LINEAR 3-ARMENDARIZ RINGS Ayoub Elshokry

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7

LINEAR 3-ARMENDARIZ RINGS

Ayoub Elshokry¹ & Eltiyeb Ali²

1,2Department of Mathematics, Northwest Normal University, Lanzhou 730070, China

1,2Department of Mathematics, University of Khartoum, Omdurman, Sudan

1e-mail: [email protected] and ²e-mail: [email protected] ABSTRACT

In the present note we study the properties of linear 3-Armendariz rings, which are generalization of 3-Armendariz rings and linear Armendariz rings and the connections among linear 3-Armendariz rings, 3-Armendariz rings and ring satisfies condition

(P )

. We prove that a right Ore ring

R

is linear 3-Armendariz if and only if so is

Q ,

^where

Q

is the classical right quotient ring of

R .

With the help of this result we can show that a commutative ring

R

is linear 3-Armendariz if and only if the total quotient ring of

R

is linear 3-Armendariz.

Keywords: Armendariz ring; 3-Armendariz ring; linear 3-Armendariz ring.

AMS classification number: 16N60, 16S36, 16U99.

1. INTRODUCTION

Throughout this paper all rings are associative rings which maybe not have identity. Given a ring

R

, the polynomial ring over

R

is denoted by

R [x ]

. The study of Armendariz ring was initiated by Armendariz [5] and Rege and Chhawchharia [8]. A ring

R

is called Armendariz if whenever polynomials,

] [

= ) ( ,

= )

( x a

₀

a

₁

x a x g x b

₀

b

₁

x b x R x

f    

_n ⁿ

   

_m ^m



satisfy

f ( x ) g ( x ) = 0

, then

0,

j

=

i

b

a

for all

i, j

. (The converse is always true.) Some properties of Armendariz rings have been studied in Rege and Chhawchharia [8], Anderson and Camillo [4], Kim and Lee [9], Huh et al. [2], and Lee and Wong [10].

Hong et al. [3], have studied a generalization of Armendariz rings, which they called



-skew Armendariz rings, where



is an endomorphism of

R .

Suiyi [12] introduced the notion of 3-Armendariz ring. He defined a ring

R

is called 3-Armendariz ring if whenever polynomials:

] [

= ) ( ,

= )

( x a

₀

a

₁

x a x g x b

₀

b

₁

x b x h x c

₀

c

₁

x c x R x

f    

_n ⁿ

   

_m ^m

   

_r ^r



satisfy

f ( x ) g ( x ) h ( x ) = 0

^{, then}

a

_i

b

_j

c

_k

= 0

, for all

i , , j k

. Due to Lee and Wong [10], a ring

R

is called weak Armendariz (or linear Armendariz) if for given

f ( x ) = a

₀

 a

₁

x

and

g ( x ) = b

₀

 b

₁

x  R [ x ],

such that

0 = ) ( ) ( x g x

f

then

a

_i

b

_j

= 0,

for all

0  i  1,0  j  1.

(the converse is obviously true.) A ring

R

is called reduced if it has no nonzero nilpotent elements. Reduced rings are Armendariz by [5, Lemma 1] and subrings of Armendariz rings are also Armendariz ring. It is obviously that Armendariz rings are linear Armendariz and that subrings of linear Armendariz rings are still linear Armendariz. There is weak Armendariz (or linear Armendariz) ring but not Armendariz by [10, Example 3.2]. The structure of linear Armendariz rings was also observed by Anderson and Camillo [4], containing the relation closely related rings. A ring is called an abelian if every idempotent is central. Weak Armendariz (or linear Armendariz) rings are abelian by [10, Lemma 3.4(3)]. Due to Jeon et al. [6], the class of weak Armendariz (or linear Armendariz) rings is closed under direct products. Motivated by results in Suiyi [12], Jeon et al. [6], Lee and Wong [10], Kim and Lee [9], and Rege and Chhawchharia [8], we investigate a generalization of linear Armendariz ring and 3-Armemdariz rings which we called linear 3-Armendariz ring.

2. LINEAR 3-ARMENDARIZ RINGS

Definition 2.1 A ring

R

is called linear 3-Armendariz if whenever polynomials

f ( x ) = a

₀

 a

₁

x

,

x b b x

g ( ) =

₀



₁ and

h ( x ) = c

₀

 c

₁

x  R [ x ],

satisfy

f ( x ) g ( x ) h ( x ) = 0

^then

a

_i

b

_j

c

_k

= 0,

for all

1. 1,0 1,0

0  i   j   k 

Condition (P): For all

a , b , c  R ,

if

( abc )

²

= 0,

then

abc = 0.

(see [12])

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8

Lemma 2.2 [11, Proposition 1]. If

R

is a reduced ring, then

R

satisfies the condition

(P )

, but the converse is not true.

Lemma 2.3 [12, Theorem 1]. If

R

(P )

, then

R

is 3-Armendariz.

Lemma 2.4 If

R

is 3-Armendariz, then for any

f

₁

( x ), f

₂

( x ),  , f

_n

( x )  R [ x ], n  3,

if

0,

= ) ( )

( )

(

₂ ₃

1

x f x f f x

f 

_n then ₍ ₎ ₍ ₎

= 0.

) 2

1(x f x fn x

f

C C

C 

Proof. Assume that

f

₁

( x ) f

₂

( x ) f

₃

( x )  f

_n

( x ) = 0,

then

f

₁

( x ) f

₂

( x )( f

₃

( x )  f

_n

( x )) = 0.

Thus

0.

)

=

( ) 2( )

1(x f x fn x

f

C C

C 

Since

R

is 3-Armendariz. So

0. = ) ( ) ( )

(

₃

2 )

1(

f x f x f x

C

_f _x



_n

Thus

(

₍ ₎ ₂

( ))

₃

( )(

₄

( ) ( )) = 0.

1

f x f x f x f x

C

_f _x



_n By the hypothesis, we have

0.

)

=

( ) 4( ) 3( ) 2( )

1(x f x f x f x fn x

f

C C C

C

_

Hence ₍ ₎ ₃

( )

₄

( )

₅

( ) ( ) = 0,

) 2

1(

C f x f x f x f x

C

_f _x _f _x



_n so

0. = )) ( ) ( )(

( )) (

(

₍ ₎ ₃ ₄ ₅

) 2

1(

C f x f x f x f x

C

_f _x _f _x



_n

Repeating this process, we can show that ₍ ₎ ₍ ₎

= 0.

) 2

1(x f x fn x

f

C C

C 

Theorem 2.5 Let

R

be a ring. Then

R

(P )

if and only if

R [x ]

) (P

^.

Proof.

 )

. Assume that

f ( x ), g ( x ), h ( x )  R [ x ]

be such that

( f ( x ) g ( x ) h ( x ))

²

= 0

where

.

= ) ( ,

= )

( x

ⁿ_i₌₀

a

_i

x

ⁱ

g x

^m_j₌₀

b

_j

x

^j

h x

^r_k₌₀

c

_k

x

^k

f   

Then

0. = ) ( ) ( ) ( ) ( ) ( )

( x g x h x f x g x h x f

Since

R

satisfies the condition

( P ), a

_i

b

_j

c

_k

a

_s

b

_t

c

_q

= 0

for any

i , j , k , s , t , q ,

by Lemma 2.3 and Lemma 2.4. In particular

( a

_i

b

_j

c

_k

)

²

= 0

for any

i , j , k .

Thus

a

_i

b

_j

c

_k

= 0

for any

i , j , k .

Since

R

(P ).

There for

f ( x ) g ( x ) h ( x ) = 0,

which implies that

R [x ]

(P )

^.

 ).

It is clear that

R

is a subring of

R [x ]

and the subrings of a ring satisfies the condition

(P )

is also satisfies the condition

(P ).

Remark 2.6 Let

R

be a ring with identity.

R

is Armendariz if and only if

R

is 3-Armendariz. (see [12]) Clearly, any 3-Armendariz ring is linear 3-Armendariz, but the converse is not true by the following example.

Example 2.7 [6, Example 1.2(1)]. Let

R = Z

₃

[ x , y ]/( x

³

, x

²

y

²

, y

³

)

where

Z

₃ is a Galois field of order 3, with identity 1,

Z

₃

[ x , y ]

is the polynomial ring with two indeterminates

x, y

over

Z

₃

,

and

( x

³

, x

²

y

²

, y

³

)

is the ideal of

Z

₃

[ x , y ]

generated by

x

³

, x

²

y

²

, y

³. Let

R [t ]

be the polynomial ring with an indeterminate

t

over

R .

Since

( x  y t )

³

= ( x  y t )( x

²

 2 x y t  y

²

t

²

) = 0

with

x y

²

 0, R

is not Armendariz, by Remark 2.6,

R

is not 3-Armendariz. But

R

is linear 3-Armendariz.

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9

Lemma 2.8 Let

S

be a subring of

R .

^If

R

is linear 3-Armendariz. Then so is

S .

Proof. Let

f ( x ) = 

¹_i₌₀

a

_i

x

ⁱ

, g ( x ) = 

¹_j₌₀

b

_j

x

^j

, h ( x ) = 

¹_k₌₀

c

_k

x

^k

 S [ x ],

be such that

f ( x ) g ( x ) h ( x ) = 0.

Then

f ( x ), g ( x ), h ( x )  R [ x ].

Since

R

is linear 3-Armendariz, then

a

_i

b

_j

c

_k

= 0.

This means that

S

Proposition 2.9 The class of linear 3-Armendariz rings is closed under finite direct products.

Proof. Let _s

s

R

R = 

 be the finite direct product of

R

_s where

 = {1,2,  , n }, R

_s is linear 3-Armendariz ring. Suppose

f ( x ) g ( x ) h ( x ) = 0

for some polynomials

f ( x ) = a

₀

 a

₁

x , g ( x ) = b

₀

 b

₁

x

and

] [

= )

( x c

₀

c

₁

x R x

h  

, where

a

_i

= ( a

_i₁

, a

_i₂

,  , a

_in

), b

_j

= ( b

_j₁

, b

_j₂

,  , b

_jn

), c

_k

= ( c

_k₁

, c

_k₂

,  , c

_kn

),

are elements of the product ring

R

. Set

f

_s

( x ) = 

¹_i₌₀

a

_is

x

ⁱ^,

g

_s

( x ) = 

¹_j₌₀

b

_js

x

^j and

h

_s

( x ) = 

¹_k₌₀

c

_ks

x

^k

 R [ x ].

Since

f ( x ) g ( x ) h ( x ) = 0

then



_i__j__k₌_l

a

_i

b

_j

c

_k

= 0,0  l  i  j  k ,

so

0,

= ) ,

, (

₁ ₁ ₁

=l i j k in jn kn

k j

i_ _

a b c  a b c



and so



_i__j__k₌_l

( a

_is

b

_js

c

_ks

) = 0,1  s  n .

Thus

f

_s

( x ) g

_s

( x ) h

_s

( x ) = 0

in

R

_s

[ x ],1  s  n .

Since

R

_s is linear 3-Armendariz rings, then we have

a

_is

b

_js

c

_ks

= 0

in

R

_s

,1  s  n .

Then it is clear that

a

_i

b

_j

c

_k

= a

_i₁

b

_j₁

c

_k₁

,  , a

_in

b

_jn

c

_kn

= 0.

Therefore

a

_i

b

_j

c

_k

= 0.

Thus

R

Proposition 2.10 If

R

satisfies condition

(P )

, then

R

is linear 3-Armendariz.

Proof. If

R

satisfies condition

( P )

it easy proof,

0. = 0

= bca cab

abc  

Let

f ( x ) = a

₀

 a

₁

x , g ( x ) = b

₀

 b

₁

x , h ( x ) = c

₀

 c

₁

x  R [ x ],

be satisfy

f ( x ) g ( x ) h ( x ) = 0.

Next we only need to Proof

a

_i

b

_j

c

_k

= 0,

for all

0  i , j , k  1

. ) ( ) (

) (

=

) )(

)(

(

= ) ( ) ( ) (

3 1 1 1 2 0 1 1 1 0 1 1 1 0 0

0 1 0 1 0 1 0 0 0 0 0

1 0 1 0 1 0

x c b a x c b a c b a c b a x c b a c b a c b a c b a

x c c x b b x a a x h x g x f



Then we have the following equation

(1) 0,

0

=

0 0

b c a

(2) 0,

0

=

0 1 0 1 0 1 0

0

b c a b c a b c

a  

(3) 0,

1

=

0 1 0 1 1 1 1

0

b c a b c a b c

a  

0.(4)

1

=

1 1

b c a

Multiplying (2) on left side by

c

₀ then we have

0(5)

0

=

0 1 0 0 1 0

0

a b c c a b c

c 

Multiplying

(5)

on left side by

b

₀ yields

0.(6)

0

=

0 1 0 0

c a b c b

Then

( a

₁

b

₀

c

₀

)

²

= 0.

Thus,

a

₁

b

₀

c

₀

= 0.

Since

R ,

satisfies condition

(P )

for all

i = 0,1.

Now

(5)

be comes

0.

0

=

1 0 0

a b c

c

Thus,

c

₀

a

₀

b

₁

= 0

, for all

j = 0,1.

Now from

(2)

it follows

a

₀

b

₀

c

₁

= 0

for all

k = 0,1.

(4)

10 Multiplying (3) on left side by

b

₀

c

₀ then we have

0.(7)

1

=

0 1 0 0 0 1 1 0

0

c a b c b c a b c

b 

Multiplying (7) on right side by

b

₀ yields

0.

0

=

0 1 1 0 0

c a b c b b

Thus,

a

₁

b

₁

c

₀

= 0

for all

i , j = 0,1.

^Since

R

satisfies condition

(P ).

Thus, (7) becomes

b

₀

c

₀

a

₁

b

₀

c

₁

= 0.

Thus,

0

1

=

0 1

b c

a

for all

i , k = 0,1.

Now (3) it follows

a

₀

b

₁

c

₁

= 0

for all

j , k = 0,1.

Therefore

a

_i

b

_j

c

_k

= 0

, for all

. , , j k

i

This means that

R

is linear 3-Armendariz ring.

The following example show that the converse of Proposition 2.10, is not true.

Example 2.11 Let

. ,

, , , , , ,

| 0 0 0 0 0

0 0 0 0

0 0

=

 





 





 





 







 





 





Z h g f e d c b h a g f e

d c b a

R

Next, we proof

R

is linear 3-Armendariz ring, but it doesn’t satisfies condition

(P ).

Proof. For any

 = 

₀

 

₁

x ,  = 

₀

 

₁

x ,  = 

₀

 

₁

x  R [ x ],

and

 = 0.

We have the following ring isomorphism

. ] [ , , , , , , ,

| 0 0 0 0 0

0 0 0 0

] [ :



































 abcd e f g h x

h g f e

d c b a

x

R Z



. 0 0

0 0

0

0 0

0 0 0 0 0

0 0 0 0

0 0

0 0 0 0 0

0 0 0 0

0 0

1 0

1 0 1

0

1 0 1 0 1 0 1 0

1 1 1 1

0 0 0 0











































x h h

x g g

x f f x

e e

x d d x c c x b b x a a

h x g f e

d c b a

h g f e

d c b a

By ring isomorphism we have

(5)

11

 





 





0 0 0 0 0

0 0 0 0

= ) ( ) ( ) (

= 0

3 2 1

B G A













with

A

₁

= a

₁₀

 a

₁₁

x , B

₂

= b

₂₀

 b

₂₁

x , G

₃

= g

₃₀

 g

₃₁

x , a

₁_i is (1,2) entry of



_i

, b

₂_j is (2,3) entry of



_j

, g

₃_k is (3,5) entry of



_k

.

By [12, Example 3],

Z

is 3-Armendariz. Therefore

Z

is linear 3-Armendariz. so

0,

3

=

2 1i

b

j

g

k

a

then



_i



_j



_k

= 0.

Thus,

R

is linear 3-Armendariz ring. But it is clear that

R

don’t satisfies condition

(P ).

Proposition 2.12 Let

R

be a ring satisfies condition

(P )

^{, then}

 



 





 



 









 









 







R d c b a a d a

c b a

S | , , ,

0 0

= 0

Proof. We notice that

0 . 0 , 0 0

0 0

2 2 2

1 1 1

a S d a

c b a

a d a

c b a





 









 









 









 







The operations of additive and multiplication are denoted by

) ,

, ,

(

= ) , , , ( ) , , ,

( a

₁

b

₁

c

₁

d

₁

 a

₂

b

₂

c

₂

d

₂

a

₁

 a

₂

b

₁

 b

₂

c

₁

 c

₂

d

₁

 d

₂

and

) ,

, ,

(

= ) , , , )(

, , ,

( a

₁

b

₁

c

₁

d

₁

a

₂

b

₂

c

₂

d

₂

a

₁

a

₂

a

₁

b

₂

 b

₁

a

₂

a

₁

c

₂

 b

₁

d

₂

 c

₁

a

₂

a

₁

d

₂

 d

₁

a

₂

respectively. So every polynomial in

S [x ]

can be expressed in the form

)) ( ), ( ), ( ), (

( p

₀

x p

₁

x p

₂

x p

₃

x

for some

p

_i

(x )

^,s in

R [x ]

. For any three polynomials of

S [x ],

);

( ), ( ), ( ), ( (

= )

( x f

₀

x f

₁

x f

₂

x f

₃

x f

);

( ), ( ), ( ), ( (

= )

( x g

₀

x g

₁

x g

₂

x g

₃

x g

).

( ), ( ), ( ), ( (

= )

( x h

₀

x h

₁

x h

₂

x h

₃

x h

Suppose that

f ( x ) g ( x ) h ( x ) = 0

then we have the following Equations

(1) 0,

= ) ( ) ( )

(

₀ ₀

0

x g x h x

f

(6)

12

(2) 0,

= ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

(

₁ ₀ ₀ ₀ ₁ ₁ ₀ ₀

0

x g x h x f x g x h x f x g x h x

f  

0, (3)

= ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (

) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (

0 0 2 0

3 1 0

2 0

3 0 1 3

1 0 2

0 0

x h x g x f x h x g x f x h x g x f



0.(4)

= ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

(

₀ ₃ ₀ ₃ ₀ ₃ ₀ ₀

0

x g x h x f x g x h x f x g x h x

f  

Since

R [x ]

^satisfies ^condition

(P ),

^by ^Theorem ^2.5, ^then ^from ^(1), ^we ^have

0. = ) ( ) ( ) (

= ) ( ) ( )

(

₀ ₀ ₀ ₀ ₀

0

x f x g x g x h x f x

h

₀

( x )

then we have

0.(5)

= ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

(

₀ ₁ ₀ ₀ ₁ ₀ ₀

0

x f x g x h x h x f x g x h x

h 

Multiplying (5) on right side by

f

₀

( x )

then we have

0.(6)

= ) ( ) ( ) ( ) ( )

(

₀ ₁ ₀ ₀

0

x f x g x h x f x

h

Thus,

f

₀

( x ) g

₁

( x ) h

₀

( x ) = 0.

From (5), we have

h

₀

( x ) f

₁

( x ) g

₀

( x ) h

₀

( x ) f

₀

( x ) = 0.

Thus

0,

= ) ( ) ( )

(

₀ ₀

1

x g x h x

f

since

R [x ]

satisfies the condition (P), again by Theorem 2.5. Now, from (2) it follows

0. = ) ( ) ( )

(

₀ ₁

0

x g x h x

f

h

₀

( x )

yields

0.(7)

= ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

(

₀ ₃ ₀ ₀ ₃ ₀ ₀

0

x f x g x h x h x f x g x h x

h 

f

₀

( x )

then we have

0. = ) ( ) ( ) ( ) ( )

(

₀ ₀ ₃ ₀

0

x h x f x g x h x

f

Thus,

f

₀

( x ) g

₃

( x ) h

₀

( x ) = 0.

Since

R [x ]

satisfies the condition (P), also Theorem 2.5, from (7) we have

0. = ) ( ) ( ) ( )

(

₀ ₃ ₀

0

x g x f x h x

h

Thus,

g

₀

( x ) f

₃

( x ) h

₀

( x ) = 0.

Now from (4) it follows

f

₀

( x ) g

₀

( x ) h

₃

( x ) = 0.

Now, multiplying (3) on left side by

h

₀

( x )

then we have

0(8)

= ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

(

₀ ₂ ₀ ₀ ₁ ₃ ₀ ₀ ₂ ₀ ₀

0

x f x g x h x h x f x g x h x h x f x g x h x

h  

g

₀

( x )

yields

0. = ) ( ) ( ) ( ) ( )

(

₀ ₂ ₀ ₀

0

x h x f x g x h x

g

Thus,

g

₀

( x ) f

₂

( x ) h

₀

( x ) = 0.

Thus

(8)

^becomes

0.(9)

= ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

(

₀ ₂ ₀ ₀ ₁ ₃ ₀

0

x f x g x h x h x f x g x h x

h 

g

₃

( x )

yields

0. = ) ( ) ( ) ( ) ( )

(

₀ ₁ ₃ ₀

3

x h x f x g x h x

g

Thus,

f

₁

( x ) g

₃

( x ) h

₀

( x ) = 0.

Thus from (9) we have

0. = ) ( ) ( ) ( )

(

₀ ₂ ₀

0

x f x g x h x

h

Thus,

f

₀

( x ) g

₂

( x ) h

₀

( x ) = 0.

f

₀

( x ) g

₂

( x ) h

₀

( x ) = 0.

Now (3) becomes

0.(10)

= ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

(

₀ ₂ ₀ ₁ ₃ ₁ ₀ ₃

0

x g x h x f x g x h x f x g x h x

f  

h

₀

( x )

yields

0.(11)

= ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

(

₀ ₁ ₃ ₀ ₁ ₀ ₃

0

x f x g x h x h x f x g x h x

h 

g

₀

( x ) h

₃

( x )

yields

(7)

13

0. = ) ( ) ( ) ( ) ( )

(

₃ ₀ ₁ ₀ ₃

0

x h x h f x g x h x

g

Thus,

f

₁

( x ) g

₀

( x ) h

₃

( x ) = 0.

Then from (11) we have

h

₀

( x ) f

₀

( x ) g

₁

( x ) h

₃

( x ) = 0.

Thus,

0. = ) ( ) ( )

(

₁ ₃

0

x g x h x

f

₀

( x ) g

₀

( x ) h

₂

( x ) = 0.

Now let

j

j j j

j i

i i i

i

x

a d a

c b a x

g a x d a

c b a x

f



 









 













 









 







 0 0

= 0 ) ( 0 ,

0 = 0 )

(

¹₌₀ ¹₌₀

and

k

k k k

k

x

a d a

c b a x

h



 









 









 0 0

= 0 ) (

¹₌₀

where

,

= ) ( ,

= )

(

¹₌₀ ₁ ¹₌₀ ₂ ¹₌₀ ₃ ¹₌₀

0

i i i i

i i i

i

a x f x b x f x c x f x d x

x

f    

j j j j

j j j

j

a x g x b x g x c x g x d x

x

g

₀

( ) = 

¹₌₀

 ,

₁

( ) = 

¹₌₀

 ,

₂

( ) = 

¹₌₀

 ,

₃

( ) = 

¹₌₀

 .

= ) ( ,

= )

(

¹₌₀ ₁ ¹₌₀ ₂ ¹₌₀ ₃ ¹₌₀

0

k k k k

k k k

k

a x h x b x h x c x h x d x

x

h        

Then we obtain that,

0,

= 0,

= _k _i _j _j _i _k _i _j _k _i _i _k _i _j _k _i _j _k _i _j _k _i _j _k _i _j _k _i _j _k _i _j _k

k j

iaa aaa aba baa aab daa ada aad aca bd a caa aac

a                       

and

b

_i

a 

_j

d

_k

 = 0.

for all

i , j , k = 0,1,

by the results of Proposition 2.10, the condition that

R

satisfies condition

(P )

, we have

0. 0 = 0 0 0

0 0 0

0 0



 









 











 









 











 









 







k k k

j j j

i i i

a d a

c b a

a d a

c b a

a d a

c b a

For all

i , , j k

and therefore

S

Given a ring

R

and a bimodule _R

M

_R

,

the trivial extension of

R

by

M

is the ring

T ⁽ R ^, M ⁾ ⁼ R  M

^with

the usual addition and the following multiplication

( r

₁

, m

₁

)( r

₂

, m

₂

) = ( r

₁

r

₂

, r

₁

m

₂

 m

₁

r

₂

).

This is isomorphic to

the ring of all matrices

0 ,

 







 





 r m r

where

r  R , m  M

and the usual matrix operations are used.

Corollary 2.13 If a ring

R

satisfies condition

(P ),

then the trivial extension

T ( R , R )

is linear 3-Armendariz ring.

Proof. Notice that

T ( R , R )

is isomorphic to

(8)

14

 



 





 



 









 









 







R b a a a b a

U | ,

0 0

0 =

and that each subring of linear 3-Armendariz ring is also linear 3-Armendariz. Thus,