A RADICAL FOR RIGHT NEAR-RINGS : THE RIGHT JACOBSON RADICAL OF TYPE-0

The notions of a right quasiregular element and right modular right ideal in a near-ring are initiated. Based on these J 0(R), the right Jacobson radical of type-0 of a near-ring R is introduced. It is obtained that J 0 is a radical map and N(R)⊆ J 0(R), where N(R) is the nil radical of a near-ring R. Some characterizations of J 0(R) are given and its relation with some of the radicals is also discussed.


Introduction
Throughout this paper, R stands for a right near-ring.The structure of matrix near-rings was studied by Rao in [3,4].It is clear from these papers that right Jacobson-type radicals have an important role to play in the study of Meldrum-van der Walt matrix near-rings.This motivated the authors to develop the right Jacobson-type radicals of near-rings.The aim of this paper is to give a good beginning in this direction of investigation.Left quasiregularity was introduced and studied in near-rings.In this paper, right quasiregularity is developed, and right modules of near-rings and the right Jacobson radical of type-0 are studied.
In Section 2, the notions of a right quasiregular element and a right modular right ideal are introduced.Using these right 0-primitive ideal, right 0-primitive near-ring, and J r 0 (R), the right Jacobson radical of R of type-0 are introduced.A nil subset of R is right quasiregular and the constant part of R is also right quasiregular.It is shown that J r 0 is a radical map and J r 0 (R) is the largest right quasiregular ideal of R.Moreover, N(R), the nil radical of R, is contained in J r 0 (R) and that P(R) ⊆ N(R) ⊆ J r 0 (R).If R is a zero-symmetric near-ring with DCC on the left R-subgroups of R, then J 0 (R) ⊆ J r 0 (R).In Section 3, a right R-group of type-0 is introduced.It is shown that G is a right Rgroup of type-0 if and only if G is right R-isomorphic to R/K, for some maximal right modular right ideal K of R. A right 0-primitive ideal of R is a prime ideal of R.An ideal P of a d.g.near-ring R is a right 0-primitive ideal of R if and only if P = (0 : G), for some right R-group G of type-0.
In Section 4, the right Jacobson radical of type-0 of a biregular near-ring is studied.For such a near-ring R, it is shown that J r 0 (R) = J 0 (R) = {0}.Moreover, if R is biregular, then an ideal P of R is right 0-primitive if and only if P is left 0-primitive if and only if P is a maximal right and left modular ideal.It is shown by an example that in general J r 0 (R) differs from the existing Jacobson-type radicals of R and the J r 0 -radical class contains almost all the classes of near-rings with trivial multiplication.

Right quasiregularity and the right J
Remark 2.2.Note that unlike in the left quasiregularity case, if R is a near-ring and a ∈ R, then the right ideal of R generated by the set Proof.Let a be a nilpotent element in R, say a n = 0 for some positive integer n.Let K be the right ideal of R generated by and hence a is right quasiregular.
One of the major differences between right and left quasiregular elements of R and hence between its right and left Jacobson-type radicals is the following.
Lemma 2.6.Let e be a nonzero distributive idempotent in R. Then e is not a right quasiregular element of R.
Therefore, e is not right quasiregular.Definition 2.7.A right ideal K 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 K is said to be right modular by e.The proof follows from the definitions.Remark 2.9.Let K be a right ideal of R. If K is right modular by e, then e ∈ K if and only if K = R. R. S. Rao and K. S. Prasad 3 Proposition 2.10.Let K be a proper right ideal of R. If K is, right modular, then K is contained in a maximal right ideal of R which is also right modular.
The proof of this result is easy and hence omitted.Proposition 2.12.If K 1 ,K 2 ,...,K n are maximal right modular right ideals of R such that n i=1 K i = {0}, then R has a left identity.The proof follows from Proposition 2.11.Definition 2.13.J r 1/2 (R) is the intersection of all maximal right modular right ideals of R and if R has no maximal right modular right ideals, then J r 1/2 (R) = R. Theorem 2.14.J r 1/2 (R) is the largest right quasiregular right ideal of R. Proof.Let q ∈ J r 1/2 (R).Suppose that q is not right quasiregular.Let K be the right ideal of R generated by the set {x − qx | x ∈ R}.Now q / ∈ K.By Zorn's lemma, we get a right ideal M, which is maximal for the property that K ⊆ M, q / ∈ M. Now M is a maximal right ideal of R. M is right modular right ideal of R as x − qx ∈ M, for all x ∈ R. As q ∈ J r 1/2 (R), q ∈ M, a contradiction.Therefore each element of J r 1/2 (R) is right quasiregular.We see now that each right quasiregular right ideal of R is contained in J r 1/2 (R).Let K be a right quasiregular right ideal of R. We claim that K ⊆ J r 1/2 (R).Suppose that K is not contained in J r 1/2 (R).We get a maximal right modular right ideal M such that For this, consider the nonabelian group G of order 6.Then M 0 (G) is a simple near-ring.Since G has only one nonzero maximal normal subgroup, M 0 (G) has only one nonzero maximal right ideal which is right modular by the identity element of M 0 (G).So, J r 1/2 (M 0 (G)) is not an ideal of M 0 (G).Remark 2.16.Let K be a right ideal of R. We show that there is a unique largest ideal of R contained in K. Now {0} is an ideal of R and {0} ⊆ K. Let I and J be ideals of R and Similarly, if I 1 ,I 2 ,...,I n are ideals of R and I j ⊆ K, for all 1 ≤ j ≤ n, then we get that 4 The right Jacobson radical of type-0 Definition 2.17.The largest ideal of R contained in J r 1/2 (R) is denoted by J r 0 (R) and is called the right Jacobson radical of R of type-0.
Theorem 2.18.R → J r 0 (R) is a radical map.Proof.(1) First suppose that R has no maximal right modular right ideal.Now R = J r 0 (R) and that R/J r 0 (R) = {0}.So J r 0 (R/J r 0 (R)) = {0}.Suppose now that R has a maximal right modular right ideal.Let {M α | α ∈ Δ} be the collection of all maximal right modular right ideals of R. Since M α is a maximal right modular right ideal of R and J r 0 (R (2) Let h be a homomorphism of the near-ring R onto a near-ring S. If S has no maximal right modular right ideal, then J r 0 (S) = S. Then clearly h(J r 0 (R)) ⊆ S = J r 0 (S).Suppose that S has a maximal right modular right ideal.Let {N α | α ∈ Δ} be the collection of all maximal right modular right ideals of S.
We denote the ideal of R generated by an element a of R by (a).The following result is obvious in view of Theorem 2.14.
is a right quasiregular ideal}.Theorem 2.20.J r 0 (R) is the largest right quasiregular ideal of R. The proof follows from Theorem 2.14.
Theorem 2.21.The nil radical N(R) of R is contained in J r 0 (R).The proof follows from Lemma 2.4 and Theorem 2.20.
Remark 2.25.If R is a ring, then J r 0 (R) is the (right) Jacobson radical of R and a right 0-primitive ideal of the near-ring R is a (right) primitive ideal of the ring R.
Theorem 2.26.If J r 0 (R) = R, then J r 0 (R) is the intersection of all right 0-primitive ideals of R.
Proof.J r 0 (R) is an ideal contained in each maximal right modular right ideal of R.So J r 0 (R) is contained in each right 0-primitive ideal of R. Hence it is contained in the intersection of all right 0-primitive ideals of R. On the other hand, the intersection of all right 0primitive ideals of R is an ideal contained in each maximal right modular right ideal of R and that it is contained in J r 0 (R).Theorem 2.27.A maximal right modular ideal of a near-ring R is a right 0-primitive ideal of R.
Proof.Let R be a near-ring and let K be a maximal right modular ideal of R. Since K is a proper right modular right ideal of R, K is contained in a maximal right modular right ideal M of R. Since K is a maximal ideal of R, K is the largest ideal contained in M. Hence, K is a right 0-primitive ideal of R.
Definition 2.29.A 0-primitive ideal of R defined in Pilz [2] is called a left 0-primitive ideal of R and similarly a left 0-primitive near-ring.
Theorem 2.30.Let P 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.
Proof.Let P be a right 0-primitive ideal of R.So we get a maximal right modular right ideal M of R such that P is the largest ideal of R contained in M. Now M/P is a maximal right modular right ideal of R/P.Since P is the largest ideal of R contained in M, the zero ideal of R/P is the largest ideal of R/P contained in M/P.Therefore R/P is a right 0-primitive near-ring.Suppose now that R/P is a right 0-primitive near-ring.So we get a maximal right modular right ideal M/P of R/P such that the zero ideal of R/P is the largest ideal of R/P contained in M/P.Clearly M is a maximal right modular right ideal of R. Since the zero ideal of R/P is the largest ideal of R/P contained in M/P, P is the largest ideal of R contained in M. Therefore, P is a right 0-primitive ideal of R.
Proof.Let R be a commutative right 0-primitive near-ring.We get a modular maximal right ideal M of R such that {0} is the largest ideal of R contained in M. Suppose that M is right modular by e. x − ex ∈ M, for all x ∈ R. Since R is commutative, M is an ideal of R. Therefore M = {0}.Since x − ex ∈ M = {0}, x = ex = xe.So e is the identity element of R. Now R is a commutative ring with identity.Since M = {0} is a maximal ideal of R, R is a field.

Right R-groups of type-0
Definition 3.1.A group (G,+) is called a right R-group if there is a mapping (g,r) → gr of G × R into G such that (1) (g + h)r = gr + hr, (2) g(rs) = (gr)s, for all g,h ∈ G and r,s ∈ R.
6 The right Jacobson radical of type-0 R is a right R-group.If K is a subgroup of (R,+) and kr ∈ K for all k ∈ K and r ∈ R, then K is a right R-subgroup of R. Every right ideal of R is an ideal of the right R-group R. Also, if K is a right ideal of R, then R/K is a right R-group, where (x + K)r = xr + K, for all x + K ∈ R/K and r ∈ R. Definition 3.2.Let G 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.We see that e + K is a generator of the right R-group R/K.Let r,s ∈ R. Now r − er ∈ K.So r + K = er + K = (e + K)r ∈ (e + K)R and hence (e + K)R = R/K.Also (r + s) − e(r + s), r − er, s − es ∈ K. Let k = r − er and let t = s − es.So r = k + er, s = t + es.Since (r + s) − e(r + s) ∈ K, we get that (k + er) + (t + es) − e(r + s) = k + (er + t − er) + er + es − e(r + s) ∈ K and that er + es − e(r + s) ∈ K. Therefore e(r + s) + K = (er + es) + K.So (e + K)(r + s) = (er + K) + (es + K) = (e + K)r + (e + K)s.This shows that e + K is a generator of R/K.So g is a generator of G and hence G is monogenic.
Proposition 3.5.Let K be a right ideal of R. Then K is right modular if and only if there is a right R-group G with a generator g such that K = (0 : g).
Proof.Suppose that K is right modular by e.As seen in the above proposition e + K is a generator of the right R-group R/K.Now r ∈ (K : e + K) Conversely suppose that g is a generator of the right R-group G and (0 : g) = K.Since gR = G, we get e ∈ R such that ge = g.Let r ∈ R. Now g(r − er) = gr − gr = 0. Therefore r − er ∈ (0 : g) = K.Hence, K is right modular by e. Definition 3.6.Let G be a right R-group.G is said to be simple if G = {0} and {0} and G are the only ideals of G. Definition 3.7.A monogenic right R-group G is said to be a right R-group of type-0 if G is simple.R. S. Rao and K. S. Prasad 7 Proposition 3.8.Let G 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 Proof.G is a right R-group.Suppose that G is of type-0.Let g ∈ G be a generator.Therefore from the proof of Proposition 3.4, G is R-isomorphic to R/K for some right modular right ideal K of R. Since G is simple, we get that R/K is also simple.Hence, K is a maximal right ideal of R. Conversely, suppose that G is R-isomorphic to R/K, where K is a maximal right modular right ideal of R. Since R/K = {K } has exactly two ideals, we get that {0} and G are the only ideals of G, where {0} = G.Let K be right modular by e.So e + K is a generator of R/K.Therefore, G is also monogenic.Hence, G is a right R-group of type-0.Definition 3.9.Let G be a right R-group.The annihilator of G denoted by (0 : G) is defined as (0 : If A and B are nonempty subsets of R, then (A : B) denotes the set {r ∈ R | Br ⊆ A}.
Corollary 3.10.Let K be a right modular right ideal of R. Then (K : R) ⊆ K.
Proof.Since K is a right modular right ideal of R, by Proposition 3.5, there ia a right Rgroup G with a generator g such that K = (0 : g).Therefore, K = (0 : g) ⊇ (0 : G) = (0 : R/K) = (K : R).Proposition 3.11.Let R be a zero-symmetric near-ring and let K be a right ideal of R right modular by e. Then (K : R) = (K : eR) and the largest ideal of R contained in K is the largest ideal of R contained in (K : R).
Proof.Since eR ⊆ R, (K : R) ⊆ (K : eR).Let x ∈ (K : eR).Now eyx ∈ K, for all y ∈ R.But yx − eyx ∈ K, for all y ∈ R. Therefore, yx ∈ K, for all y ∈ R, that is, x ∈ (K : R).So (K : eR) ⊆ (K : R).Therefore, (K : R) = (K : eR).Let J be the largest ideal of R contained in K.For x ∈ J, Rx ⊆ J ⊆ K. Therefore, J ⊆ (K : R).Let I be an ideal of R contained in (K : R).By Corollary 3.10, (K : R) ⊆ K. So, I ⊆ K. Therefore, I ⊆ J. Hence, J is the largest ideal of R contained in (K : R).Proposition 3.12.Let P 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.
Proof.Let P be an ideal of a zero-symmetric near-ring R. Suppose that P is a right 0-primitive ideal of R.So we get a maximal right modular right ideal K of R such that P is the largest ideal of R contained in K. Now by Proposition 3.8, R/K is a right R-group of type-0.By Proposition 3.11, P is the largest ideal of R contained in (K : R) = (0 : R/K).Conversely, suppose that P is the largest ideal of R contained in (0 : G), where G is a right R-group of type-0.Now G is R-isomorphic to R/K for some maximal right modular right ideal K of R.So (0 : G) = (0 : R/K) = (K : R).Since P is the largest ideal of R contained in (0 : G) = (K : R), by Proposition 3.11, P is the largest ideal of R contained in K. Hence, P is a right 0-primitive ideal of R.
We recall some of the definitions and results of [3] which are required to observe that right Jacobson radicals are relevant for the study of near-rings in terms of matrix nearrings.
Matrix near-rings were introduced in [ Suppose that any two minimal right ideals of R are isomorphic as right R-groups.Then, a direct sum of minimal right ideals K i and is (isomorphic to) a matrix near-ring M n (S).
Proof.Since J r 1/2 (R) = {0} and R has DCC on right ideals of R, we get that the intersection of a finite number of maximal right modular right ideals of R is zero.So, R is a direct sum of a finite number of minimal right ideals K 1 ,K 2 ,...,K n of R. By Proposition 2.12, R has a left identity as the intersection of a finite number of maximal right modular right ideals of R is zero.Since R is a simple near-ring with left identity, it has an identity.Also, since by our assumption any two minimal right ideals of R are isomorphic as right Rgroups, by Proposition 4.11, R has a set of matrix units {e i j | 1 ≤ i, j ≤ n}.Therefore, by Theorem 4.12, R is (isomorphic to) a matrix near-ring M n (S).
Example 4.14.We give an example of a nonring which satisfies the hypothesis of Theorem 4.13.Let G be a finite simple nonabelian additive group.By [3,Corollary 19], E(G 2 ) is isomorphic to the matrix near-ring M 2 (E(G)).As mentioned soon after [3,Corollary 19], E(G 2 ) = M 0 (G 2 ).So, M 0 (G 2 ) is a simple d.g.near-ring with DCC on right ideals.Let i ∈ {1, 2}.Let G 1 = G × {0} and let G 2 = {0} × G. Since G i is a maximal (minimal) normal subgroups of G 2 , K i = (G i : G 2 ) = {m ∈ M 0 (G 2 ) | m(a) ∈ G i , for all a ∈ G 2 } is a maximal right ideal of M 0 (G 2 ).Moreover, K 1 ∩ K 2 = {0}.Thus J r 1/2 (M 0 (G 2 )) = {0}.This shows that M 0 (G 2 ) = K 1 ⊕ K 2 , where K i is a minimal right ideal of M 0 (G 2 ).Define e i : G 2 → G i by e i ((a 1 ,a 2 )) = (b 1 ,b 2 ), where b j = a i if j = i and 0 if j = i.Now e i is a group homomorphism and hence it is a distributive idempotent in M 0 (G 2 ) and e i M 0 (G 2 ) ⊆ K i .Since e 1 and e 2 are orthogonal distributive idempotents in M 0 (G 2 ) and e 1 + e 2 = 1, by [3, Proposition 2], we get that e i M 0 (G 2 ) is a right ideal of M 0 (G 2 ).Thus, K i = e i M 0 (G 2 ).R. S. Rao and K. S. Prasad 13 The mapping e 12 : G 2 → G 2 defined by e 12 ((a 1 ,a 2 )) = (a 2 ,0) is an endomorphism of G 2 .So, e 12 is a distributive element in M 0 (G 2 ).It is an easy verification that the mapping h : e 2 M 0 (G 2 ) → e 1 M 0 (G 2 ) defined by h(e 2 m) = e 12 (e 2 m) is an isomorphism of the right M 0 (G 2 )-groups.So, K 1 and K 2 are isomorphic as right M 0 (G 2 )-groups.Since a minimal right ideal K of M 0 (G 2 ) is isomorphic to K j for some j ∈ {1, 2} as right M 0 (G 2 )-groups, we get that any two minimal right ideals of M 0 (G 2 ) are isomorphic as right M 0 (G 2 )groups.So, M 0 (G 2 ) satisfies the hypothesis of Theorem 4.13.
Example 4.15.Let G be a finite simple nonabelian additive group.Now by Pilz [2, Corollary 7.48], E(G) = M 0 (G).So, M 0 (G) is a finite simple d.g.near-ring with identity.Moreover, J 2 (M 0 (G)) = {0} and each minimal left ideal of M 0 (G) is isomorphic to G as left M 0 (G)-groups.Since each distributive element of M 0 (G) is an endomorphism of (G,+),0 and the automorphisms of (G,+) are the only distributive elements of M 0 (G).Therefore, M 0 (G) has no nontrivial matrix units.Hence, M 0 (G) is not isomorphic to a matrix nearring M n (S), where n > 1.

Proposition 2 . 8 .
Let e ∈ R. Then e is right quasiregular if and only if no proper right ideal of R is right modular by e.

Proposition 2 .
11. Let K and L be right modular right ideals of R and R = K + L. Then K ∩ L is also a right modular right ideal of R. Proof.Suppose that K is right modular by e 1 and L is right modular by e 2 .Let e 1 = b 11 + b 12 and let e 2 = b 21 + b 22 , where b 11 ,b 21 ∈ K and b 12 ,b 22 ∈ L. Let e = b 21 + b 12 .Let r ∈ R; r − er = r − (b 21 + b 12 )r = r − b 12 r − b 21 r = (r − e 1 r) + (b 11 r − b 21 r) ∈ K; r − er = r − (b 21 + b 12 )r = r − b 12 r − b 21 r = (r − b 12 r + b 22 r − r) + (r − e 2 r) ∈ L. Therefore, r − er ∈ K ∩ L, and hence K ∩ L is a right modular right ideal.
1].Definition 4.10.Let R be a zero-symmetric near-ring with identity.A subset {e i j | 1 ≤ i, j ≤ n} of distributive elements in R is said to be a set of matrix units in R if and only if e 11 + e 22 + ••• + e nn = 1 and e rs e pq = δ sp e rq , where Proposition 4.11.Let R be a zero-symmetric near-ring with identity.R = K 1 ⊕ K 2 ⊕ ••• ⊕ K n ,a direct sum of n pairwise isomorphic right ideals K i of R as right R-groups if and only if R has a set of matrix units {e i j | 1 ≤ i, j ≤ n}.In this case K i = e ii R, for all 1 ≤ i ≤ n.
Theorem 4.12.Let R be a simple and d.g.near-ring with identity.Then R is isomorphic to a matrix near-ring M n (S) if and only if R has a set of matrix units {e i j | 1 ≤ i, j ≤ n}.Theorem 4.13.Let R be a simple d.g.near-ring with DCC on right ideals of R and J r 1/2 = {0}.