. The
calculations of the polarization given in Fig. 7 are obtained for
$\phi=$$80^{o}$.  Figure 7:  Linear polarization degree versus the height from
the solar surface $h$. We put together the results obtained in the zero-field
case and these obtained for (1) a purely dipolar magnetic field (2) the sum of
a dipolar field and a non-dipolar field associated to a current sheet with
$\gamma=$ 0.25.

###  4.2 Perturbed dipolar field: $\gamma\neq$ 0

To the dipolar part of the magnetic field we now add the contribution
resulting from an equatorial current sheet. We adopt a ratio $\gamma$=0.25
between the current sheet and the dipolar background field. In order to
highlight the Hanle effect of the equatorial current sheet, we calculate the
degree of polarization in a position located at $\phi$=$80^{o}$, and we vary
the height above the limb $h$ (see Fig. 6). Figure 7 shows the difference
between the linear polarization in the zero-magnetic field case and the one in
the presence of the magnetic field, $\Delta p$=$|p_{B=0}-p_{B\neq 0}|$. A
polarimetric sensitivity smaller than $\Delta p$ is needed in order to apply
the Hanle effect as a technique of magnetic field investigations. Our results
show that $\Delta p$ $\sim$ 5 %, i.e. well within the typical measurement
sensitivities of a new generation of instruments such as LYOT (see Sect. 5).
We point out that by using the UV SUMER spectrometer aboard SoHO, Raouafi et
al. (1999) measured the linear polarization of the O vi $\lambda$1032 line
with a polarimetric precision equal to 1.7 %. We note in passing that such an
accuracy is reached although SUMER was not initially designed to measure the
polarization.

##  5  Measurement of the linear polarization degree and its direction

###  5.1  Principle of the measurement

Raouafi et al. (1999) used the rotation of the SUMER spectrometer to measure
the linear polarization of the D2 component of the O vi $\lambda$1032 line.
They extracted the polarization of the D2 line from a ratio of the intensities
of the non-polarizable D1 line and of the D2 line (see the Fig. 3 of Raouafi
et al. 1999). This technique was possible because the wavelengths of the two
components D1 and D2 are sufficiently different to be resolved (1031.93 Å for
the D2 line and 1037.62 Å for the D1). However, the wavelengths of the D1 and
D2 lines of the L$\alpha$ line cannot be resolved since they are very close:
in the vacuum $\lambda$(D1)= 1215.668 Å and $\lambda$(D2)= 1215.674 Å. As a
result, the technique presented by Fig. 3 of Raouafi et al. (1999) cannot be
applied to measure the linear polarization of the D2 line of L$\alpha$.

Using the so-called Poincaré representation444A suitable graphical
representation of polarized light conceived by Henri Poincaré in 1892., one
can demonstrate that the intensity observed when the instrument is placed at
an arbitrary position referred by an angle $\beta$ around the LOS, is

$\displaystyle I(\beta)=\frac{1}{2}(Q\cos 2\beta+U\sin 2\beta+I)$ (7)

In this expression the Stokes $V$ is assumed to be zero. The quantity $I$
denotes the unpolarized part of the intensity of the D1 and D2 L$\alpha$
lines. $I(\beta)$ represents the “real” (polarized and unpolarized) observed
intensity of the two resonance lines. In addition, taking into account that
$\alpha_{0}$ corresponds to the direction of linear polarization (i.e.
privileged direction of the electric field), Eq. (7) becomes

$\displaystyle I(\beta)=\frac{1}{2}(\sqrt{Q^{2}+U^{2}}\cos
2(\beta-\alpha_{0})+I)$ (8)

Note that $\cos 2\alpha_{0}=\frac{Q}{\sqrt{Q^{2}+U^{2}}}$ and $\sin
2\alpha_{0}=\frac{U}{\sqrt{Q^{2}+U^{2}}}$.

On the other hand, generally speaking, the linear polarization is defined as

$\displaystyle p=\frac{I_{max}-I_{min}}{I_{max}+I_{min}}$ (9)

where $I_{max}$ and $I_{min}$ are the maximum and minimum intensities. Using
Eqs. (8) and (9), one finds that

$\displaystyle p=\frac{\sqrt{Q^{2}+U^{2}}}{I}$ (10)

and

$\displaystyle\frac{I(\beta)}{I}=\frac{1}{2}(p\cos 2(\beta-\alpha_{0})+1)$
(11)

In Eq. (11) we have three unknowns: $I$, $p$, and $\alpha_{0}$. Then,
theoretically, the linear polarization state is fully obtained through only
three measurements of the $I(\beta)$ which corresponds to three rotations of
the polarizer-spectrometer. One takes for example $\beta=0,\frac{\pi}{4},$ and
$\frac{\pi}{2}$, therefore:

$\displaystyle\tan 2\alpha_{0}$ $\displaystyle=$
$\displaystyle\frac{2I(\frac{\pi}{4})-I}{I(0)-I(\frac{\pi}{2})}$ (12)
$\displaystyle p$ $\displaystyle=$
$\displaystyle\frac{I(0)-I(\frac{\pi}{2})}{\cos 2\alpha_{0}\times
I}=\frac{2I(\frac{\pi}{4})-I}{\sin 2\alpha_{0}\times I}$

where the intensity of the unpolarized light is given by:

$\displaystyle I$ $\displaystyle=$ $\displaystyle I(0)+I(\frac{\pi}{2})$ (13)

Obviously, more than three measurements of $I(\beta)$ are welcome in order to
increase the accuracy.

###  5.2  The LYOT project

The LYOT project is a L$\alpha$ coronagraph combined with a L$\alpha$ disk
imager (see Vial et al. 2002, Millard et al. 2006, and Vial et al. 2008). In
addition, it is planned to implement a simple polarizer system. The polarizing
measurements will be performed by rotating the polarizer or the whole
instrument to obtain the intensity of the L$\alpha$ light at different $\beta$
angles (previous section). The choice of the L$\alpha$ line is well justified
by its sensitivity to the coronal magnetic field (as demonstrated in this
paper) and by the fact that in the corona the L$\alpha$ emission is very
intense. In fact, a high signal to noise ratio is needed since in the very low
corona the anisotropy of the light is small, which in turn means that the
polarization degree is small (smaller than 5 %, see Fig. 7). We note that no
coronagraph observing as low as 1.15 R☉ is envisaged beyond 2012 except for
LYOT images which should be obtained with an excellent signal to noise ratio.

##  6 Conclusion

Measurement and interpretation of the scattering polarization of UV coronal
lines provide a largely unexplored diagnostic of the coronal magnetic field.
The greatest difficulty facing the UV coronal spectropolarimetry is that the
polarization measurements integrate radiation along the LOS over structures
with different properties but also that the observations of these lines are
impossible from ground-based telescopes; they can only be observed with the
help of high-sensitivity instruments flown on space missions.

We have performed a forward modeling of the coronal Hanle effect on the
polarization of the L$\alpha$ line generated by anisotropic scattering of
chromospheric light. The main feature of this modeling consists in integrating
the effect of the LOS. We show that the information about the coronal magnetic
field is not lost through LOS integration. To confirm these results, we plan
to work with different families of maps of magnetic fields and to add small
scale magnetic perturbations. For instance, one can think of a set of active
loops whose field determination could be compared with field extrapolations.
One should however keep in mind that (1) a realistic thermodynamic model is
required in order to integrate along the LOS and that (2) our modeling is
limited to the case of optically thin plasmas in the L$\alpha$ line.

Finally, we notice that it is suitable to combine measurements in L$\alpha$
with measurements in polarized lines like the Fe xiv $\lambda$5303 which have
a different sensitivity to the magnetic field. An analysis of combined
measurements should give more information data to constrain the magnetic field
topology and strength (an example of the Hanle effect in a multi-line approach
is given in Landi Degl’Innocenti 1982). It is also of interest to remark that
the ground level ${}^{2}S_{J}$ of L$\alpha$ is non polarizable by radiation
anisotropy, but that this is no longer true in the presence of hyperfine
structure and if the depolarizing effect of the isotropic collisions is
negligible. Because the sensitivity to the Hanle effect depends on the level
life-time, the hyperfine polarization of such a long-lived level is much more
vulnerable to very weak magnetic fields than the short-lived upper levels
${}^{2}P$. Consequently, one could distinguish a very small perturbation of
the magnetic field (smaller than 1 Gauss) which corresponds for instance to a
current sheet with a very small $\gamma$ and a background field of the order
of 10 Gauss or larger. In particular, this could be the key to distinguish
potential magnetic field structures from non potential ones.

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