# On the Jacobson Radical of an(m,n)-Semiring

Authors: Yongwen Zhu
Journal: Algebra
 (2013)

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## Abstract

The notion ofn-ary semimodules is introduced so that the Jacobson radical of an (m,n)-semiring is studied and some well-known results concerning the Jacobson radical of a ring (a semiring or a ternary semiring) are generalized to an (m,n)-semiring.


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## Body

## 1. Introduction

The concept of semigroups [1] was generalized to that of ternary semigroups [2], that of n-ary semigroups [3–6], and even to that of (n,m)-semigroups [7]. Similarly, it was natural to generalize the notion of rings to that of ternary semirings, that of n-ary semirings, and even that of (m,n)-semirings.Indeed, there were some research articles on semirings, (see, for example, [8–14]), specially on the radical of a semiring; see [15–18]. Semigroups over semirings were studied in [19] and semimodules over semirings were studied in [14]. The notion of semirings can be generalized to ternary semirings [20] and Γ-semirings [21], even to (m,n)-semirings [22–24]. The radicals of ternary semirings and of Γ-semirings were studied in [20, 21], respectively. The concept of (m,n)-semirings was introduced and accordingly some simple properties were discussed in [22–24], where the concept of radicals was not mentioned.The notion of the Jacobson radicals was first introduced by Jacobson in the ring theory in 1945. Jacobson [25] defined the radical of R, which we call the Jacobson radical, to be the join of all quasi-regular right ideals and verified that the radical is a two-sided ideal and can also be defined to be the join of the left quasi-regular ideals.The concept of the Jacobson radical of a semiring has been introduced internally by Bourne [15], where it was proved that the right and left Jacobson radicals coincide; thus one could say the Jacobson radical briefly. These and some other results were generalizations of well-known results of Jacobson [25].In 1958, by associating a suitable ring with the semiring, Bourne and Zassenhaus defined the semiradical of the semiring [16]. In [18] it was proved that the concepts of the Jacobson radical and the semiradical coincide.Iizuka [17] considered the Jacobson radical of a semiring from the point of view of the representation theory [15] without reducing it to the ring theory. The external notion of the radical was proved to be related to internal one; at the same time, it was shown that the radical defined in [17] coincides with the Jacobson radical and with the semiradical of the semiring.In the present paper, we investigate(n,m)-semirings by means of n-ary semimodules so that we can define externally the Jacobson radical of an (n,m)-semiring, and then we establish the radical properties of the Jacobson radical of an (n,m)-semiring. Some necessary notions such as irreducible n-ary semimodules over an (n,m)-semiring are adequately defined. All results in this paper generalize the corresponding ones concerning the radical of a ring [25], of a semiring [15–18], or of a ternary semiring [20].


## 2. Preliminaries

We used following convention as followed by [4]: The sequence xi,xi+1,…,xm is denoted by xim. Thus the following expression
(1)f(x1,…,xi,yi+1,…,yj,zj+1,…,zm)
is represented as
(2)f(x1i,yi+1j,zj+1m).
In the case when yi+1=⋯=yj=y, then (2) is expressed as
(3)f(x1i,y(j-i),zj+1m).
If x1=⋯=xi=yi+1=⋯=yj=zj+1=⋯=zm=f(a1m), then (2) can be written as f(f(a1m)(m)).Recall that ann-ary semigroup (S,f) is defined as a nonempty set S with an n-ary associative operation f:Sn=S×⋯×S︸n→S; that is,
(4)f(x1i-1,f(xin+i-1),xn+i2n-1)=f(x1j-1,f(xjn+j-1),xn+j2n-1)
for all x12n-1∈S and all 1≤i<j≤n. Whence we may denote f(f(x1n),xn+12n-1) by f(x12n-1) briefly. Generally, we have the notation f(x1t(m-1)+1) for each positive t and all xj∈S. Thus for positive integer k, f(x1k) is well defined if and only if k≡1mod(n-1); see [7, Lemma 1.1]. An n-ary semigroup (S,f) is called cancellative if
(5)f(x1i-1,a,xi+1n)=f(x1i-1,b,xi+1n)⟹a=b
for all a,b,xj∈S.The next definition is a generalization of the concept of ternary semirings in [20] and similar to the notion of the (m,n)-semirings in [24].Definition 1.
A nonempty setR together with an m-ary operation f, called addition, and an n-ary operation g, called multiplication, is said to be an (m,n)-semiring if the following conditions are satisfied. (1)
(R,f) is an m-ary semigroup and (R,g) is an n-ary semigroup.(2)
g is distributive with respect to operation f; that is, for every a1i-1, b1m, ai+1n∈R,
(6)g(a1i-1,f(b1m),ai+1n)=f(g(a1i-1,b1,ai+1n),…,g(a1i-1,bm,ai+1n)).(3)
(R,f) is commutative; that is, for every permutation p of {1,2,…,m} and all x1m∈R,
(7)f(x1,x2,…,xm)=f(xp(1),xp(2),…,xp(m)).(4)
There is an element0, called the zero of (R,f,g), satisfying the following two properties:
(4A)
0 is an f-identity; that is, for every x∈Rf(0(m-1),x)=x;(4B)
0 is a g-zero; that is, for all x1m∈R, g(x1n)=0 whenever there exists i such that xi=0.It is clear that the zero of an(m,n)-semiring R is necessarily unique.Definition 2.
An(m,n)-semiring (R,f,g) is called additively cancellative if the m-ary semigroup (R,f) is cancellative and multiplicatively cancellative if the n-ary semigroup (R,g) is cancellative.Recall that for ann-ary semigroup (S,f), a nonempty subset T of S is called a subsemigroup of (S,f) if f(a1n)∈T whenever all a1n∈T. For i≤n, we call T an i-ideal of S if f(a1n)∈T whenever ai∈T. T is called an ideal of S if and only if it is an i-ideal of S. See, for example, [7, Definition 1.6].Definition 3.
A nonempty subsetT of an (m,n)-semiring (R,f,g) is called an n-ary subsemiring of (R,f,g) if T is a subsemigroup of (R,f) as well as a subsemigroup of (R,g) and an (i-)ideal of (R,f,g) if T is a subsemigroup of (R,f) as well as an (i-)ideal of (R,g) (where i≤n). An 1-ideal is also called a right ideal and an n-ideal is also called a left ideal. An ideal T of (R,f,g) is called a k-ideal if f(x,y2m)∈T;x∈R and y2m∈T imply that x∈T. An ideal T of (R,f,g) is called an h-ideal if f(x,y2m,z2m)=f(y1,z2m);x,z2m∈R and y1m∈T imply that x∈T.LetA be an ideal of (R,f,g). Then the k-closure of A, denoted by A¯, is defined by A¯={x∈R:f(x,a2m)=a1forsomea1m∈A}. Similarly, the h-closure of A, denoted by A^, is defined by A^={x∈R:f(x,a2m,y2m)=f(a1,y2m) for some a1m∈A and some y2m∈R}. One can show that A¯ is a k-ideal and A^ is an h-ideal. Furthermore, it is shown that an ideal A of (R,f,g) is a k-ideal if and only if A¯=A and that A is an h-ideal if and only if A^=A.Definition 4.
An equivalence relationρ on an (m,n)-semiring (R,f,g) is said to be a congruence relation or simply a congruence of (R,f,g) if the following conditions are satisfied:(1)
a1mρb1m⇒f(a1m)ρf(b1m) for all a1m,b1m∈R,(2)
a1nρb1n⇒g(a1n)ρg(b1n) for all a1n,b1n∈R.LetI be a proper ideal of an (m,n)-semiring (R,f,g). Then the congruence on (R,f,g), denoted by kI, and defined by setting xkIy if and only if f(x,a2m)=f(y,b2m) for some a2m,b2m∈I, is called the Bourne congruence on (R,f,g) defined by the ideal I. We denote the Bourne congruence class of an element x by x/I and denote the set of all such congruence classes of (R,f,g) by R/I. If the Bourne congruence kI is proper, that is, 0/I≠R, then we can define two operations, m-ary addition and n-ary multiplication on R/I by f¯(a1m/I)=f(a1m)/I and g¯(b1n/I)=g(b1n)/I for all a1m,b1n∈R. Then (R/I,f¯,g¯) is an (m,n)-semiring and is called the Bourne factor(m,n)-semiring.Similarly, the congruence on(R,f,g), denoted by hI, and defined by setting xhIy if and only if f(x,a2m,y2m)=f(y,b2m,y2m) for some a2m,b2m∈I and some y2m∈R, is called the Iizuka congruence on (R,f,g) defined by the ideal I. We denote the Iizuka congruence class of an element x by x[/]I and denote the set of all such congruence classes of (R,f,g) by R[/]I. If the Iizuka congruence hI is proper, that is, 0[/]I≠R, then we can define two operations, m-ary addition and n-ary multiplication on R[/]I by f^(a1m[/]I)=f(a1m) and g^(b1n[/]I)=g(b1n) for all a1m,b1n∈R. Then (R[/]I,f^,g^) is an (m,n)-semiring and we call it the Iizuka factor(m,n)-semiring.The next definition is a generalization of [20, Definition 2.13].Definition 5.
A commutativem-ary semigroup (M,f0) with an identity 0M (operation f0 to be called addition) is called a rightn-ary semimodule over an (m,n)-semiring (R,f,g) or simply an n-ary R-semimodule if there exists a mapping M×R×⋯×R︸n-1→M (images to be denoted by ar2⋯rn or briefly by ar2n for all a∈M and r2n∈R) satisfying the following conditions: (1)
f0(a1m)r2⋯rn=f0(a1mr2⋯rn) for all a1m∈M and all r2n∈R;(2)
ar2⋯(f(s1m))⋯rn=f0(ar2⋯s1m⋯rn) for all a∈M and all r2n,s1m∈R;(3)
(ar2⋯rn)rn+1⋯r2n-1=ar2⋯g(rk+1k+n)⋯r2n-1 for all a∈M, k∈{1,…,n-1}, and rk+1k+n∈R;(4)
0Mr2⋯rn=0M for all r2n∈R;(5)
ar2⋯rn=0M whenever a∈M, r2n∈R, and ri=0 for some i.Definition 6.
A nonempty subsetN of a right n-ary semimodule (M,f0) over an (m,n)-semiring (R,f,g) is called an n-ary subsemimodule of M if (i) f0(a1m)∈N and (ii) a1x2n∈N for all a1m∈N and x2n∈R.
Ann-ary subsemimodule N of M is called an n-ary k-subsemimodule if f0(a,b2m)∈N;a∈M and b2m∈N imply that a∈N. An n-ary subsemimodule N of M is called an n-ary h-subsemimodule if f(a,b2m,c2m)=f(b1,c2m);a,c2m∈R and b1m∈N imply that a∈N.For example, an(m,n)-semiring (R,f,g) can be regarded as a right n-ary R-semimodule naturally. Then if I is a k-ideal (an h-ideal) of the (m,n)-semiring (R,f,g), then I is also an n-ary k-(h-)subsemimodule of this right n-ary R-semimodule R.Definition 7.
A rightn-ary R-semimodule (M,f0) is said to be cancellative if (M,f0) is a cancellative m-ary semigroup.Definition 8.
An equivalence relationρ on right n-ary R-semimodule (M,f0) is said to be a congruence relation or simply a congruence of (M,f0) if the following conditions are satisfied:(1)
a1mρb1m⇒f0(a1m)ρf0(b1m) for all a1m,b1m∈M,(2)
aρb⇒ar2nρbr2n for all a,b∈M and all r2n∈R.We say that a congruenceρ of (M,f0) admits the cancellation law (of addition) if(3)
f0(x,a2m)ρf0(y,b2m) and a2mρb2m imply xρy.LetN be an n-ary subsemimodule of an n-ary right semimodule (M,f0) over an (m,n)-semiring R. Then the congruence on (M,f0), denoted by kN, and defined by setting
(8)xkNyifff0(x,a2m)=f0(y,b2m)forsomea2m,b2m∈N,
is called the Bourne congruence on M defined by the n-ary subsemimodule N. We denote the Bourne congruence class of an element x by x/N and denote the set of all such congruence classes of M by M/N. Define two operations, m-ary addition and n-ary scalar multiplication on M/N, by f0¯(a1m/N)=f0(a1m)/N and (a1/N)r2n=(a1r2n)/N for all a1m∈M and all r2n∈R. With these two operations, M/N is an n-ary right semimodule over R and we call it the Bourne factorn-ary semimodule.Similarly, we can define theIizuka congruence  hN and the Iizuka factorn-ary semimodule  M[/]N. It is easy to show that M[/]N is cancellative.In what follows, we always assume that then-ary right semimodule is cancellative.


## 3. Primitive(m,n)-Semirings

Definition 9.
Let(R,f,g) be an (m,n)-semiring with zero 0. Then the zeroid of R, denoted by Z(R), is defined as
(9)Z(R)={x∈R:f(x,y2m)=f(0,y2m)forsomey2m∈R}.Clearly, the zero element0 of R belongs to Z(R). Furthermore, we have the following.Lemma 10.
The zeroidZ(R) of an (m,n)-semiring (R,f,g) is the smallest h-ideal of (R,f,g).Proof.
It is easily verified thatZ(R) is an ideal of R. To show Z(R) is an h-ideal of R, we suppose f(x,y2m,z2m)=f(y1,z2m), where x,z2m∈R and y1m∈Z(R). For each i∈{1,…,m} there exist u(i)2m∈R such that f(yi,u(i)2m)=f(0,u(i)2m), so we have
(10)f(f(x,y2m,z2m),u(1)2m,…,u(m)2m)=f(f(y1,z2m),u(1)2m,…,u(m)2m),
that is,
(11)f(x,u(1)2m,f(y2,u(2)2m),…,f(ym,u(m)2m),z2m)=f(f(y1,u(1)2m),u(2)2m,…,u(m)2m,z2m).
It follows that(12)f(x,u(1)2m,f(0,u(2)2m),…,f(0,u(m)2m),z2m)=f(f(0,u(1)2m),u(2)2m,…,u(m)2m,z2m).
Hence we obtain(13)f(x,u(1)2m,u(2)2m,…,u(m)2m,z2m)=f(0,u(1)2m,u(2)2m,…,u(m)2m,z2m),
which shows that x∈Z(R), so that Z(R) is an h-ideal of R.
At last, suppose thatI is an arbitrary h-ideal of R. We aim to show Z(R)⊆I. For this, let x∈Z(R). Then there exist y2m∈R such that f(x,y2m)=f(0,y2m), so f(x,0m-1,y2m)=f(0,y2m). It follows that x∈I since I is an h-ideal and 0∈I. Thus Z(R)⊆I.Definition 11.
LetM be a right n-ary R-semimodule. The annihilator of M in R, denoted by (0:M) or AR(M), is defined as the subset
(14){r∈R:ar2m=0Mwhenevera∈M,r2m∈Randri=rforsomei}.Lemma 12.
A
R
(
M
) is an h-ideal of R.Proof.
It is obvious thatAR(M) is an ideal of R. To show that it is an h-ideal, suppose f(x,y2m,z2m)=f(y1,z2m), where x,z2m∈R and y1m∈AR(M). Then for all ri∈R,
(15)ar2⋯f(x,y2m,z2m)⋯rn=ar2⋯f(y1,z2m)⋯rn,
that is,
(16)f0(ar2⋯x⋯rn,ar2⋯y2m⋯rn,ar2⋯z2m⋯rn)=f0(ar2⋯y1⋯rn,ar2⋯z2m⋯rn),
which deduces that
(17)f0(ar2⋯x⋯rn,0m-1,ar2⋯z2m⋯rn)=f0(0,ar2⋯z2m⋯rn)
since ar2⋯yj⋯rn=0 for each j∈{1,…,m}. Thus we have
(18)f0(ar2⋯x⋯rn,ar2⋯z2m⋯rn)=f0(0,ar2⋯z2m⋯rn).
By cancellation law ofM, ar2⋯x⋯rn=0. Hence x∈AR(M), as required.Definition 13.
A rightn-ary R-semimodule M is said to be faithful if Z(R)=AR(M).One of difficulties when studying the radical of an(m,n)-semiring R is how to give an appropriate definition of the irreducibility of a right n-ary R-semimodule. The next definition is a generalization of [20, Definition 3.9].Definition 14.
A rightn-ary R-semimodule M is said to be irreducible if for every arbitrary fixed pair u2m,v2m∈M with uj≠vj for some j and for any x∈M, there exist a(i)2m,b(i)2m∈R with i=2,…,m such that
(19)f0(x,u2a(2)2n,…,uma(m)2n,v2b(2)2n,…,vmb(m)2n)=f0(0M,u2b(2)2n,…,umb(m)2n,v2a(2)2n,…,vma(m)2n).Remark 15.
SinceM is cancellative, it is easily seen that a right n-ary R-semimodule M is irreducible if and only if for every arbitrary fixed pair u2m,v2m∈M with uj≠vj for all j and for any x∈M, there exist a(i)2m,b(i)2m∈R with i=2,…,m such that equality (2) holds.Lemma 16.
LetI be an h-ideal of an (m,n)-semiring R. If M is an irreducible right n-ary R/I-semimodule, then M is an irreducible right n-ary R-semimodule.Proof.
LetM be an irreducible right n-ary R-semimodule. Then we can define an n-ary action on M by ar2⋯rn=a(r2/I)⋯(rn/I) for all a∈M and for all r2n∈R, and this makes M into an irreducible right n-ary R-semimodule.The converse of Lemma16 is not necessarily true. But in particular we have the following lemma.Lemma 17.
IfM is a right n-ary R-semimodule then M is a right n-ary R/AR(M)-semimodule, where R/AR(M) is the Bourne factor semiring. Moreover, if M is an irreducible right n-ary R-semimodule, then M is also an irreducible right n-ary R/AR(M)-semimodule.Proof.
SupposeM is a right n-ary R-semimodule. We define an n-ary action on M as follows: ar2/I⋯rn/I=ar2⋯rn where I=AR(M), for all a∈M and for all r2n∈R. We now show that this definition is well-defined. If for each i=2,…,n, ri/I=si/I, then rikIsi, that is, there exist x(i)2m,y(i)2m∈I such that f(ri,x(i)2m)=f(si,y(i)2m). It follows that af(r2,x(2)2m)⋯f(rn,x(n)2m)=af(s2,y(2)2m)⋯f(sn,y(n)2m), which implies that ar2⋯rn=as2⋯sn since x(i)2m,y(i)2m∈AR(M). Thus ar2/I⋯rn/I=as2/I⋯sn/I, as required. It is easy to see that the above definition makes M into a right n-ary R-semimodule.
Moreover, ifM is an irreducible right n-ary R-semimodule then it is routine to verify that M is also an irreducible right n-ary R/AR(M)-semimodule by (2).Lemma 18.
LetM be a right n-ary R-semimodule. Then AR/AR(M)(M)={0/AR(M)}.Proof.
Letx/I∈AR/I(M), where I=AR(M). Then for any a∈Mar2/I⋯rn/I=0M whenever r2n∈R and ri=x for some i∈{2,…,n}. It follows that ar2⋯rn=0M where ri=x for some i∈{2,…,n}. This shows that x∈I and so that x/I=0/I. Consequently, AR/AR(M)(M)={0/AR(M)}.Lemma 19.
Any rightn-ary R-semimodule M is a faithful R/AR(M)-semimodule.Proof.
LetM be a right n-ary R-semimodule. Then in view of Lemma 17, M is an R/AR(M)-semimodule. On the one hand, by Lemma 12, AR/AR(M)(M) is an h-ideal of R/AR(M). On the other hand, by Lemma 10, Z(R/AR(M)) is the smallest h-ideal of R/AR(M). Thus Z(R/AR(M))⊆AR/AR(M)(M). According to Lemma 18, AR/AR(M)(M)={0/AR(M)}. So Z(R/AR(M))={0/AR(M)}=AR/AR(M)(M), which means that M is a faithful R/AR(M)-semimodule.Lemma 20.
IfI is an h-ideal of an (m,n)-semiring R then Z(R/I)={0/I} where R/I is the Bourne factor semiring.Proof.
Supposex/I∈Z(R/I). Then we have f(x/I,y2m/I)=f(0/I,y2m/I) for some y2m/I∈R/I. Thus we have f(x,y2m)/I=f(0,y2m)/I which implies that f(x,y2m,a2m)=f(0,y2m,b2m) for some a2m,b2m∈I. Hence f(x,a2m,y2m)=f(0,b2m,y2m). This shows that x∈I since I is an h-ideal of R. Consequently, x/I=0/I. Thus Z(R/I)={0/I}.Definition 21.
An(m,n)-semiring R is said to be primitive if it has a faithful irreducible cancellative n-ary R-semimodule. An ideal P is said to be primitive if the Bourne factor semiring R/P is primitive.Evidently, an(m,n)-semiring R is primitive if and only if {0} is a primitive ideal of R. The following theorem characterizes primitive ideals of an (m,n)-semiring.Theorem 22.
Anh-ideal P of (m,n)-semiring R is primitive if and only if P=AR(M) for some irreducible right n-ary R-semimodule M.Proof.
LetP be an h-ideal of R such that P=AR(M) for some irreducible right n-ary R-semimodule M. Then by Lemmas 17 and 19  M is a faithful irreducible n-ary R/P-semimodule. This shows that R/P is primitive and hence P is a primitive h-ideal of R.
Conversely, letP be a primitive h-ideal of R. Then R/P is a primitive (m,n)-semiring. So there exists a faithful irreducible n-ary R/P-semimodule M. Now by Lemma 16  M is an irreducible n-ary R-semimodule. It remains to show that P=AR(M). Now x∈AR(M)⇔ for all a∈M and r2n∈R, ar2⋯rn=0M whenever x=ri for some i∈{2,…,n}⇔ar2/P⋯rn/P=0M/P whenever x/P=ri/P for some i∈{2,…,n}⇔x/P∈AR/P(M)=Z(R/P) since M is a faithful n-ary R/P-semimodule ⇔x/P∈AR/P(M)={0/P}, by Lemma 20  ⇔x/P=0/P⇔x∈P. Thus P=AR(M) as desired.


## 4. Jacobson Radical of an(m,n)-Semiring

Let us begin this section by defining the semi-irreducibility of a rightn-ary R-semimodule.Definition 23.
A rightn-ary R-semimodule M is said to be semi-irreducible if MRn-1≠{0M}; that is, ar2⋯rn≠0M for some a∈M and some r2n∈R, and M does not contain any n-ary k-subsemimodule other than {0M} and M.Lemma 24.
LetI be a subset of an (m,n)-semiring R and M a right n-ary R-semimodule with MRi-2IRn-i≠{0M} for some i∈{2,…,n}. In the case where i=2, we assume further that I is a left ideal of R. Then the following statements are true:(1)
IfM is semi-irreducible and a∈M, then a=0 if and only if aRi-2IRn-i={0M};(2)
IfM is irreducible and a,b∈M, then a=b if and only if ar2i-1sri+1n=br2i-1sri+1n for all rj∈R and all s∈I.Proof.
Suppose that(M,f0) is a semi-irreducible right n-ary semimodule over an (m,n)-semiring (R,f,g), and that I is a subset of R such that MRi-2IRn-i≠{0M} for some i∈{2,…,n}. In the case where i=2, we further assume that I is a left ideal of R.
(
1
) Assume that M is semi-irreducible. Let a∈M be such that
(20)aRi-2IRn-i={0M}.
Set(21)M0={x∈M:xRi-2IRn-i={0M}}.
It is clear thata∈M0, and it is easy to show that M0 is a subsemimodule of M. Let f0(x,y2m)∈M0 and y2m∈M0. Then f0(x,y2m)Ri-2IRn-i={0M} and y2mRi-2IRn-i={0M}. Thus xRi-2IRn-i={0M}; that is, x∈M0. This shows that M0 is a k-subsemimodule of M. Since MRi-2IRn-i≠{0M}, M0≠M. Since M is semi-irreducible, M0={0M} and therefore a=0.
The converse part is obvious.
(
2
) Assume that M is irreducible. Let a,b∈M be such that a≠b. Set uj=a, vj=b for j=2,…,m. Since MRi-2IRn-i≠{0M}, we have xr2i-1sri+1n-i≠0M for some x∈M,s∈I and rj∈R. Since M is irreducible, according to Definition 14, there exist a(i)2m,b(i)2m∈R with i=2,…,m such that
(22)f0(x,u2a(2)2n,…,uma(m)2n,v2b(2)2n,…,vmb(m)2n)=f0(0M,u2b(2)2n,…,umb(m)2n,v2a(2)2n,…,vma(m)2n).
Hence(23)f0(xr2i-1sri+1n-i,u2a(2)2nr2i-1sri+1n-i,…,uma(m)2nr2i-1sri+1n-i,v2b(2)2nr2i-1sri+1n-i,…,vmb(m)2nr2i-1sri+1n-i)=f0(0Mr2i-1sri+1n-i,u2b(2)2nr2i-1sri+1n-i,…,umb(m)2nr2i-1sri+1n-i,v2a(2)2nr2i-1sri+1n-i,…,vma(m)2nr2i-1sri+1n-i)=f0(0M,u2b(2)2nr2i-1sri+1n-i,…,umb(m)2nr2i-1sri+1n-i,v2a(2)2nr2i-1sri+1n-i,…,vma(m)2nr2i-1sri+1n-i).
SinceM is cancellative and xr2i-1sri+1n-i≠0M, at least one of the following 2(m-1) equalities does not hold:
(24)uja(j)2nr2i-1sri+1n-i=vja(2)2nr2i-1sri+1n-iwherej=2,…,m;ujb(j)2nr2i-1sri+1n-i=vjb(2)2nr2i-1sri+1n-i,wherej=2,…,m.
So we conclude that ifar2i-1sri+1n=br2i-1sri+1n for all rj∈R and all s∈I, then a=b.
The converse part follows easily.Lemma 25.
LetM be a right n-ary R-semimodule and M≠{0M}. Then M is semi-irreducible if and only if for every nonzero a∈M, aRn-1¯=M.Proof.
Assume thatM is a semi-irreducible right n-ary R-semimodule and M≠{0M}. Let a∈M be such that a≠0M. Then by Lemma 24  aRn-1≠{0M}. Since aRn-1¯ is an n-ary k-subsemimodule of M, aRn-1¯=M.
Conversely, suppose that for any nonzeroa∈M, aRn-1¯=M. Let N≠{0M} be an n-ary k-subsemimodule of M. Then there exists b∈N such that b≠0M. So by hypothesis, bRn-1¯=M. Hence for any x∈M, there exist a1m∈bRn-1 such that f0(x,a2m)=a1. Since bRn-1⊆N, we have a1m∈N. Since N is an n-ary k-subsemimodule, f0(x,a2m)=a1 implies that x∈N. This shows that N=M. Now if MRn-1={0M} then aRn-1={0M} for all a∈M. Hence aRn-1¯={0M}. So we have M={0M}, a contradiction. Therefore, MRn-1≠{0M}. Thus M is semi-irreducible.Corollary 26.
If a rightn-ary R-semimodule M is irreducible, then it is semi-irreducible and MRn-1¯=M.Proof.
Assume thatM is an irreducible right n-ary R-semimodule. Then M≠{0M} and, consequently, there exists a nonzero y∈M. In view of (2) with uj=y and vj=0M for j=2,…,m, we obtain that for any x∈M there exist a(j)2n∈R with j=2,…,m such that
(25)f0(x,ya(2)2n,…,ya(m)2n,0M(m-1))=f0(0M,yb(2)2n,…,yb(m)2n,0M(m-1)),
so that
(26)f0(x,ya(2)2n,…,ya(m)2n)=f0(0M,yb(2)2n,…,yb(m)2n)∈yRn-1.
It follows thatx∈yRn-1¯. Thus yRn-1¯=M. By Lemma 25, M is semi-irreducible.
Furthermore,MRn-1≠{0M}, which implies that MRn-1¯≠{0M}. Since MRn-1¯ is an n-ary k-subsemimodule of M, MRn-1¯=M as required.Now we can define the Jacobson radical of an(m,n)-semiring in an external way.Definition 27.
LetR be an (m,n)-semiring and Δ be the set of all irreducible right n-ary R-semimodules. Then J(R)=⋂M∈Δ‍AR(M) is called the Jacobson radical of R. If Δ is empty then R itself is considered as J(R); that is, J(R)=R, and in this case, we say that R is a radical(m,n)-semiring. An (m,n)-semiring R is said to be Jacobson semisimple or J-semisimple if J(R)={0}.By Lemma12, AR(M) is an h-ideal of R. Note that the intersection of any family of h-ideals is again an h-ideal. Consequently, we obtain the following.Lemma 28.
J
(
R
) is an h-ideal of R.Lemma 29.
IfM is a right n-ary R-semimodule then M is a right n-ary R/J(R)-semimodule, where R/J(R) is the Bourne factor semiring. Moreover, if M is an irreducible right n-ary R-semimodule, then M is also an irreducible right n-ary R/J(R)-semimodule.Proof.
This lemma can be proved by the same method as in proving Lemma17.Theorem 30.
IfR is an (m,n)-semiring, then the Bourne factor semiring R/J(R) is Jacobson semisimple.Proof.
ByΔ and Λ, we denote the set of all irreducible right n-ary R-semimodules and the set of all irreducible right n-ary R/J(R)-semimodules, respectively. Then according to Lemmas 28, 16, and, 29, we obtain that Δ=Λ. For any x∈J(R/J(R)) and any M∈Δ, we have x∈AR/J(R)(M), which means that for any a∈M, ar2/J(R)⋯rn/J(R)=0M whenever r2n∈R and ri=x for some i∈{2,…,n}. Thus ar2⋯rn=0M whenever r2n∈R and ri=x for some i∈{2,…,n}, so x∈AR(M) for all M∈Λ. That is, x∈J(R). Hence x/J(R)=0/J(R). We have shown that J(R/J(R))={0/J(R)}. By Definition 27, R/J(R) is Jacobson semisimple.The next theorem is a direct corollary of Theorem22, giving an internal characterization of the Jacobson radical of an (m,n)-semiring.Theorem 31.
J
(
R
) is the intersection of all primitive h-ideals of R.Definition 32.
LetP be an i-ideal of an (m,n)-semiring R for some i∈{1,…,n}. Then P is said to be strongly seminilpotent if there exists a positive integer t such that (PRn-2)t-1P⊆Z(R), where Rn-2=R⋯R︷n-2, (PRn-2)t-1=(PRn-2)(PRn-2)⋯t-1 times, (PRn-1)0P=P. P is said to be strongly nilpotent if there exists a positive integer t such that (PRn-2)t-1P={0}.Theorem 33.
IfP is a strongly semi-nilpotent left ideal of R, then P⊆J(R).Proof.
Suppose on the contrary that(27)P⊈J(R)=⋂M∈Δ‍AR(M),
where R is an (m,n)-semiring and Δ is the set of all irreducible right n-ary R-semimodules. Then there exists an M∈Δ such that P⊈AR(M). Thus there exists i∈{2,…,n} such that
(28)MRi-2PRn-i≠{0M}.
SinceP is strongly semi-nilpotent, there exists a positive integer t such that (PRn-2)t-1P⊆Z(R). By Lemmas 10 and 12, Z(R)⊆AR(M). It follows that (PRn-2)t-1P⊆AR(M), which implies that
(29)MRi-2(PRn-2)t-1PRn-i={0M}.
If (29) holds for all positive integers t’s, then in particular it is true for t=1 and in this case we have MRi-2PRn-i={0M}, a contradiction to (28). If (29) does not hold for all t, then there exist x∈M and positive s such that
(30)xRi-2(PRn-2)s-1PRn-1≠{0M},xRi-2(PRn-2)sPRn-i={0M}.
Thus x≠0M and there exists a∈xRi-2(PRn-2)s-1PRn-i such that a≠0M. It follows that
(31)aRi-2PRn-i⊆xRi-2(PRn-2)s-1PRn-iRi-2PRn-i=xRi-2(PRn-2)sPRn-i={0M},
so we have(32)aRi-2PRn-i={0M}.
By Lemma24, we obtain a=0M, again a contradiction. This completes the proof.The next result is a direct corollary of Theorem33.Corollary 34.
If an(m,n)-semiring R is Jacobson semisimple then R does not contain any non-zero strongly semi-nilpotent left ideal and hence R does not contain any nontrivial strongly nilpotent left ideal.Lemma 35.
IfM is a (semi-)irreducible right n-ary R-semimodule and N≠{0M} is an arbitrary R-subsemimodule (and NRn-1≠{0M}), then N is (semi-)irreducible, and for any x2n,y2n∈R the following statement is true: the equality ux2n=uy2n holds for all u∈M if and only if the same equality holds for all u∈N. Furthermore, AR(M)=AR(N).Proof.
AssumeM is an irreducible right n-ary R-semimodule. Then from (2), it follows that N is irreducible. If M is a semi-irreducible and NRn-1≠{0M}, then N is semi-irreducible by Definition 23 since any subsemimodule of N is clearly a subsemimodule of M.
Letx2n,y2n∈R be such that the equality ux2n=uy2n holds for all u∈M. Since M is semi-irreducible, for any a(≠0M)∈M and any b(≠0M)∈N, there exist positive p, q and x(i)2p,y(j)2q∈R such that p≡q≡0mod(m-1) and
(33)f0(a,bx(2)2n,…,bx(p)2n)=f0(0M,by(2)2n,…,by(q)2n).
Thus we have the following two equalities:(34)f0(ax2n,bx(2)2nx2n,…,bx(p)2nx2n)=f0(0M,by(2)2nx2n,…,by(q)2nx2n),f0(ay2n,bx(2)2ny2n,…,bx(p)2ny2n)=f0(0M,by(2)2ny2n,…,by(q)2ny2n).
It follows that(35)f0(0M,ax2n,bx(2)2nx2n,…,bx(p)2nx2n,by(2)2ny2n,…,by(q)2ny2n)=f0(0M,ay2n,bx(2)2ny2n,…,bx(p)2ny2n,by(2)2nx2n,…,by(q)2nx2n).
Observing thatbx(i)2n,by(j)2n∈N, since N is a submodule, we have bx(i)2nx2n=bx(i)2ny2n and by(2)2nx2n=by(2)2ny2n for all i,j by the assumption. Hence by cancellation law, (35) deduces that ax2n=ay2n. The converse implication is clear.
Furthermore, lettingyi=0 for some i, we get that the equality ux2n=0M holds for all u∈M if and only if the same equality holds for all u∈N. Thus AR(M)=AR(N).Lemma 36.
LetI be an ideal of an (m,n)-semiring R. (1)
IfM is an (semi-)irreducible right n-ary R-semimodule (and MRn-1≠{0M}), then M is an (semi-)irreducible right n-ary I-semimodule.(2)
IfM is an irreducible right n-ary I-semimodule, then there exists an irreducible right n-ary R-semimodule M*, which can be regarded as an I-subsemimodule N of M.Proof.
(
1
) Let M be an irreducible R-semimodule and u2m,v2m∈M be such that uj≠vj for some j. Without loss of generality, we suppose that M≠{0M}. From (2) we deduce that MRn-1≠{0M}. By Lemma 24, ujc2n≠vjc2n for some c2n∈I. Since M is irreducible, by (19) there exist a(i)2m,b(i)2m∈R with i=2,…,m such that
(36)f0(x,(u2c2n)a(2)2n,…,(umc2n)a(m)2n,(v2c2n)b(2)2n,…,(vmc2n)b(m)2n)=f0(0M,(u2c2n)b(2)2n,…,(umc2n)b(m)2n,(v2c2n)a(2)2n,…,(vmc2n)a(m)2n),
that is,
(37)f0(x,u2(c2na(2)2n),…,um(c2na(m)2n),v2(c2nb(2)2n),…,vm(c2nb(m)2n))=f0(0M,u2(c2nb(2)2n),…,um(c2nb(m)2n),v2(c2na(2)2n),…,vm(c2na(m)2n)),
which means that M is an irreducible I-semimodule by (19) again since for all j∈{2,…,n}c2na(j)2n,c2nb(j)2n∈In-1.
Assume thatM is a semi-irreducible R-semimodule and MIn-1≠{0M}. According to Lemma 24, for any u(≠0M)∈M there exist b2n∈R such that ub2n≠0M. By Lemma 25, (ub2n)Rn-1¯=M, so for any x∈M there exist positive integers p,q and x(i)2n,y(j)2n∈R such that p≡q≡0mod(m-1) and
(38)f0(x,(ub2n)x(2)2n,…,(ub2n)x(p)2n)=f0(0M,(ub2n)y(2)2n,…,(ub2n)y(q)2n),
which shows that
(39)f0(x,(ub2nx(2)2n),…,u(b2nx(p)2n))=f0(0M,u(b2nx(2)2n),…,u(b2nx(q)2n)).
Note that for alli, j, (b2nx(i)2n),(b2nx(j)2n)∈In-1. Thus we obtain uIn-1¯=M. By Lemma 25 again, M is a semi-irreducible right n-ary I-semimodule.
(
2
) Let M be an irreducible right n-ary I-semimodule, and let N=MIn-1. Then N≠{0M} and N is an I-subsemimodule of M. Thus by Lemma 35, N is irreducible and for any x2n,y2n∈R the following conclusion is true: the equality ux2n=uy2n holds for all u∈M if and only if the same equality holds for all u∈N. If f0(a(1)x(1)2n,…,a(p)x(p)2n)=f0(b(1)y(1)2n,…,b(q)y(q)2n) for some x(i)2n,y(j)2n∈I and a(i),b(j)∈N, then for any r2n∈R and any z2n∈I,
(40)f0(a(1)(x(1)2nr2n),…,a(p)(x(p)2nr2n))z2n=f0(a(1)(x(1)2nr2nz2n),…,a(p)(x(p)2nr2nz2n))=f0(a(1)x(1)2n,…,a(p)x(p)2n)(r2nz2n)=f0(b(1)y(1)2n,…,b(q)y(q)2n)(r2nz2n)=f0(b(1)(y(1)2nr2nz2n),…,b(q)(y(q)2nr2nz2n))=f0(b(1)(y(1)2nr2n),…,b(q)(y(q)2nr2n))z2n,which implies that
(41)f0(a(1)(x(1)2nr2n),…,a(p)(x(p)2nr2n))=f0(b(1)(y(1)2nr2n),…,b(q)(y(q)2nr2n))
by Lemma 24 since M is an irreducible right n-ary I-semimodule. Thus we can define an operation on NRn-1 into N by setting
(42)f0(a(1)x(1)2n,…,a(p)x(p)2n)r2n=f0(a(1)(x(1)2nr2n),…,a(p)(x(p)2nr2n)),
where x(i)2n,y(j)2n∈I and a(i),b(j)∈N. Thus N with the addition and the above operation becomes a right n-ary R-semimodule M* which, as a right n-ary I-semimodule, is isomorphic to the right n-ary I-semimodule N. It is clear that M* is an irreducible right n-ary R-semimodule.Now we are ready to generalize [17, Theorem 2].Theorem 37.
IfI is an ideal of an (m,n)-semiring R, then J(I)=J(R)∩I.Proof.
LetR be an (m,n)-semiring and let Δ be the set of all irreducible right n-ary R-semimodules. Then by Definition 27  J(R)=⋂M∈Δ‍AR(M). If I is an ideal of an (m,n)-semiring R, then J(I)=⋂M∈Λ‍AI(M), where Λ is the set of all irreducible right n-ary I-semimodules.
For anyM∈Δ, according to Lemma 36, we have M∈Λ. It is evident that AI(M)=AR(M)∩I. This shows that J(I)⊆J(R)∩I.
For anyM∈Λ, according to Lemma 36, we have that M*∈Δ and that M* can be regarded as an I-subsemimodule N of M. By Lemma 35, we have that AI(M)=AI(N)=AI(M*)=AR(M*)∩I. This shows that J(I)⊇J(R)∩I.
Summarizing the above, we obtain thatJ(I)=J(R)∩I.Consequently, we have.Theorem 38.
For an(m,n)-semiring R, J(R) is a radical (m,n)-semiring; that is, J(J(R))=J(R).



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*Source: 272104-2013-10-07.xml*