# Kurosh-Amitsur Right Jacobson Radical of Type 0 for Right Near-Rings

Authors: Ravi Srinivasa Rao; K. Siva Prasad; T. Srinivas
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

---

## Abstract

By a near-ring we mean a right near-ring.J0r, the right Jacobson radical of type 0, was introduced for near-rings by the first and second authors. In this paper properties of the radical J0r
are studied. It is shown that J0r is a Kurosh-Amitsur radical (KA-radical) in the variety of all near-rings
R, in which the constant part Rc
of R
is an ideal of R. So unlike the left Jacobson radicals of types 0 and 1 of near-rings, J0r is a KA-radical in the class of all zero-symmetric near-rings. J0r is not s-hereditary and hence not an ideal-hereditary radical in the class of all zero-symmetric near-rings.


---

## Body

## 1. Introduction

R denotes a
right near-ring and all near-rings considered are right near-rings and not
necessarily zero-symmetric.In [1, 2], the first author studied the structure
of near-rings in terms of right ideals, and showed that as in rings, matrix
units determined by right ideals identify matrix near-rings. To show the
importance of the right Jacobson radicals of near-rings in the extension of a
form of the Wedderburn-Artin theorem of rings involving the matrix rings to
near-rings, the right Jacobson radicals of type ν were introduced
and studied by the first and second authors in [3–6], ν∈{0,1,2,s}. In
[6], Wedderburn-Artin theorem was extended to near-rings, and some
generalizations of it were presented.In this paper, properties of the right Jacobson
radical of type 0 are studied. It is known that the left Jacobson radicals of
types 0 and 1 are not KA-radicals in the class of all zero-symmetric
near-rings, and only the left Jacobson radicals of types 2 and 3 are
KA-radicals in the class of all zero-symmetric near-rings. Surprisingly,J0r, the right Jacobson radical of
type 0, is a KA-radical in the class of all zero-symmetric near-rings. It is
also shown that J0r is a KA-radical even in a
bigger class of near-rings, namely, in the variety of all near-rings R, in
which the constant part of R is an ideal of R. Moreover, J0r is not s-hereditary, and hence
not an ideal-hereditary radical in the class of all zero-symmetric near-rings.


## 2. Preliminaries

Near-rings
considered are right near-rings and not necessarily zero-symmetric. Unless
otherwise specified,R stands for a right near-ring. Near-ring notions not
defined here can be found in [7].R0 and Rc denote the zero-symmetric part
and the constant part of R, respectively.ℱ denotes the
class of near-rings R, in which the constant part Rc of R is an ideal of R.
In [8],
Fuchs has shown that the class of near-rings ℱ is a variety.
Obviously, ℱ contains all
zero-symmetric, constant, and abstract affine near-rings. Now we give here some
definitions and results of [3], which will be used later.An elementa∈R is called  right
quasiregular if and only if the right ideal of R generated by the set {x−ax∣x∈R} is R. A right
ideal (left ideal, ideal, subset) K of R is called a  right quasiregular
right ideal (left ideal, ideal, subset) of R if each element of K is right
quasiregular.A right idealK of R is called  right modular if
there is an element e∈R such that x−ex∈K  for all x∈R. In this
case, we say that K is  right modular bye.A maximal right modular right ideal ofR is called a  right
0-modular right ideal of R.J1/2r(R) is the intersection of all right 0-modular right
ideals of R, and if R has no right 0-modular right ideals, then J1/2r(R)   =  R.The largest ideal ofR contained in J1/2r(R)  is denoted by J0r(R) and called the  right Jacobson radical ofR of type
0.The largest ideal contained in a right 0-modular right
ideal ofR is called a
right 0-primitive ideal of R. R is called a
right 0-primitive near-ring if {0} is a
right 0-primitive ideal of R.A group (G, +) is called a  rightR-group if
there is a mapping ((g, r) →gr) of G×R into G  such
that (i) (g+h)r=gr+hr and (ii) g(rs)=(gr)s  for all g,h∈G and r,s∈R. A subgroup
(normal subgroup) H of a right R-group G is called an  R-subgroup (ideal) of G if hr∈H for all h∈H and r∈R.LetG be a right R-group. An element g∈G is called a  generator of G if gR=G and g(r+s)=gr+gs for all r,s∈R. G is said to
be  monogenic if G has a generator.G is said to be  simple if  G≠{0}, and G and {0}
are the only ideals of G.A monogenic rightR-group G is said to be a  rightR-group of type 0
if G is simple.Theannihilator of a right R-group G, denoted
by (0:G), is defined as (0:G)={a∈R∣Ga={0}}.Lemma 2.1.
The constant part ofR is right quasiregular.Lemma 2.2.
A nilpotent element ofR is right quasiregular.Theorem 2.3.
J1/2r(R) is the largest right
quasiregular right ideal of R.Theorem 2.4.
J0r(R)  is the largest right
quasiregular ideal of R.Theorem 2.5.
J0r(R) is the intersection of all
right 0-primitive ideals of R.Theorem 2.6.
LetP be an ideal of R. P is a right 0-primitive ideal
of R if and only if R/P is a right 0-primitive near-ring.Proposition 2.7.
LetG be a right R-group of type 0 and g0 a generator of G. Then (0:g0):={r∈R∣g0r=0}  is a right 0-modular
right ideal of R.Proposition 2.8.
LetG be a right R-group. G is a right R-group of type
0 if and only if there is a maximal right modular right ideal K of R such that
G is R-isomorphic to R/K.Proposition 2.9.
LetP be an ideal of
a zero-symmetric near-ring R. P is right
0-primitive if and only if P is the largest
ideal of R contained in (0:G) for some
right R-group G of type 0.LetQ be a mapping
which assigns to each near-ring R an ideal Q(R) of R. Such mappings are called ideal-mappings. We consider
the following properties which Q may satisfy:(H1)h(Q(R))⊆Q(h(R)) for all
homomorphisms h of R;(H2)Q(R/Q(R))={0} for all R;Q is r-hereditary if I∩Q(R)⊆Q(I) for all ideals I of R;Q is  s-hereditary if Q(I)⊆I∩Q(R) for all ideals I of R;Q is  ideal-hereditary if it is both r-hereditary and s-hereditary,
that is, if Q(I)=I∩Q(R) for all ideals I of R;Q is idempotent if Q(Q(R))=Q(R) for all R;Q is complete if Q(I)=I and I is an ideal of R that implies  I⊆Q(R).WithQ we associate
two classes of near-rings ℝQ and 𝕊Q defined by ℝQ:={R∣Q(R)=R}, 𝕊Q:={R∣Q(R)={0}}, and are called a Q-radical class
and a Q-semisimple
class, respectively.An ideal-mappingQ is a Hoehnke radical (H-radical) if it
satisfies conditions (H1) and (H2).An ideal-mappingQ is a Kurosh-Amitsur radical (KA-radical) if it is a complete idempotent H-radical.Let𝕄 be a class of
near-rings. Classes of near-rings are always assumed to be abstract, that is,
they contain the one element near-ring and are closed under isomorphic copies.
With every near-ring R, we associate
two ideals of R, depending on 𝕄. These ideals are defined by the following:(2.1)𝕄(R):=Σ{I∣IisanidealofR,I∈𝕄},(R)𝕄:=∩{I∣IisanidealofR,R/I∈𝕄}. The mapping P defined by P(R):=(R)𝕄 is always an H-radical and is
called the H-radical corresponding
to 𝕄.From Theorems2.5 and 2.6, we have the following.Proposition 2.10.
J0r is an H-radical
corresponding to the class of all right 0-primitive near-rings.


## 3. Properties of the RadicalJ0r

If(A,+) is a group and T is a subset of A, then the subgroup (normal subgroup) of A generated by T is denoted by 〈T〉s(〈T〉n).Remark 3.1.
LetG be a right R-group. It is
clear that H={g∈G∣gR={0}} is an ideal of G. So if G is simple and gR={0}, then g=0 provided GR≠{0}.Theorem 3.2.
LetG be a right R-group of type  0. Suppose that S is an invariant subnear-ring and a
right ideal of R. If GS≠{0}, then G is also a right S-group of
type 0.Proof.
Suppose thatGS≠{0}. Clearly, G is a right S-group. Let g∈G and gS:={gs∣s∈S}⊆G. Consider the normal subgroup 〈gS〉n of (G,+). Let r∈R, h∈〈gS〉n. Now h=(x1+δ1(gs1)−x1)+(x2+δ2(gs2)−x2)+⋯+(xk+δk(gsk)−xk), si∈S, xi∈G, δi∈{1,−1}. Since SR⊆S, hr=(x1r+δ1(g(s1r))−x1r)+(x2r+δ2(g(s2r))−x2)+⋯+(xkr+δk(g(skr))−xkr)∈〈gS〉n. So 〈gS〉n is an ideal of
the right R-group G,  and hence it is
also an ideal of the right S-group G. Let 0≠h∈G. Suppose that hS={0}. Since hR≠{0}, 〈hR〉n is a nonzero
ideal of the right R-group G. Since G is a simple
right R-group, 〈hR〉n=G. So GS=〈hR〉nS⊆〈hS〉n={0}, a contradiction to GS≠{0}. Therefore, hS≠{0}. Let g0 be a generator
of the right R-group G. So g0 is a
distributive element of the right R-group G and g0R = G. Clearly, g0 is a
distributive element of the right S-group G and hence g0S is a subgroup
of (G,+). We have (g0S)R=g0(SR)g0S. So g0S is an R-subgroup of G. Let g ∈G and s ∈S. Since g0R=G, g=g0r for some r∈R. So g+g0s−g=g0r+g0s−g0r=g0(r+s−r)∈g0S, as S is a normal
subgroup of (R,+). Therefore, g0S is an ideal of
the right R-group G and hence g0S=G. So g0 is also a
generator of the right S-group G. Let K be a nonzero
ideal of the right S-group G. Let 0≠y∈K. As seen above, 〈yS〉n is a nonzero
ideal of the right R-group G, and hence 〈yS〉n=G. Since G=〈yS〉n⊆K, G=K. Therefore, {0} and G are the only
ideals of the right S-group G and hence G is a right S-group of
type 0.Proposition 3.3.
LetG be a right R-group of
type 0 and let T be a right quasiregular invariant subnear-ring of R. If T is a right ideal of R, then GT={0}.Proof.
Suppose thatT is a right ideal
of R and g0 is a generator
of G. So g0(r+s)=g0r+g0s for all r,s∈R and g0R=G. Now L:=(0:g0)={r∈R∣g0r=0} is a right
0-modular right ideal of R. Therefore, L contains the
largest right quasiregular right ideal of R. Since T is a right
quasiregular right ideal of R, T⊆L, that is, g0T={0}. Let g∈G and t∈T. Now g=g0r for some r∈R. gt=g0(rt)=0, as rt∈T. Therefore, GT={0}.SinceRc is right
quasiregular in R, we have the following.Corollary 3.4.
IfRc is a normal
subgroup of (R,+), then GRc={0} for all right R-groups G of type 0.Corollary 3.5.
LetR∈ℱ. If G is a right R-group of
type 0, then GJ0r(R)={0}.Proof.
LetG be a right R-group of
type 0. We have that I:=J0r(R) is the largest
right quasiregular ideal of R. Since Rc is a right
quasiregular ideal of R,Rc⊆I. So I is an invariant
ideal of R. Therefore, by Proposition 3.3, GI={0}.Proposition 3.6.
LetR∈ℱ. Let  I be an ideal of R  and K:=I+Rc. If G is a right K-group of
type 0, then G is a right I-group of
type 0.Proof.
Suppose thatG is a right K-group of
type 0 and g0 is a generator
of G. So g0 is distributive
over K and g0K=G. Let Kc be the constant
part of K. Since Kc=Rc is a normal
subgroup of K, by Corollary 3.4, GRc={0}. Clearly, G is a right I-group. Now G=g0K=g0(I+Rc)=g0I, and hence g0 is a generator
of the right I-group G. Let H be a nonzero
ideal of the right I-group G. Let h∈H and k∈K. k=i+rc, i∈I, rc∈Rc and h=g0t, t∈I. hk=g0t(i+rc)=g0((t(i+rc)−ti)+ti)=g0(t(i+rc)−ti)+g0(ti)=0+(g0t)i=hi∈H. Therefore, H is a nonzero
ideal of the right K-group G and hence H=G. So G is a right I-group of
type 0.We show now
that the Hoehnke radicalJ0r is complete in
the variety ℱ.Theorem 3.7.
LetR∈ℱ. If I is an ideal of R and J0r(I)=I, then I⊆J0r(R).Proof.
LetI be an ideal of R and J0r(I)=I. Suppose that I⊈J0r(R). So K:=I+Rc is an ideal of R and K⊈J0r(R). We get a right R-group G of type 0 such
that GK≠{0}. Since K is an invariant
ideal of R, by Theorem 3.2, G is a right K-group of
type 0. Therefore, by Proposition 3.6, G is a right I-group of
type 0. This is a contradiction to the fact that J0r(I)=I. Therefore, I⊆J0r(R).Theorem 3.8.
J0r is a complete
Hoehnke radical in the variety ℱ.Theorem 3.9.
J0r is a complete
Hoehnke radical in the class of all zero-symmetric near-rings.Theorem 3.10.
Suppose thatS is an invariant
subnear-ring of R. If G is a right S-group of
type 0, then G is also a right R-group of
type 0.Proof.
Suppose thatG is a right S-group of
type 0 and g0 is a generator.
We have that g0 is distributive
over S and g0S=G. For g∈G and r∈R, define gr:=g0(sr), if g=g0s, s∈S. We show now that this operation is well defined.
Suppose that g=g0s=g0t, s,t∈S. Let r∈R and h:=g0(sr)−g0(tr). Now hk=(g0(sr)−g0(tr))k=g0((sr)k)−g0((tr)k)=g0(s(rk))−g0(t(rk))=g(rk)−g(rk)=0 for all k∈S. Therefore, hS={0}, and hence h=0, that is, g0(sr)=g0(tr). We show that G is a right R-group of
type 0. It is clear that G is a right R-group. g0=g0e for some e∈S. Now G⊇g0R=g0(eR)⊇g0(eS)=g0S=G. So g0R=G. Let  p,q∈R and x=g0(p+q)−(g0p+g0q). xs=(g0(p+q)−(g0p+g0q))s=(g0(p+q))s−((g0p+g0q))s=g0(ps+qs)−(g0ps+g0qs)=(g0(ps)+g0(qs))−(g0(ps)+g0(qs))=0 for all s∈S. Therefore, x=0, and hence g0 is a generator
of the right R-group G. It can be easily verified that the action of R on G is an extension
of the action of S on G. So an ideal of the right R-group G is also an ideal
of the right S-group G. Since the right S-group G has no
nontrivial ideals, the right R-group G also has no
nontrivial ideals. Therefore, G is also a right R-group of
type 0.We show now
that the Hoehnke radicalJ0r is idempotent
in the variety ℱ.Theorem 3.11.
LetR∈ℱ. Then J0r(J0r(R))=J0r(R).Proof.
LetI:=J0r(R). I is the largest
right quasiregular ideal of R. Since Rc is a right
quasiregular ideal of R, Rc⊆I. So I is an invariant
ideal of R. Suppose that J0r(I)≠I. So there is a right I-group G of type 0. By
Theorem 3.10, G is an R-group of
type 0. Now, by Corollary 3.5, GI=GJ0r(R)={0}. This is a contradiction to the fact that G is an I-group of
type 0. Therefore, J0r(I)=I, that is, J0r(J0r(R))=J0r(R).From Theorems3.7 and 3.11, we have the following.Theorem 3.12.
J0r is a
Kurosh-Amitsur radical in the variety ℱ.Theorem 3.13.
J0r is a
Kurosh-Amitsur radical in the class of all zero-symmetric near-rings.Theorem 3.14.
J0r is not
s-hereditary in the class of all zero-symmetric near-rings.Proof.
ConsiderG:=Z8, the group of integers under addition modulo 8. Now T: G→G, defined by T(g)=5g, for all g∈G, is an
automorphism of G. T fixes 0, 2, 4, and
6, and maps 1 to 5 and 3 to 7. A:={I,T} is an
automorphism group of G. {0}, {2}, {4}, {6}, {1,5}, and {3,7} are the orbits.
Let R be the
centralizer near-ring MA(G), the near-ring of all self maps of G which fix 0 and
commute with T. An element of R is completely
determined by its action on {1,2,3,4,6}. An element f∈R maps 2G into 2G and f(1) and f(3) are arbitrary
in G. This example was considered in [9] and showed that P:=(0:2G)={f∈R∣f(h)=0,forallh∈2G} is the only
nontrivial ideal of R. Let f0 be the element
of P which fixes all
the elements in G−2G. Clearly f−f0f∈(2G:G)={t∈R∣t(G)⊆2G} for all f∈R. Since (2G:G) is a proper
right ideal of R, f0 is not right
quasiregular in R. So P is not a right quasiregular ideal of R. Since R is a near-ring
with identity, it is not right quasiregular. Therefore, {0} is the largest
right quasiregular ideal of R, and hence J0r(R)={0}. So R is J0r-semisimple. It
is shown in [9] that K:=(4G:G)P is a nonzero
ideal of P and K2={0}. Since a nil ideal is right quasiregular, K is a right
quasiregular ideal of P. Therefore, {0}≠K⊆J0r(P) and hence P is not J0r-semisimple.
So J0r is not
s-hereditary in the class of all zero-symmetric near-rings.Corollary 3.15.
J0r is not s-hereditary in
the class of all near-rings.Theorem 3.16.
J0r is not an
ideal-hereditary radical in the class of all zero-symmetric near-rings.It is not known to the authors whetherJ0r is a KA-radical
in the class of all near-rings. J0r may fail to be
idempotent and thus Kurosh-Amitsur in the class of all near-rings.



---

*Source: 741609-2008-02-21.xml*