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Contents
Introduction 1
1 NEAR RINGS 41.1 Basic Definitions and Examples . . . . . . . . . . 41.2 Ordered near rings . . . . . . . . . . . . . . . . . 11
2 N-GROUPS 252.1 Semisimple N-groups . . . . . . . . . . . . . . . . 252.2 Chain conditions an N-groups . . . . . . . . . . . 41
3 Generalized derivations an near rings 493.1 α - derivations . . . . . . . . . . . . . . . . . . . 493.2 Generalized derivations . . . . . . . . . . . . . . . 54
Bibliography
i
IntroductionRing theory is a show piece of mathematical unification, bring-
ing together several branches of the subject and creating a pow-
erful machine for the study of problems of considerable histori-
cal and mathematical importance. Near - rings are generalized
rings. Near - rings arise in a natural way; take the set M(Γ)
of all mapping of a group (Γ,+) into itself, define addition (+)
pointwise and o as composition. Then (M(Γ),+, ◦) is a near
ring. Even if Γ is a abelian, only one distributive law is always
fulfilled. (f + g) ◦ h = f ◦ h + g ◦ h holds by the definition of
f + g, while for f ◦ (g + h) = f ◦ g + f ◦ h we would have to
assume that f is a homomorphism. Another example is supplied
by the polynomials w.r.t addition and substitution.
Near - ring provide non-linear theory of group mappings his-
torically, the first step towards near - rings was an axiomatic
research is done by Dickson in 1905. He showed that there
do exist fields with only one distributive law. Some years later
these field theory proved be useful in coordinating certain impor-
tance classes of geometric planes (Descartes plane and pauppian
planes). A part from the applications concerning axiomatics and
geometry mentioned above, the special classes of finite near-
rings give new and highly efficient classes of balanced in com-
plete block designs the characterize Frobenius group and hence
1
also finite groups with fixed point free automorphism groups.
The notation of the ring with derivation is quite and plays a
significant role in the integration of analysis, algebraic geome-
try and algebra. In 1940s it was found that the Galois theory
of algebraic equations can be transferred to the theory of or-
dinary differential equations. The study of derivations in rings
through initiated long back, but got impetus only after posner
who in 1957 established two very striking results on derivations
in prime rings. The notion derivation in rings has also been
generalised in various directions such as generalized derivation,
Jorchon derivations, generalized Jorden derivation also there has
been considerable intergest in investigating commutativity of
rings more often that of prime and semi prime rings. Analo-
gus to be concept of derivation of rings, so on the concept of
derivation on near - rings was intiated by H.E. Bell and G. Ma-
son in 1987 [BM]. Since then only a few paper appeared in this
topic [A], [BA], [H]. In 1991 Bresar. M introduced the notation
of generalized derivations corresponding to the derivation on a
ring R [B]. In [OG] generalized derivation of Prime near rings is
discussed. This motivated as to study the concepts of near rings
and generalized derivation of prime near rings. Our work is di-
vided into 3 Chapters. The first chapter on Near - Rings is
divided into two sections the first section deals with definitions
and examples and the second section deals with Near - ring.
2
The Second Chapter on -groups is divided into two sections the
first section deals with semisimple N- groups and the sec-
ond section deals with chain conditions on N-groups. The
third chapter discuses Generalized derivations on near -
rings. This chapters divided into two sections. The first sec-
tion deals with α derivations and the second section deals
with generalized derivations. Our work ends with a detailed
bibliography.
—————————————————
3
Chapter 1
NEAR RINGS
1.1 Basic Definitions and Examples
Definition 1.1.1 A near ring N is a set N together with two
binary operations, addition and multiplication, such that
i (N,+) is a group (not necessarily abelian)
ii (N, .) is a semigroup (not necessarily with an iden-
tity element) and
iii For all x, y, z in N , (x + y)z = xz + yz (Right
distributive law
If the above conditions are satisfied for all elements in N , then
N is called a right near ring. Similarly we can define a left near
ring. Through out our work we will consider only right near
rings.
Example 1.1.2 Let (G,+) be any group. Define a multiplica-
tion ’·’ on G as follows for all x, y in G , x · y = x.
4
Then (G,+, ·) is a near ring.
Proof
Given that (G,+) is a group. we have to check the remaining
conditions
i. (G, ·) is a semigroup.
ii. If x1, x2, x3 ∈ G then (x1 + x2) · x3 = x1x3 + x2x3
For x, y ∈ G, given that
x · y = x
⇒ ’·’ is colosed.
claim
(x · y)z = x · (yz)
L.H.S = (x · y) · z
= x · z
= x −−− (1)
R.H.S = x · (y · z)
= x · y
= x −−− (2)
From(1) and (2)
Therefore,
(x.y).z = x.(y.z) ∀ x, y, z ∈ G
5
⇒ ’·’ in G is associative.
Therefore (G, ·) is a semigroup.
Let x, y, z ∈ G.
Then x · z = x and y · z = y
∴ x · z + y · x = x+ y.
Also (x+ y) · z = x+ y by definition of ’·’ .
Hence (x+ y).z = xz + yz
Therefore (G,+, ·) is a near ring.
Example 1.1.3 Let N = R[x], the set of all polynomials in
x over the field of real numbers R. Define addition ’+’ and
composition ’◦’ on N as follows: For p(x), q(x) ∈ N ,
let
p(x) = a0 + a1x+ a2x2 + . . .
q(x) = b0 + b1x+ b2x2 + . . .
Then p(x) + q(x) = (a0 + b0) + (a1 + b1)x+ (a2 + b2)x2 + . . .
Definition 1.1.4 Near ring with identity
Let N be a near ring, N is said to be near ring with identity, if
there is an element 1 ∈ N such that 1.n = n.1 = n∀n ∈ N
Example 1.1.5 The set N of all polynomials over the field of
real numbers, with constant term o togethers, with constant term
6
o together with the usual addition and composition of polynoo-
mials, is a near ring.
Since (N, .) is a semigroup with an identity element 1, we call
N a near ring with an identity element 1.
Definition 1.1.6 Let N be a near ring and let S be a non-
empty subset of N . Then S is said to be subnear ring of N , if
S itselt is a near ring under the operations in N .
Example 1.1.7 Set of continuous function from R→ R forms
a subnear ring under usual addition and compsition of mapping
Definition 1.1.8 Constant element Let N be a near ring.
Let k ∈ N . Then k is said to be constant in N if kx = k ∀x ∈N
Definition 1.1.9 A near ring N is said to be constant near
ring if every element of N is constant.
Example 1.1.10 Example 1.1.2, (G,+) is group and ’.’ is de-
fined as xy = x ∀x, y ∈ G clearly each element of G is constant
in G. Therefore (G,+, .) is a constant near ring.
Example 1.1.11 Consider example 1.1.3, we observe that 0.p(x) =
0 ∀p(x) ∈ IR[x] = N ,
Therefore 0(zero) is ther only constant term of N .
7
Lemma 1.1.12 An element x in N is a constant if and only if
x = x0
proof
Assume that x in N is a constant.
claim x = x0
X is constant ⇒ xy = x∀y ∈ N — (1)
Equation (1) is true for all y ∈ NTherefore equation (1) is true for y=0
Therefore x0 = x which is required.
Conversely assume that x = x0 — (2)
claim x is a constant in N
ie., we have to prove xy = x ∀y ∈ NGiven x = x0
For any y ∈ N .
⇒ xy = (x0)y
= x(0y) Associative law holds inN
= x0 0is the constant[0y = 0]
= x using (2)
⇒ xy = x ∀y ∈ N which is required
Lemma 1.1.13 Let N be a near ring. Let k(N) be the collec-
tion of all constant element in N . Then K(N) is a subnear ring
8
of N
Proof
K(N) = {k ∈ N/ k is constant inN}
= {k ∈ N/kx = k ∀x ∈ N}
claim
k(N) is a subnear ring of N
we have to prove
(i) (k(N),+) is a group
(ii) (k(N), .) is a semigroup
(iii) ((x + y).z = xz + yz ∀x, y, z ∈ K(N) (Right dis-
tributive law)
(k(N),+) is a group of N .
Therefore 0.x = 0 ∀x ∈ N, 0 ∈ k(N)⇒ k(N) 6= ø Therefore
x.z = x, y.z = y ∀z ∈ NN is near ring,
x.z − y.z = x− y ∈ N
ie.x− y = (x.z − y.z) = (x− y).z
⇒ for forx, y ∈ k(N), x− y ∈ k(N)
(k(N),+) is a group.
To prove (k(N), .) is a semi group
ie., we have to prove
9
(i) If x, y ∈ k(N) then x.y ∈ k(N)
(ii) x(yz) = (xy)z ∀x, y, z ∈ k(N)
(i) Let x, y ∈ k(N)
claim
xy ∈ k(N)
(xy)z = xy∀z ∈ N
(x.y)z = x.(y.z)
= x.y∀z ∈ N
ie.,(xy.z = xy∀z ∈ N
⇒ x.y ∈ k(N),
’.’ in N is associative,and k(N) ≤ N ,
’.’ is associative in k(N) Since each element of k(N) is an
element of N . associative lawa is true for k(N).
(k(N), .) is a semigroup.
Right distributive law
Let x, y, z ∈ k(N)
claim
(x+ y).z = xz + yz
L.H.S = (x+ y).z
= (x+ y) −−− (1)
10
x+ y ∈ k(N)
R.H.S
xz + yz = x+ y −−− (2)
From, (1) and (2), L.H.S = R.H.S
(x+ y).z = xz + yz
distributive law is true
(k(N),+, .) is a subnear of N , since each element of k(N) is
constant.
(k(N),+, .) is a constant subnear ring of N
Definition 1.1.14 Let N be a near ring. Let k(N) be the col-
lection of all constant elements in N . The subnearring of N .
k(N) is called the constant subnear ring
1.2 Ordered near rings
Definition 1.2.1 Linear order (or) simple order (or) order re-
lation. Let N be a any non-empty set. Let C be any relation on
N . Then C is said to be order relation if the following conditions
are true,
(i) Comparability
For each x, y ∈ N for which x 6= y; either xCy or yCx
11
(ii) Non reflexivity
There is no x in N such that xCx holds (iii) Transitivity
If xCy and yCz then xCz, ∀x, y, z ∈ N
Example 1.2.2
let R be a real number system, ’<’ be the usual ordering of num-
bers,
We define C as follows;
C = {(a, b)/a < b, a, b ∈ R}
Note x < y means ”‘x ≤ y and x 6= y ”‘
Definition 1.2.3 Ordered near ring
Let N be a near ring.
Let 5 be a linear order on N . Then N is said to be an ordered
near ring if
(i) x 5 y ⇒ z + x 5 z + y and x+ z 5 y + z
(ii) x < y and 0 < z ⇒ xz < yz
Example 1.2.4 The set of all real numbers with linear order
< forms an ordered near ring
Definition 1.2.5 Positive element
Let N be an ordered nearing.Let x ∈ N . Then x is said to be a
positive element if o < x.
12
Notation
Let P denote the set of all positive elements in N .
p = {x ∈ N/0 < x}
Lemma 1.2.6 Let N be an ordered near ring. Let P denote
the set of all positive elements in N . Then the following are
true
(i) P is not empty if N 6= O
(ii) If a ∈ P , then for all x in N − x+ a+ x ∈ p
(iii) If a, b ∈ P , then a+ b ∈ P and ab ∈ P
(iv) For any element x in N , exactly one of the following
conditions holds:
x = 0, x ∈ p,−x ∈ p
Proof (i) P is not empty if n 6= 0
Assume that n 6= 0
There is an element x in N with x 6= 0
Since N is an ordered near ring, either
O < x or O > x hold. If O < x, x ∈ p⇒ P is non-empty
If O > x,O < −x ⇒ x ∈ p ⇒ P is non-empty Hence (i) is
proved. To Prove (ii) If a ∈ p , then for all x in N −x+a+x ∈p.
13
Let a ∈ p. Then a > 0
Let x ∈ N
a > 0 ⇒ −x+ a > −x+ 0
⇒ (−x+ a) + x > −x+ x
⇒ −x+ a+ x > 0
⇒ −x+ a+ x ∈ p
Hence (ii) is proved
To Prove
(iii) Let a, b ∈ P, a > 0 and b > 0, a > 0
⇒ a+ b > 0 + b’¡’ linear order
⇒ a+ b > b > 0
⇒ a+ b > 0⇒ a+ b ∈ p
a > 0
ab > 0.b
ab > 0⇒ ab ∈ p
Hence (iii) is proved.
To Prove (iv) Let x ∈ Nwe know that N 6= 0
There is an element x ∈ N such that x 6= 0. since N is an
14
ordered near ring, only one of the following hold
1. x > 0
2. x < 0
If x > 0⇒ x ∈ PIf x < 0,−x > 0,−x ∈ pHence (iv) is proved.
The following lemma proves that the converse of the above lemma
is also true.
Lemma 1.2.7 Let N be a near ring
Let P = {x ∈ N ;x > 0} be subset of N , with the properties
1. P is not empty if N 6= 0
2. If a ∈ p, then for all x in N − x+ a+ x ∈ p
3. If a, b ∈ P then a+ b ∈ P , and ab ∈ p, and
4. For any element xin N , exactly one of the fol-
lowing conditions holds x = 0, x ∈ P,−x ∈ P
Then N is an ordered near ring.
proof To Prove the Lemma
We have to prove that there is a linear order in N which satisfies
the following conditions
15
(i) x 5 y ⇒ z + x ≤ z + y and x+ z ≤ y + z
(ii) x < y and 0 < z ⇒ xz < yz
we define linear order on N as follows, For all x, y in N , x ≤ y
iff either
x = y or − x+ y ∈ p
we have to prove that this order 5 satisfies the above two con-
ditions.
Now assume that x ≤ y ⇒ x = y or −x+ y ∈ p
If x = y, z + x = z + y
and x+ z = y + z
⇒ z + x ≤ z + y and x+ z ≤ y + z
If − x+ y ∈ p then
z + x 5 z + y
and x+ z 5 y + z
−x+ y ∈ p
⇒ −x+ y > 0
⇒ −z − x+ y > −z
⇒ (−z − x) + y + z > −z + z
⇒ −(z + x) + (z + y) > 0
⇒ −(z + x) + (z + y) ∈ p
⇒ −z + x ≤ z + y
16
Simillarly we can Prove x+ z ≤ y + z
Which Proves (i)
Assume x < y and 0 < z
Let x < y and 0 < z
0 < z ⇒ z ∈ p
x < y means x = y (or) −x+ y ∈ p
if x = y, xz = yz
⇒ xz ≤ yz
if− x+ y ∈ p, (−x+ y) > 0
(−x+ y)z > 0.3
⇒ −xz + yz > 0
⇒ −xz + yz ∈ p
⇒ xz ≤ yz0
This is true the cases ’5’ is an ordered relation in N . (N,+, .)
is an ordered near ring.
Theorem 1.2.8 Let N be an ordered near ring with and iden-
tity element 1. If for a, b in N, ab = 0, then either a = 0
(or)b = 0
17
Proof Let N be an ordered near ring For a, b ∈ N assume that
ab = 0
Claim Either a = 0 or b = 0
Suppose that,a 6= 0 and b 6= 0 case (i) let b ∈ p Then b > o
a ∈ N such that 6= o, either a ∈ P or −a ∈ PIf a ∈ p, ab ∈ p(b ∈ p)⇒ ab > 0
⇒⇐ to ab = 0
Our assumption is wrong.
If − a ∈ p, −a > o
(−a)b = −(ab)
= −(0)
= 0
(−a)b = 0 −−−−−−(1)
−a ∈ p = b ∈ p
= −ab ∈ p
= −ab > 0 −−−−(2)
From (1) and (2) use get a contradiction
case (ii)
18
Let −b ∈ p,
Let b′ = −b
⇒ −b′ = b
= a(−1)b′
= a(−1)b′
o = a′b′where a = a(−1)
a′b′ = 0 and b′ ∈ P
by case (i) once again we get a contradiction.
unless a′ = 0
suppose a′ = 0
⇒ a(−1) = 0 ———————(i)
Hence 0 = 0(−1)
= [a(−1)(−1)]
= a(1)
0 = a
⇒ 0 = 0
This is a contradiction to a 6= o our assumption is wrong,
If ab = 0 then either a = 0 or b = 0
holds
Hence the Proof.
19
Lemma 1.2.9 Let H be a subgroup of (N,+), where N is an
ordered near ring with P as its set of Positive elements. Then
H = 0 if and only if H ∩ P = ø
Proof Let N be an ordered near ring. Let H be a subgroup of
N .
Assume H = 0
Claim H ∩ p = ø
Suppose that H ∩ p = ø
⇒ Letx ∈ H ∩ p
⇒ Letx ∈ Handx ∈ p
⇒ Letx = 0andx > 0
Which is a contraciction.
Our assumption is wrong.
H ∩ p = ø
Conversely assume that H ∩ p = ø
Claim H = 0
Suppose that, H 6= {o}There is an element x ∈ H with x 6= 0
Since H is a subgroup of N , by lemma 1.2.7
x > 0 or x < 0
If x > 0 thenx ∈ p
x ∈ Handx ∈ p ⇒ x ∈ H ∩ p
20
⇒⇐ to H ∩ p = ø
If x < 0,−x > 0,−x ∈ pSince H is a subgroup of N , and x ∈ H,−x ∈ H and −x ∈ p⇒ −x ∈ H ∩ Pwhich is a⇒⇐ to H ∩ p = ø
our assumption is wrong.
H = {0} holds.
Hence the Proof.
Theorem 1.2.10 Let N be a non trivial ordered near ring.
Then N = k(N) if and only if k(N) 6= {0}
proof Let N be a nontrivial ordered near ring.
Assume N = k(N)
To Prove k(N) 6= {0}Since N is not trivial N 6= 0
⇒ k(N) 6= {0}
conversely assume that k(N) 6= {0}To Prove N = k(N)
We have to prove that N = Set of constants in N
(i,e)., we have to prove (i) N ≤ k(N) (ii) k(N) ≤ N
k(N) = {x ∈ N : xy = x∀y ∈ N} is subset of N
21
Condition (ii) is satisfied.
We have to prove condition (i) only N ≤ k(N)
We know that by the Lemma 1.2.9 H = 0 ⇒⇐ h ∩ p = ø ”
Since k(N) 6= 0 k(N) ∩ p 6= ø
Let k ∈ k(N) ∩ p
⇒ k ∈ k(N)andk ∈ p
⇒ kx = k∀x ∈ Nand k > 0
⇒ kx = k > 0⇒ kx > 0⇒ kx ∈ p
and also x ∈ p, k ∈ p⇒ xk ∈ pLet n = −xk + x ∈ NSince N is an ordered near ring one of the following hold :
(i)n = 0 (or) (ii)n > o (or) (iii)n < 0
xk + x = 0,−xk + x > 0,−xk + x < 0
If n > 0
−xk + x > 0
⇒ (−xk + x)k > 0k
⇒ (−xk)k + xk > 0
⇒ (−x)(kk) + xk > 0
⇒ −xk + xk > 0 [k ∈ k(N)]
⇒ 0 > 0
22
which is a ⇒⇐If n < 0
−xk + x < 0
xk − x > 0
(xk − x)k > 0.k
⇒ x(kk)− xk > 0
⇒ xk − xk > 0
⇒ 0 > 0
which is a ⇒⇐n = 0 is hold.
⇒ −xk + x = 0
⇒ −xk = x
x0 = (xk)0
= x(k0)0
= xk
= x
⇒ x0 = x
p ∈ k(N) thus N ∈ k(N)
Hence the Proof.
23
Corollary 1.2.11 If N is an ordered near ring which is not a
constant near ring then k(N) = {0}
proof Let N be an ordered near ring with k(N) 6= N . But we
know that ” The above theorem 1.2.10 N = K(N) if and only if
K(N) 6= {0}” gives us N 6= K(N)⇒ K(N) = 0
Hence the Proof.
Corollary 1.2.12 If N is an ordered near ring with identity
element 1. different from 0(zero), then k(N) = 0
Proof Let N be an ordered near ring with identity element 1.
different from Zero
Claim K(N) = 0
Suppose that k(N) 6= 0
By the above theorem 1.2.10 we observe that
”K(N) 6= 0⇒ k(N) = N”
Given 1 ∈ N and 0 ∈ N1 ∈ k(N)⇒ 1 is a constant.
However 1.x = x∀x ∈ N, 1.x = 1∀x ∈ N
⇒⇐
K(N) = 0
Hence the Proof.
—————————————————————
24
Chapter 2
N-GROUPS
2.1 Semisimple N-groups
Notation
To distinguish additive identity of two groups, N and M , we
denote 0N and 0M
Definition 2.1.1 Let N be a near ring. Let (M,+) is an
additive group. M is said to be an N group, if the mapping
N × M → M (given by (x,m) → xm)satisfies the following
conditions.
i ((x+ y)m = xm+ ym∀x, y ∈ N and m ∈M .
ii (xy)m = x(ym)∀x, y ∈ N and m ∈M
Example 2.1.2 Taking M = N in the definition, the near ring
N has the structure of N-group (n,+).
For condition(i) is know as right distributive law.
condtion (ii) associative law which are holds in N
25
Definition 2.1.3 Let M be a N- group. Let H be a non-empty
subset of M . Then H is called N-subgroup of M if H is itself
form a N-group with operations on M .
Proposition 2.1.4
Let N be a near ring, x an element of N which is a constant
and M an N-group. Then
i For all m in M,xm = xoM .
ii If xoN = oN , then xoM = oM
Proof Let N be a near ring.
Let x ∈ NLet M be a N-group.
Assume that x is a constant in N .
claim
xm = xoM∀x ∈ N and m ∈M
x is a constant in N ⇒ xy = x∀y ∈ NTake y = oN xoN = X —————(1)
x.m = (x+ oN).M
ie., xm = oM = xm+ oM , each term above is an element in M
and M is an additive group.
Using Left cancellation law. oM = oN .m
xm = (x.oN)m
26
= x(oN .m)
= xoM
xm = xoM∀m ∈M
Hence (i) is proved.
(ii) Assume that xoN = oN
claim
xoM = oM
xoM = (xoN)oM
= x(NoM)
= (x.oN)oM {x is a constant}
= xoM = oM
xoM = oM
Hence (ii) proved .
Definition 2.1.5 Let N be a near ring.
Let M,M ′ be an two N-groups.
Then the mapping f : M →M ′ is said to be N- homomorphism
if
i f(a+ b) = f(a) + f(b)∀a, b ∈M
ii f(xa) = xf(a)∀x ∈ N and a ∈M
27
Definition 2.1.6 Let M be a N-group. Let k be a subset of
M . Then k is called an N-submodule of M . if
i k is a normal subgroup of (M,+)
ii x(m+ a)− xm ∈ k∀x ∈ N,m ∈M,a ∈ k
Theorem 2.1.7 If f : M → M ′ is an N-homomorphism from
M onto M ′ and k′ is an N-submodule of M ′ then f induces
an N-isomophism of N-groups f ′ : m/k → M ′/k′ where k =
f−1(k′)
Proof Since f : M →M ′ is an N-homomorphism from M onto
M ′ it satisfies the following conditions,
i f(a+ b) = f(a) + f(b)
ii f(xa) = xf(a)∀∀a, b ∈Mx ∈ N
since k is an submodule of M ′ it satisfies the following conditions
i k is an normal subgroup of M ′
ii x(m+a)−xm ∈ k where x ∈ N,m ∈M ′ and a ∈ k
claim
f induces an N-homomorphism of N-groups
f ′ : m/k →M ′/k′ where k = f−1(k′)
since f is onto, f−1(k′) exists.
Let k = f−1(k′) = {b ∈M : f(b) ∈ f(k)}.
28
Then k is a N-submodule of M .
M/k exists and M ′/k′ exists.
In a mapping f ′ : m/k → M ′/k′ which is defined f ′(ck) =
c′k′∀c ∈M, c′ ∈M ′
claim f is an isomorphism.
we know that f : M → M ′ is onto homomorphism k and k′
are subgroups of M and M ′ respectively. Therefore f ′ : M/k →M ′/k′ is an onto N-homomorphism.
Let f ′(c, k) = f ′(c′1k′)
⇒ c′1k′ = (c′1)
′(k′)′
⇒ c1k = c′1k′
f ′(c1k) = f(c′1k′)⇒ c1k = c′1k
′
⇒ f is one - one.
f ′ : M/K →M ′/K ′ satisfies all conditions of isomorphism .
f is an isomorphism
Hece proved
Theorem 2.1.8 Let M be an N-group and let M = M1 ⊕M2
(direct sum of subgroups) where Mii = 1, 2 is an N submodules
of M . Then if for any m in M , M = m1 +m2,miinMi, i = 1, 2
then xm = xm1 + xm2, x ∈ N
proof Consider −xm 6= x(m1 +m2)
Let M be an N-group
29
Let M = M1⊕M2. therefore any element m in M can be written
in a unieque way as m1 +m2
where m1 ∈M1 and m2 ∈M2
since each Mi is an N-submodule of M , it satisfies the following
conditions
i Mi is normal subgroup of M, i = 1, 2
ii for each x ∈ N,m ∈ M,mi ∈ Mi the expression
x(m+mi)− xm ∈ mi(i = 1, 2)
consider −xm1 + x(m1 +m2)
we know that M2 is N submodule of M ,
−xm1 + x(m1 +m2) = l2 (say) ∈M2
⇒ −xm1 + x(m1 +m2) = l2
⇒ −xm1 +−xm1 + x(m1 +m2) = xm1 + l2
⇒ x(m1 +m2) = xm1 + l2 −−−−− (1)
consider x(m1 +m2)− xm2
we know that M1 is an N- submodule of M .
Therefore x(m1 +m2)− xm2 ∈M1
⇒ x(m1 +m2)− xm2 = l1 (say)
⇒ x(m1 +m2)− xm2 + xm2 = l1 + xm2
⇒ x(m1 +m2) = l1 + xm2 −−−−− (2)
30
From (1) and (2)
x(m1 +m2) = xm1 + l2 = l1 + xm+ 2 −−−−− (3)
since M = M1 ⊕M2 every element in M can be written as a
unique way, from (3)
We obtain
l1 = xm1, l2 = xm2 −−−−− (4)
put (4) in (3)
x(m1 +m2) = xm1 + xm2
xm = xm1 + xm2
Hence the proof.
Theorem 2.1.9 Let M = M1⊕M2 be a direct decomposition of
a N-group M as a direct sum of its N-submodule M1,M2. Then
any N-submodule of Mi(i = 1, 2) is an N-submodule of M .
Proof
Let M = M1 ⊕ M2, where M is an N-group Mi is an N-
submodule of M(i = 1, 2)
To prove
any N-submodule of Mi of Mi is an N-submodule of M
Let L be an N-submodule of M1
claim
L is an N-submodule of M1. since Lis a normal subgroup of
M1, L is a normal subgroup of M .
31
Let x ∈ N,m ∈M and a ∈ L claim
x(m+ a)− xm ∈ Lm = m1 +m2 where m1 ∈M1 and m2 ∈M2
x(M + a)− xm = x(m1 +m2 + a)− x(m1 +m2) −−− (1)
x(m1 +m2 + a)− x(m1 +m2) = x((m1 + a) +m2)− x(m1 +m2)
= x(m1 + a)xm2 − xm1 − xm2
x(m1 +m2 + a)− x(m1 +m2) = x(m1 + a) + xm2 − xm1 − xm2
x(m1 +m2 + a)− x(m1 +m2) = x(m1 + a)− xm1
⇒ from(1), x(m+ a)− xm = x(m1 + a)− xm1 ∈ LL is an N-submodule of M1
Similarly we can prove (L is an N submodule of M2)
L is an N-submodule of M .
Hence the Proof.
Theorem 2.1.10 Let M1,M2,M3 be N-subgroups of an N group
M . If M1 ⊇ M2 and M2 and M3 are normal subgroups of
(M,+), then M1 ∩ (M2 +m3) = M2 +m1) ∩M3.
Proof
Let x ∈M1 ∩ (M2 +M3)
⇒ x ∈M1 and x ∈M2 +m3 and x = m2 +m3
m2 ∈M2,m3 ∈M3
32
⇒ x = m2 +m3 ∈ m1
⇒ x = m2 +m1andx ∈M3
⇒ x ∈ m2 +m1andx ∈M3
⇒ x ∈ (m2 +m1) ∩M3
⇒ M1 ∩ (M2 +M3) ⊆ (M2 +m1) ∩M3
Similarly
(M2 +m1) ∩M3 ⊆M1 ∩ (M2 +m3)
M1 ∩ (M2 +M3) = M2 + (M1 ∩M3)
Hence the Proof
Definition 2.1.11 A family {mi/i ∈ I} of N-submodules of
an N-group M is said to be independent if for each J in I,Mj ∩∑i6=ji∈I Mi) = 0
Remark 2.1.12 If {mi/i ∈ I} is an independent family of
Nsubmodules, them Mi ∩Mj = 0 for all i, j in I, i 6= j. since
Mi are normal subgroups of (M,+), it follows that when i 6= j,
the elements of Mi commute with elements of Mj, with respect
to addition in M .
33
Remark 2.1.13
Any subfamily of an independent family of N submodules of an
N-group is itself independent.
Remark 2.1.14 Since a representation of an element of the
sum of a family of N-submodules involves only a finite subfamily,
a family of N submodules is independent if anf only if every finite
subfamily is independent.
Proposition 2.1.15
Let {mi/i ∈ I} be an idependent family of N-submodules of an
N group M and L an N submodule of M such that L∪∑
i∈I Mi.
Then L∪Mi/i ∈ I is an independent family of Nsubmodules of
M .
Proof To prove {L} ∪ {Mi/i ∈ I} is an independent family.
i.e., we have to prove
Mj ∩ (L+
i 6=j∑i∈I
Mi) = 0∀i, j
Let x ∈Mj ∩ (L+∑i 6=j
i∈I Mi)
⇒ x ∈Mjandx ∈ L+∑i∈I
Mi)
⇒ x ∈Mjandx = y + x1 + x2 + ...+ xn wherey ∈ Landxi ∈Mi
⇒ x = y + x1 + x2 + ....+ xn ∈Mj
⇒ y = x− x1 − x2 − ....−Xn ∈i6=j∑i∈I
Miandy ∈ L
34
also
y ∈ L ∩i6=j∑i∈I
Mi
But we are given that
L ∩∑i∈I
Mi = 0
⇒ y = 0
ie., x− x1 − x2 − ...−Xn = 0
⇒ x = x1 = x2 = .... = xn
⇒ 0 ∈Mj ∩ (L
i 6=j∑i∈I
Mi)
⇒ Mj ∩ (L
i6=j∑i∈I
Mi)
{L} ∩ {mi/i ∈ I} is an independent family of N submodules
of M
Hence the proof.
Definition 2.1.16 Let M be an N group and {mi/i ∈ I} be a
family of N-submodule of M . Then M is said to be the direct
sum of the family {mi/i ∈ I}, if
i M =∑Mi and
ii The family {mi/i ∈ I} is independent where M =∑
i∈I ⊕Mi
35
Definition 2.1.17 A non trivial N group M is said to be ir-
reduciable if it has no N submodule other than (o) and M
Theorem 2.1.18 Let {mi/i ∈ I} be a family of irreducible N-
submodules of on N group M such that M =∑
i∈I Mi. If L is
an N submodule of M then there exists a subfamily {mi/i ∈ I}with J ⊆ I such that M = L⊕
∑j∈J ⊕Mj
Proof
Given that {mi/i ∈ I} be a family of irreduciable N submodule
of an N -group M .
M =∑
i∈I Mi, L is an N-submodule of M .
To prove that M = L⊕∑
j∈J ⊕Mj
Consider the family E of independent sets of N submodules of
the form {L} ∪ {Mk/k ∈ K} where K ⊆ I. Partially order
this family of independent sets of N-submodules by inclusion.
Every chain of such sets in this family is bounded above, by
its union which belongs to this family so that this family E is
inductive by ”An inductive partially ordered set has a maximal
element” there exists in this family E a maximal element say
{L} ∪ {Mj/j ∈ J} with J ∈ I. Now for each i in I, since Mi is
irreducible, Mi ∩ (L+∑
j∈J Mj = 0 or Mi
In the first case we can augment the family {L}∪{Mj/j ∈ J}with Mi to obtain an independent set in the family contradicting
the maximality of the element {L} ∪ {Mj/j ∈ J}
36
Hence Mi ∩ {L+∑Mj} = Mi ∪ {Mj/j ∈ J} ie.,
Mi ⊆ L+∑
j∈J Mj, i in I being arbitrary,
M = L⊕∑j∈J
⊕Mj
Hence the proof �
Definition 2.1.19 An N group M is said to be semisimple if
M =∑
i∈I Mi where {Mi/i ∈ I} is a family of irreduciable N
submodules of M
Remark 2.1.20
A homomorphic image of an irreducible N group it either 0 or
else irreducible so that homomorphic images of semisimple N
groups are semisimple.
Theorem 2.1.21 Every N submodule L of a semisimple N
group M is a direct summand of M , ie., there exists an N sub-
module L′ of M such that M = L⊕ L′
Proof
Given that every N submodule L of a semisimple N-group M is
a direct summand of M ,
Let N submodule L′ of M
To prove that M = L⊕ L′
In the proof of theorem 2.1.18
37
Put L′ =∑
j∈J Mj, where {L} ∪ {Mj/j ∈ J} is the maximul
element of the family E
Hence M = L⊕ L′ �
Theorem 2.1.22 Let M =∑
i∈I Mi be a sum of irreducible N
submodules of M . Then there exists a subset J of I such that
M =∑
j∈J ⊕Mj
Proof Given that M =∑
i∈I Mi be a sum of irreducible N -
submodules of M .
A subset J of I
To prove that M =∑
j∈J ⊕Mj
In the proof of theorem 2.1.18
Put L = 0
M =∑j∈J
⊕Mj
Hence the proof. �
Theorem 2.1.23 Let M be an N group such that every N sub-
module of M is a direct summand of M . Then M is semisimple.
Proof
Let M be an N group. Let every N submodule of M is a direct
summand of M .
To prove that M is semisimple.
Let L be an N submodule of M and L1 an N submodule of L.
Then L being a direct summand of M , by using theorem 2.1.18.
38
L1 is an Nsubmodule of M and so by the hypothesis in the the-
orem, L1 is a direct summand of M .
Thus M = L1 ⊕ L′1 for a N submodule L′1 of M .
by using Proposition 2.1.10. L−L1⊕(L∩L′1) every N-submodule
of L is a direct summand of L. Now if M is an irreducible N
group, it is semisimple.
Assume therefore that M has a proper N-submodule L with
L 6= 0 and L 6= M .
Let m ∈M and m 6= L. consider the family F of all N submod-
ule L′ such that, (i) L′ ⊇ L and (ii) m 6= L′. since L ∈ F , F
is notempty. Also any chain of the elements in F has an upper-
bound in F the union of the elements in the chain thus by using
” An inductive partially ordered set has a maximul element.”
F has a maximal element say k′. k′ being a direct summand,
M = k⊕ k′, for N-submodule k. certainly k 6= 0, since m ∈M .
but m 6= k′. we contend k is an irreducible N submodule. For if
not, the proof k = k1 ⊕K2 where ki, i = 1, 2 are non-trivial N-
submodules of k. If m ∈ k′+k1 (or) m 6= k′+k2 contradiction to
the maximality of k′. Hence m ∈ (k′+k1)∩(k′+k2). by using the
Proposition 2.1.10, (k′+k1)∩(k; +k2) = (k′+k1)∩k2 = k′+0 = k′
a contradiction. Thus k is irreducible. Let now M ′ be the sum
of all irreducible N submodule of M . Tthen M = M ′ + M ′1. if
M ′1 6= 0, since M ′
1inherits the property asserted in the hypoth-
esis there exists in M ′1 an irreducible N submodule M”1 of M ′
1
39
which is simultaneously an irreducible N-subodule of M . But
M”1 ⊆M ′, a contradiction. So M = M ′
Hence M is semisimple. �
Theorem 2.1.24 For a non-trivial N group M , the following
conditions are equivalent.
1. M is a sum of family of irreducible N submodules of M
2. M is a direct sum of a family of irreducible N-submodules
of M .
3. Every N submodule of M is a direct summand of M
Proof
1. M is a sum of family of irreducible N submodules of
M ⇒ M is a direct sum of a family of irreducible N sub-
module of M . By using theorem 2.1.22
2. M is a direct sum of a family of irreducible N-submodules
of M ⇒ every N -submodule of M is a direct -summand of
M . By using theorem 2.1.21.
3. Every N submodule of M is a direct summand of M ⇒M is a sum of a family of irreducible N submodules of M .
by using theroem 2.1.23
Hence the proof �
40
2.2 Chain conditions an N-groups
Definition 2.2.1
(i) M = M1 ⊇M2 ⊇M3 ⊇ . . . ⊇Mr ⊇ ... and
(ii) M = M1 ⊇ M2 ⊇ M3 ⊇ ... ⊇ Mr ⊇ ...are chains of
N- subgroups such that each N-sub group of the chain is an
N - sub module of the Preceding N - subgroup.
(i) is said to be refinement of (ii) if for each i = 1, 2..., the term
m′1 of (ii) occurs Mj in (i) for some j.
Definition 2.2.2 A desending normal chain of N - subgroups
of an N-group M is a chain of N - sub groups of M of the
form M = M0 ⊇ M0 ⊇ M0 ⊇ ...... where each Mi+1 is an
N-Submodule of mi, i = 0, 1
Definition 2.2.3 Let M be an N-group. Let {M−i}i ∈ I be a
normal chain of N-groups. It is said to be a strictly descending
if for each iMi )Mi+1
Definition 2.2.4 An N-group M satisfies the maximal condli-
tion for N-Submodules if for every non-empty set of N-Sub mod-
ules of M has maximal element.
Definition 2.2.5 An N-group M is said to be notherian if for
any N- Subgroup L of M such that L = L1 ⊆ L2 ⊆ ... ⊆ Lk = M
41
with li and N-Submodule of Li+1, i = 1, 2...k− 1. L Satisfies the
maximal condition for the N-Submodules of L.
Definition 2.2.6 An N-group M is said to be artination if its
satisfies the descending chain condition.
M = M0 ⊃M1 ⊃M2 ⊃ ...Mn = (0)
Where each Mi+1 is an N - Submodule of Mi for each i
Proposition 2.2.7 Let M1,M2,M′1,M
′2 be N-Subgroups of an
N-group M such that M ′1 is an N-Submodule of Mi, i = 1, 2
Then M ′1 +M1 ∩M ′
2 and M ′1 +M1 ∩M2 are N subgroups of M
such that M ′1 +M1 ∩M ′
2 is an N submodule of M ′1 +M1 ∩M2
Proof Since M ′1 is an N-Submodule of M1 and M1 ∩M2 an
N-Subgroup of M ′i
By using Theorem 2.2.15
Similarly M ′1+ <1 ∩M2. Let a ∈M ′
1 +M1 ∩M2
n ∈M ′1 +M1 ∩M ′
2 and x ∈ NLet b = b1 + b2 with b1 ∈M ′
1 and b2 ∈M1 ∩M ′2 Thus x(a+ b)−
xa = x(a+ b1 + b2)− x(a+ b1) + x(a+ b1)− xaSince a+ b1 is in m1 and b2 ∈M1 ∩M ′
2 is in M1 ∩M ′2
so x(a + b1 + b2) − x(a + b1) ∈ M1 ∩M ′2 as M1,M
′2 is an N-
submodule of M1
By using Theorem 2.2.15
Also x(a+ b1 + b2)− xa is in M ′1,M
′1 being an N-Submodule of
42
M ′1. Thus x(a+ b)− xa belongs to M ′
1 +M1 ∩M ′2
Hence the Proof. �
Proposition 2.2.8 Let M1,M2,M′1,M
′2 be N-Subgroups of an
N-group M such that M ′i is an N-Submodule of Mi, i = 1, 2 Then
M ′2 +M2∩M ′
1 and M ′2 +M2∩M1 are N-Subgroups M such that
M ′2 +M2 ∩M ′
1 and N-Submodule of M ′2 +M2 ∩M1
Proof Since M ′2 is an N-Submodule of M2 and M2 ∩ M ′
1 an
N-Subgroup of Mi
By using Theorem 2.2.15
Similarly for M ′2 + M2 ∩M ′
1. Let a ∈ M ′2 + M2 ∩M ′
1, b ∈ M ′2 +
M2 ∩M ′1 and x ∈ N
Let b = B1 + b2 with b1 ∈M ′2 and b2 ∈M2 ∩M ′
1
a, b1, b2 are all in M2 and b2 ∈M2 ∩M ′1
Thus x(a+ b)− xa = x(a+ b1 + b2)− x(a+ b1 + x(a+ b1)− xaSince a+ b1 is in m2 and b2 is inM2 ∩M ′
1
so x(a + b1 + b2) − x(a + b1) ∈ M2 ∩M ′1 as M2,M
′1 is an N-
submodule of M2
By using Theorem 2.2.15
Also x(a+ b1)− xa is in M ′2,M
′2 being an N-Submodule of M ′
2.
Thus x(a+ b)− xa belongs to M ′2 +M2 ∩M ′
1
Hence the Proof.
Proposition 2.2.9 Let M1,M2,M′1,M
′2 be N- Subgroups of an
N-group M with M ′1 ⊆M1, and M ′
2 ⊆M2.
43
If for i = 1, 2. Mi is an N-Submodule of Mi.
then M ′1+M1∩M ′
2
M ′2+M1∩M ′
2
∼= M ′2+M2∩M1
M ′2+M2∩M ′
1
Proof Let K = M1 ∩M2 and L = M ′1 +M1 ∩M ′
2
K ∩L = (M1∩M2)∩ (M ′1 +M1∩M ′
2). M ′1 is normal in M1 and
M1M′2 an Subgroup of M1
k ∩ L = (M1 ∩M2) ∩ (M ′1 +M1 ∩M ′
2)
= M1 ∩M ′2 +M1 ∩M1 ∩M ′
1)
= M1 ∩M ′2 +M2 ∩M ′
1
Also M ′1 is normal in M ′
1 and M1 ∩M ′2 is a Subgroup of M1
k + L = M1 ∩M2 +M1 ∩M ′2 +M ′
1
= M1 ∩M2 +M ′1
So L is an N-Submodule of K + L
by using proposition : 2.2.7 consider the map L → (L + K) k
mapping x on to (x+K) for every x in L.
The isomorphism (L+K)/L ∼= K/L ∩K.
Thus (M ′1 +M ′
1 ∩M2)/(M′1 +M ′
1 ∩M ′2)
= (M1 ∩M2)/(M1 ∩M2 +M2 ∩M ′1)
Hence the Proof. �
Definition 2.2.10 Let M be an N-group and
44
i M = M1 ⊇ . . . ⊇Mr = (o) and
ii M = M ′1 ⊇ 1̇0 ⊇Mr = (o) be two narmal chains of
N-subgroups of M . We say that (1) and (2) are equivalent,
if r = s and there exists a permutation t of indices i =
1, 2 . . . r − 1, i→ t(i) such that Mi/Mi+1∼= Mt(i)/Mt(i)+1for
each i
Theorem 2.2.11 Let M be an N-group. Then any two normal
chains of N-Subgroups of M given by M = M1 ⊇ . . . ⊇Mr = (o)
and M = M ′1 ⊇ . . . ⊇Ms = (o) have equivalent refinements.
Proof Let the chains be
M = M1 ⊇M2 ⊇ . . . ⊇Mr = (o)and
M = M ′1 ⊇M ′
2 ⊇ . . . ⊇M ′s = (o)
Define for i = 1, . . . r − 1 and j = 1, . . . s,
Mij = Mi+1 +Mi∩M ′j
and for j = 1, . . . s− r and i = 1, . . . r − 1
M ′ij = M ′
j+1 +M ′j∩M ′
j
Mis = Mi+1 and M ′jr = Mj+1 for i = 1, . . . r − 1
and j = 1 . . . , s− 1
By using Proposition 2.2.9,
we have clearly for i = 1, . . . r − 1 and j = 1 . . . , s− 1
Mij/Mi,j+1∼= M ′
ji/M′j,i+1
Hence the Proof. �
45
Definition 2.2.12 An N-group M is said to have a compo-
sition series, if there exists a normal chain M = M1 ) . . . )
Mr = (0) such that each factor N - group Mi/Mi+1 is irreducible.
i = 1, . . . , r − 1
Theorem 2.2.13 Any two composition series of an N-group
M are equivalent.
Proof Let the two composition series
M = M1 ⊇M2 ⊇ . . . ⊇Mr = (o)and
M = M ′1 ⊇M ′
2 ⊇ . . . ⊇M ′s = (o)
Then the refinement Mij for (1) is such that exactly for one in-
dex j.
Mij/Mi,j+1∼= Mi/Mi+1 : Mij/Mi,j+1 being trivial for others.
similarly for the other composition series. But since Mij/Mi,j+1∼=
M ′ji/M
′j,i+1 there are as many non-trivial factors of the compo-
sition series M = M1 ⊇ M2 ⊇ . . . ⊇ Mr = (o) as there are for
are these are M = M ′1 ⊇ M ′
2 ⊇ . . . ⊇ M ′s = (o) are these are
isomorphic in some order.
Hence the proof. �
Definition 2.2.14 Let M = M1 ) M2 ) . . . ) Mn+1 = (o) be
an N-group M , we have a composition series. The length of the
composition series is defined as n. The n is uniquely determined
by M .
46
Theorem 2.2.15 Let M be a non-trivial N-group. Then M is
artinian as well as notherian if and only if M has a composition
series.
Proof
Let M be both artinian and notherian. Since M = M1 (say)
is noterian, the set of all N-Submodule L such that L 6= M
has a maximal element say M2. Now the notherian condition
of M implies that M2 satisfies the maximal condition for its N-
submodules. So there exists a maximal element say M3 in the
set of all proper N-submodule of M2 and so on.
We obtain then strictly descending chain of N-subgroups of M ,
M = M1 ⊇M2 ⊇ . . . with the properties
M = M1 ⊇ M2 ) . . . ) Mr = (o),Mi+1 is an N - submod-
ule of Mi and Mi/Mi+1 is an irreducible N-group, i = 1, 2 . . .
since M is artinian. This chain must terminate necessarily at
(o), i.e., there exists a positive integer n such that M = M1 ⊇M2 ) . . . ) Mn+1 = (o), with Mi+1 an N-submodule of Mi and
Mi/Mi+1 an irreducible N-group i = 1, 2 . . . n
M = M1 ⊇M2 ) . . . )Mr = (o) is a composition series for M .
conversely let M have a composition series of length n given
M = M1 ⊇ M2 ) . . . ) Mr = (o) so by comparing any normal
chain with M = M1 ⊇ M2 ) . . . ) Mn+1 = (o) by using Theo-
rem 2.2.11, one concludes that any normal chain of N-subgroups
47
of M without repetition can be refined to a composition series of
length n, So, no chain occuring in the definition of an artinian
and notherian N-group M can have length greater than n, ie.,
if L does not satisfy the maximal condition on its N - Submod-
ules, we obtain a contradiction by producing a normal chain of
N-subgroups of M containing more than n district terms. Hence
the notherian condition holds, similarly one proves that the ar-
tinian conditionn holds.
Hence the proof. �
48
Chapter 3
Generalized derivations an nearrings
3.1 α - derivations
Definition 3.1.1 Let R be an associative ring. A mapping
d : R→ R is called a derivation if
(i) d(x+ y) = d(x) + d(y)
(ii) d(x+ y) = d(x)y + xd(y)∀x, y ∈ N
Example 3.1.2 Let R be a ring of 2X2 matrices. Define d :
R→ R by
d
((x yz w
))=
(x −yz 0
)then d is a non-zero derivation on R
Example 3.1.3 Let R =
{x yz w
wherex,y,z ∈ Z
Let S be a fiexed non-zero element of Z
49
Define d by d
((x y0 z
))=
(0 −ys0 0
)then d is a derivation on R
Definition 3.1.4 Let R be an associative ring. A mapping
FSR → R is called a generalized derivation with associated
derivation d on R if
(i) F (x+ y) = F (x) + F (y)
(ii) F (xy) = F (x)y + xd(y)∀x, y, z ∈ R
Example 3.1.5 Let R =
{(x y
0 z
)where x, y, z ∈ Z
DefineR be a fixed non-zero element of Z Define d : R→ R by
F
(x y
0 z
)=
(0 mx+mz
0 0
)then F is a generalized derivation with associated derivation d
given by
F
(x y
0 z
)=
(0 mx−mz0 0
)Definition 3.1.6 An additive endomorphism d : N → N is
said to be an d-derivation if
d(xy) = d(x)d(y) + d(x)y∀x, y ∈ N
Although the underlying group of the near ring N is not neces-
sarily commutative, the first result gives an equivalent definition
of d - derivation on N which involves a sort of commutativity
on N .
50
Example 3.1.7 Prosposition 3.1.7 Let d be a an additive en-
domorphism of a near-ring N . then d is an α derivation if and
only if d(xy) = d(x)y + d(x)d(y) for all x, y ∈ N
Proof
By definition, if d is an α derivation then forall x, y ∈ N
d(xy) = d(x)d(y) + d(x)y
then
d(xy) = d(x)d(y + y) + d(x)(y + y)
= 2α(x)d(y) + 2d(x)y
and
d(xy + xy) = 2d(xy) = 2(d(x)d(y) + d(x)y)
So that
d(x)d(y) + d(x)y = d(x)y + d(x)d(y)
Example 3.1.8 Prosposition 3.1.8 Let d be a an α derivation
on a near-ring N . then for all x, y ∈ N
(i) (d(x)d(y) + d(x)y)z = d(x)d(y)z + d(x)yz
(ii) (d(x)y + d(x)d(y)z = d(x)yz + d(x)d(y)z
proof (i)
Let x, y, z ∈ N . then
d(z(yz)) = α(x)d(yz) + d(x)(yz)
51
= d(x)(d(yz)d(z) + d(y)z + d(x)(yz)
= (d(x)d(y))d(z) + d(x)d(y)z + d(x)(yz)
= d(xy)d(z) + d(x)d(y)z + d(x)y)z −−− (1)
d((xy)z) = d(xy)d(z) + d(xy)z
= d(xy)d(z) + (α(x)d(y) + d(x)y)z −−− (2)
From(1) and (2), we get
(d(x)d(y) + d(x)y)z = α(x)d(y)z + d(x)yz
(ii) by using the proposition 3.1.7
Hence the Proof.
Example 3.1.9 Prosposition 3.1.9 Let d be a an d - derivation
of a prime near-ring N and a such that ad(x) = 0(or d(x)a = 0)
for all x ∈ N . Then a = 0 or d = 0
proof for all x, y ∈ N
0 = ad(xy) = a(d(x)d(y) + d(x)y)
= aα(x)d(y) + ad(x)y
= ad(x)d(y) + 0
= ad(x)d(y)
Then aNd(y) = 0. Since N is prime, we get a = 0 or d = 0.
when d(x)a = 0,
52
by using the proposition 3.1.8. so if d(x)a = 0 for all x, then for
all x, y ∈ N , we have
0 = d(yx)a = (α(y)d(x) + d(y)x)a
= α(y)d(x)a+ d(y)xa [by proposition 3.1.8]
= 0 + d(y)xa
Thus d(y)Na = 0. Now the primeness of N .
Implies that d = 0 or a = 0
Hence the Proof.
Example 3.1.10 Prosposition 3.1.10 Let N be a 2-torision free
prime near ring. Let d be an α - derivation on N. Such that dα.
Then d2 = 0 implies d = 0
proof suppose that d2 = 0. Let x, y ∈ N . Then
d2(xy) = 0 = d(d(xy))
= d(d(x)d(y) + d(x)y)
= d(d(x)d(y)) + d(d(x)y)
= α2(x)d2(y) + d(d(x)d(y) + α(d(x)d(y) + d2(x)y
= d(α(x)d(y) + α(d(x)d(y)
= 2d(α(x))d(y)
Hence, 2d(d(x))d(y) = 0. Since N is 2-torision free, we have
d(α(x))d(y) = 0
53
Since α is onto, we get d(x)d(y) = 0.
by using the proposition 3.1.9
Hence the Proof.
3.2 Generalized derivations
Definition 3.2.1 LetP be a prime. A near ring N is called a
p-near ring ∀x ∈ N : xp = x and px = 0
Definition 3.2.2 LetN be a near ring andd a derivation of N .
An additive maping f : N → N is said to bea a right generalized
derivation of N associatted with d if
f(xy) = f(x)y + xd(y)∀x, y ∈ R
Definition 3.2.3 Let N be a near ring and d a derivation of
N . An additive mapping f : N → n is said to be a left general-
ized derivation of N assiciated with d if
f(xy) = d(x)y + xf(y)∀x, y ∈ R
Definition 3.2.4 Note 3.2.4
f is said to be a generalized derivation of N associated with d if
it is both a left and right generalized derivation of N associated
with d.
54
Lemma 3.2.5 Letd b an arbitrory derivation on a near ring
N . Then N satisfies the following partial distributive law
(ad(b) + d(a)b)c = ad(b)c+ d(a)bc
and
(d(a)b+ ad(b))c = d(a)bc+ ad(b)c∀a, b, c ∈ N
Proof
Since a, b, c, d(a), d(b), d(c) ∈ N . These are satisfies right dis-
tributive law
(ad(b) + d(a)b)c = a(d(b)c+ d(a)bc −−− (1)
and
(d(a)b+ ad(b)c = d(a)bc+ ad(b)c −−− (2)
Equation (1) and (2) are hold.
Theorem 3.2.6 Let(f, d) be a generalized derivation of N . If
f(px, y]) = 0∀x, y ∈ N , then N is commulative ring.
Proof
Assume that f([x, y]) = 0 for all x, y ∈ N substitute xy instead
of y.
f([x, xy]) = 0
⇒ f([x, xy]) = d(x)[x, y] = 0
f([x, y]) = 0
55
⇒ f([x, xy]) = d(x)[x, y] = 0
= d(x)(xy − yx) = 0
= d(x)(xy − yx) = 0
= d(x)(xy)− d(x)(yx)
d(x)(xy) = d(x)(yx)
d(x)(xy) = d(x)(yx)∀x, y ∈ N
Replacing y by yz in equation (1)
d(x)N [x, z] = 0∀x, z ∈ N
Hence either x ∈ z (or) d(x) = 0.
Let K = {x ∈ N/x ∈ z} and L = {x ∈ N/d(x) = 0}Then K and L are two additive subgroups of
(N,+) = K ∪ L,
since a group cannot be the union of proper subgroups, either
N = k(or) N = L. But d 6= 0
This proves that N is commutative ring.
Theorem 3.2.7 Let (f, d) be a generalized derivation of N . If
f([x, y]) = ±[x, y] for all x, y ∈ N , then N is Cummutative ring.
proof Assume that
f([x, y] = ±[x, y] −−− (1)
56
for all x, y ∈ N . substitute y by xy in (1)
f([x, xy]) = ±[x, xy]
= ±(x(xy)(xy)x)
= ±(x2y − xyx)
= ±x[x, y]
f([x, xy]) = f(x[x, y]) = d(x)[x, y] + xf([x, y]
= d(x)[x, y] + x(±[x, y])
f([x, y]) = 0
f([x, xy]) = d(x)[x, y] = 0
⇒ d(x)(xy − yx) = o
⇒ d(x)(xy)− d(x)(yx) = o
⇒ d(x)(xy) = d(x)(yx)
d(x)(xy) = d(x)(yx) forall x, y ∈ NBy using the above theorem 3.2.6
N is commutative ring.
Theorem 3.2.8 Let (f, d) be a nonzero generalized derivation
of N . If f acts as a homomorphism on N , then f is the identity
map.
proof assume that f acts as a homomorphism on N
f(x, y) = f(x)f(y)
= d(x)y + xf(y) forall x, y ∈ N −−− (1)
57
Replacing y by yz in (1)
f(x)f(yz) = d(x)yz + xd(y)z + xyf(z)
By using the lemma
”Let f be a left generalized derivation of the near ring N with
associated d. Then (d(x)y + xf(y))z = d(x)yz + xf(y)z forall
x, y ∈ N” We get
d(x)yf(z) + xf(y)f(z) = d(x)yz + xd(y)z + xy + f(z)
= d(x)yf(z) + xf(yz)
= d(x)yz + xd(y)z + xyf(z)
ie.,
d(x)yf(z) + zd(y)z + xyf(z) = d(x)yz + xd(y)z + xyf(z)
Hence d(x)y(f(z)− z) = 0 for all x, y, z ∈ NN is prime and d 6= 0, we have f(z) = z for all z ∈ Z f is the
identity map.
Hence the Proof.
Theorem 3.2.9 Let (f, d) be a non-zero generalized derivation
of N . If f acts as am amto-homomorphism on N , the f is the
identity map.
proof
f(xy) = f(y)f(x) = d(x)y + xf(y) for all x, y ∈ N — (1)
58
Replacing y by xy in (1)
f(xy)f(x) = d(x)xy + xyf(xy)
since(f, d) is a generalized derivation and facts as an anti-
homomorphism on N
(d(x)y + xf(y))f(x) = d(x)(xy) + xf(y)f(x)
By using the lemma ”Let f be a generalized derivation of the
near ring N with associated d. Then (d(x)y+xf(y)z = d(x)yz+
xf(y)z for all x, y ∈ N ”
d(x)yf(x) + xf(y)f(x) = d(x)(xy) + xf(y)f(x)
d(x)yf(x) = d(x)(xy)∀x, y ∈ N −−− (2)
Replacing y by yz in (2)
d(x)N [f(x), z] = 0∀x, z ∈ N
d(x) = 0 (or) f(x) ∈ Z forall x ∈ N hold for x∈ N.Since d6= 0. since N is commutative ring by using the lemma
”Let N be a prime near ring with a nonzero generalized deriva-
tion f associated with d. if F (N)CZ, then (N,+) is abelian, if
N is 2-torision free, then N is commutative ring.”’
f is the identity map. By using the theorem 3.2.8.
Hence the Proof.
59
Theorem 3.2.10 Let(f, d) be a generalized derivation of N
Theorem 3.2.11 Let (f, d) be a generalized derivation of N .
such that d(z) 6= 0, and a ∈ N . If [f(x), a] = 0 for all x ∈ N ,
then a ∈ Z
Let [f(x), a] = 0 for all x ∈ N, a ∈ Z —(1)
Since d(z) 6= 0, c ∈ Z, such that d(c) 6= 0, as d is a derivation,
d(c) ∈ Z.Replacing x by cx in (1)
[f((x), a)] = 0 and using the lemma ”Let f be a left generalized
derivation of the near ring N with associated d. Then (d(x)y +
xf(y)z = d(x)yz + x+ f(y)z for all x, y ∈ N”’
We have
f(cx)a = af(cx)
d(c)xa+ cf(x)a = ad(c)x+ a(f(x))
since c ∈ Z and d(c) ∈ Z, we get
d(c)N [y, a]∀y ∈ N
N and 0 6= d(c) ∈ Z, a ∈ Z.
Hence the Proof.
Theorem 3.2.12 Let (f, d) be a generalized derivation of N ,
and 0 ∈ N . If[fx),a] = 0 for all x ∈ N , then d(a) ∈ Z
60
proof Let[f(x), a] = 0 —1
Assume that a 6= 0. Replacing x by ax in (1)
f(ax)a = af(ax)
d(a)xa+ af(x)a = ad(a)x+ aaf(x)
Using f(x)a = af(x),we have
d(a)xa = ad(a)x∀x ∈ N −−− 2
Taking xyinstead of xin (2)
d(a)N [a, y] = 0for ally ∈ N
Since N is a prime near ring, either d(a) = 0 (or) a ∈ z. If
0 ∈ Z, then (N,+) is abelian by using lemma ”Let N be a 3-
prime near ring. If Z/{0} contains an elements Z for which
z + z ∈ Z, then (N,+) is abelian”
f(xa) = f(ax)
f(x)a+ xd(a) = d(a)x+ af(x)
d(a),x
= 0∀a ∈ N
ie., d(a) ∈ Z
Hence the Proof.
Theorem 3.2.13 Let(f, d) be a generalized derivation of N . If
61
N is a 2-torision free near ringf 2(N) ∈ Z, then N is a commu-
tative ring.
proof
Suppose that f 2(N)CZ. Then we get,
f 2(xy) = f 2(x)y + 2f(x)d(y) + xd2(y)] ∈ X∀x, y ∈ N
f 2(x)c+ 2f(x)d(c)xd2(c) ∈ Z∀x ∈ N, c ∈ Z
since the first summ and is an element of Z, we have
2f(x)d(c)+xd2(C0 ∈ Z∀x ∈ N, c ∈ z−−−−−−−−−−−−(1)
Taking f(x)instead of x in equation(1)
2f 2(x)d(c) + f(x)d2(c) ∈ Z∀x ∈ N, c ∈ Z
since d(c) ∈ Z, f 2(x) ∈ Z, f 2(x)d(c) ∈ Zallx ∈ N, c ∈ Z
f(x)d2(c) ∈ Z∀x ∈ N, c ∈ Z
since N is prime, we get d2(X) = 0 or f(N) ≤ Z
If f(N) ≤ Z then N is a cummutative ring.
By using the lemma ” Let N be a prime near ring with a non-zero
generalized derivation f associated with d. If f(N) ≤ Z, then
(N,+) is abelian, if N is 2-torsion free, then N is commutative
ring.”
We assume d2(x) = o by equation (1),
2f(x)d(c) ∈ Z∀x ∈ N, c ∈ Z
62
Since N is a 2-torsion free near ring and d(c) ∈ Z, either
f(N) ∈ Z. If f(N) ∈ Z, we may assume that d(z) = 0
Then
f(cx) = f(xc)
f(c)x+ cd(x) = f(x)c+ xd(c)
f(c)x+ cd(x) = f(x)c∀x, c ∈ Z −−−−−−(2)
Now substitute x by f(x) in equation (2)
f 2(N) ∈ Z, we get
f(c)f(x) + cd(f(x)) = f 2(x)c∀x ∈ N, c ∈ Z
ie, f(c)f(x) + cd(f(x)) ∈ Z for all x ∈ N, c ∈ ZTaking f(x) instead of x in equation (3)
f(c)f 2(x) + cd(f 2(x)) ∈ Z∀x ∈ N, c ∈ Z
Since d(z) = 0
f(c)f 2(x) ∈ Z∀x ∈ N, c ∈ Z
since f 2(N)CZ, we get f 2(N) = 0 or f(z)CZ
If f 2(N) = 0, by using the lemma ” Let f be a generalized
derivation of N associated with the non-zero derivation. If N is
a 2-torsion free near ring and f 2 = 0, then f = 0”
If f(z) ∈ Z, then f(x)f(c) = f(f(c)x) for all x ∈ N, c ∈ Z
d(x)f(c) = f(c)f(x)∀x ∈ N, c ∈ Z
63
Using f(c) ∈ Z
f(c)(d(x)− f(x)) = 0∀x ∈ N, c ∈ Z
Since f(z)CZ, either f(z) = 0 (or) d = f . If d = f , then
f is a derivation of N and N is a commutative ring. ”Let f
be a generalized derivation of N associated with the non-xero
derivation of N associated with the non-zero derivation d. If N
is a 2-torsion free near ring and f 2 = 0, then f = 0.”
Now assume that f(x) = 0 by using equation (2)
c(d(x)− f(x)) = 0 ∀x ∈ Nc ∈ Z
since c ∈ X, either d = f or z=0. clearly d = f . If z = 0, then
f 2(N) = 0, N is a commulative ring by using lemma. ”If N is
2-torsion free and d is a derivation on N such that d2 = 0, then
d = 0 ”‘
Hence the Proof.
Lemma 3.2.14 Let (f, d) and (g, h) be two generalized deriva-
tions of N . If h is a non-zero derivation on N and f(x)h(y)−g(x)d(y) for all x, y ∈ N then (N,+) is abelian.
Proof suppose that f(x)h(y) + g(x)d(y) = 0 for all x, y ∈ NSustitute y + z for y in equation (1)
f(x)h(y) + f(x)h(2) + g(x)d(y) + g(x)d(g) = 0
64
we get
f(x)h(y, z) = 0∀x, y, z ∈ N
By using the lemma ” Let N be a prime near ring, f a nonzero
generalized derivation of N associated with the non-zero deriva-
tional, and a ∈ N . If f(N)a = 0, then a = 0.”
h(y, z) = 0∀y, z ∈ N . For any w ∈ N ,
h(wy,wz) = h(w(y, z)) = h(w)(y, z) + wh(y, z) = 0
h(w)(y, z) = 0 for all w, y, z ∈ NBy using the lemma ”‘ let N be a 3-prime near ring. If let d be
a non-zero derivation on N . Then xd(N) = {0} implies x=0,
and
d(N)x = {0} imples x = 0”
(N,+) is abelian
Hence the Proof.
65
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