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© 2018 Pushpa Publishing House, Allahabad, India http://www.pphmj.com

http://dx.doi.org/10.17654/MS103071215

Volume 103, Number 7, 2018, Pages 1215-1227 ISSN: 0972-0871

Received: October 4, 2017; Accepted: January 10, 2018 2010 Mathematics Subject Classification: 16W55.

Keywords and phrases: monoid homomorphism, strictly ordered monoid, skew generalized power series ring.

THE IMPACT OF THE MONOID HOMOMORPHISM ON THE STRUCTURE OF SKEW GENERALIZED POWER

SERIES RINGS

Ahmad Faisol1,2, Budi Surodjo1 and Sri Wahyuni1

1Department of Mathematics Universitas Gadjah Mada Indonesia

2Department of Mathematics Universitas Lampung Indonesia

Abstract

Let R be a ring, (S,) be a strictly ordered monoid and ω:S ( )R

End be a monoid homomorphism. In this paper, we study the properties of monoid homomorphism ω and its impact on the structure of skew generalized power series ring R[[S,ω]]. We show that: if

( )1 ~ω( )2,

ω then R1[[S,ω( )1]]~= R2[[S,ω( )2 ]], and R1R2[[S, ( ) ( )]]~ [[ , ( )]] [[ , ( )2 ]].

1 2 2 1

1 ω = ω ω

ω R S R S

1. Introduction

In 2007, Mazurek and Ziembowski [1] constructed a new ring which is the generalization of generalized power series rings (GPSR) R

[ ] [ ]

S that was

(2)

constructed by Ribenboim [2] by using a monoid homomorphism ω:S

( )

R

End to change the convolution product on GPSR R

[ ] [ ]

S . Furthermore, this new ring is known as skew generalized power series ring (SGPSR) denoted by R

[ [

S, ω

] ]

or R

[ [

S, ω, ≤

] ]

. Now we will give the definition and some examples of SGPSR R

[ [

S, ω

] ]

.

Regarding ordered sets, ordered monoids, artinian and narrow set, we will follow the terminology used in [2-6]. Now, we recall the construction of SGPSR [1]. Let

(

S, ≤

)

be a strictly ordered monoid, R be a commutative ring with an identity element and ω:SEnd

( )

R be a monoid homomorphism. For any sS let ωs denote the image of s under ω, i.e.,

( )

s = ωs. ω

Define RS =

{

f | f :SR

}

and R

[ [

S, ω

] ]

=

{

fRS|supp

( )

f is artinian and narrow where

}

, supp

( )

f =

{

sS| f

( )

s ≠ 0

}

.

Under pointwise addition and skew convolution, multiplication defined by

( )( ) ( ) ( ( ) )

( )

( )

χ

ω

=

g f y x

x

s

y g x f s

fg

, ,

, (1)

for all f, gR

[ [

S

] ]

, where

(

f g

) { (

x y

)

supp

( )

f supp

( )

g xy s

}

s = ∈ × | =

χ , ,

is finite, R

[ [

S, ω

] ]

is a ring which is known as skew generalized power series ring (SGPSR).

Some special cases of SGPSR R

[ [

S, ω

] ]

are given by the following example.

Example 1.1. Let R be a ring, idR be an identity map in End

( )

R , N0 be a set of positive integers, Z be a set of integers and

(

S, ≤

)

be a strictly ordered monoid.

(3)

(1) If S = N0 with usual addition, trivial order ≤ and ωs =idR, for all ,

S

s∈ then SGPSR R

[ [

S, ω

] ]

is polynomial ring R

[ ]

X .

(2) If S = Z with usual addition, trivial order ≤ and ωs = idR, for all ,

S

s∈ then SGPSR R

[ [

S, ω

] ]

is Laurent polynomial ring R

[

X, X1

]

. (3) If S = N0 with usual addition, trivial order ≤ and ω0 = σ, for some endomorphism ring σ∈End

( )

R , then SGPSR R

[ [

S

] ]

is skew polynomial ring R

[

X

]

.

(4) If S = N0 with usual addition, usual order ≤ and ωs = idR, for all ,

S

s∈ then SGPSR R

[ [

S, ω

] ]

is power series ring R

[ ] [ ]

X .

(5) If S = Z with usual addition, usual order ≤ and ωs =idR, for all ,

S

s∈ then SGPSR R

[ [

S, ω

] ]

is Laurent series ring R

[[

X, X1

]]

.

(6) If S = N0 with usual addition, usual order ≤ and ω0 = σ, for some endomorphism ring σ∈End

( )

R , then SGPSR R

[ [

S, ω

] ]

is skew power series ring R

[ [

X

] ]

.

(7) If ωs = idR, for all sS, then SGPSR R

[ [

S, ω

] ]

is generalized power series ring

[[

R(S,)

]]

= R

[ ] [ ]

S .

2. Main Results

In this section, we give the definition and some properties of monoid homomorphism ω and its impact on the structure of SGPSR R

[ [

S, ω

] ]

. First, we give the definition of equivalency of two monoid homomorphism.

Definition 2.1. Let R1 and R2 be rings,

(

S,≤

)

be a strictly ordered monoid, and ω( )1 :SEnd

( )

R1 and ω( )2 :SEnd

( )

R2 be monoid homomorphisms. Then ω( )1 and ω( )2 are said to be equivalent if there exists an isomorphism ϕ:R1R2 such that ω( )s2 = ϕω( )s1ϕ1 for all sS. In this case, we write ω( )1 ~ ω( )2 .

(4)

Example 2.2. Let S = N0, R1 = Q×Q = Q2 and R2 = Z×Z = Z2. With operation

(

x, y

) (

+ m, n

) (

= x +m, y +n

)

and

(

x, y

) (

m,n

) (

= xm, yn

)

,

R1 and R2 become rings and S becomes a strictly ordered commutative monoid with pointwise addition and usual order. For any sS,

(

p, q

)

R1 and

(

x, y

)

R2, we define monoid homomorphism

( )1 :SEnd

( )

R1 , ω

where ω( )s1

(

p, q

) (

= 0, q

)

, and

( )2 :SEnd

( )

R2 , ω

where ω( )s2

(

x, y

) (

= x,0

)

. Next, we define a map

2

:R1R ϕ

with ϕ

(

p, q

) (

= q, p

)

for all

(

p, q

)

R1. Since for any

(

p, q

) (

, m,n

)

R1, imply

( ) ( )

(

p q + m n

)

( (

p+m q+ n

) )

ϕ , , ,

(

q+n p +m

)

= ,

(

q, p

) (

+ n, m

)

=

( )

(

p, q

)

( (

m, n

) )

ϕ

= and

( ) ( )

(

p, q m,n

)

= ϕ

( (

pm, qn

) )

ϕ

(

qn, pm

)

=

(

q, p

) (

n, m

)

=

( )

(

p, q

) (

ϕ

(

m, n

) )

, ϕ

= ϕ is a ring homomorphism.

(5)

Furthermore, if ϕ

( (

p, q

) )

= ϕ

( (

m, n

) )

, then

(

q, p

) (

= n, m

)

, which is n

q = and p = m. In other words, we have

(

p, q

) (

= m,n

)

. Hence, ϕ is an injective homomorphism. For any

(

x, y

)

R2, there exists

(

p, q

)

R1 with

y

p= and q=x such that ϕ

( (

p,q

) ) (

= q, p

) (

= x,y

)

. Then, ϕ is a surjective homomorphism. In other words, ϕ:R1R2 is a ring isomorphism.

Moreover, since

( )s2ϕ

( (

p,q

) )

= ω( )s2

(

ϕ

( (

p, q

) ) )

ω

( )

( (

q p

) )

s2 ,

ω

=

(

q,0

)

=

( )

(

0, q

)

ϕ

=

(

ω( )s1

( (

p, q

) ))

ϕ

=

( )s1

( (

p, q

) )

, ϕω

= ( )1 ~ ω( )2 .

ω

Based on Definition 2.1, the impact of equivalency of two monoid homomorphisms on the structure of SGPSR R

[ [

S

] ]

is given by the following proposition.

Proposition 2.3. Let R1 and R2 be rings,

(

S, ≤

)

be a strictly ordered monoid, and ω( )1 :SEnd

( )

R1 and ω( )2 :SEnd

( )

R2 be monoid homomorphisms. If ω( )1 ~ ω( )2 , then R1

[[

S, ω( )1

]]

~= R2

[[

S, ω( )2

]]

.

Proof. Suppose ω( )1 ~ ω( )2 . Then by Definition 2.1, there exists an isomorphism ϕ:R1R2 such that ω( )s2 = ϕω( )s1ϕ1 for all sS. Next, we define a map

[[

, ( )

]] [[

, ( )

]]

,

: 1 ω12 ω2

ψ R S R S

where ψ

( )

f = f = ϕD f for all fR1

[[

S, ω( )1

]]

.

(6)

For all sS and f, gR1

[[

S( )1

]]

, we have

(

f + g

) ( )

s = ϕ

( (

f + g

) ( )

s

)

ϕD

( ) ( ) (

f s + g s

)

ϕ

=

( )

(

f s

)

(

g

( )

s

)

ϕ

=

(

ϕD f

) ( ) (

s + ϕDg

) ( )

s

= and

( )

(

ϕD fg

) ( )

s = ϕ

( ( ) ( )

fg s

)

( )

( )

( ( ) )

⎜⎜

⎛ ω

ϕ

=

=xy s

x g y

x

f 1

( ( )

( )

( ( ) ))

=

ω ϕ

=

xy s

x g y

x

f 1

( )

( ) (

( )

( ( ) ))

=

ω ϕ ϕ

=

xy s

x g y

x

f 1

( ) ( ) (

( )

) ( ) ( )

=

ω ϕ ϕ

=

xy s

x g y

x

f D 1

D

( ) ( ) (

( )

) ( ) ( )

=

ϕ ω ϕ

=

xy s

x g y

x

f D

D 2

( ) ( )

( )

( ( ( ) ) )

=

ϕ ω ϕ

=

xy s

x g y

x

f 2

D

( ) ( )

( )

( ( ) ( ) )

=

ϕ ω ϕ

=

xy s

x g y

x

f D

D 2

( ) ( )

(

ϕD f ϕDg

) ( )

s .

=

(7)

Since supp

( )

fsupp

( )

f , fR2

[[

S, ω( )2

]]

. Then, we have

(

f + g

)

= f + g

ψ

(

f + g

)

ϕ

= D

(

ϕD f

) (

+ ϕDg

)

= g f +

=

( )

f

( )

g ψ

= and

( )

fg = fg ψ

( )

fg ϕD

=

(

ϕD f

) (

ϕDg

)

= g

= f

( ) ( )

f ψ g , ψ

=

for all f, gR1

[[

S, ω( )1

]]

. In other words, the map ψ:R1

[[

S, ω( )1

]]

[[

( )2

]]

2 S, ω

R is a ring homomorphism.

Now, we will show that ψ is injective. Let fKer

( )

ψ . Then

( )

= 0.

ψ f Then, for all sS, we have

(

ϕD f

) ( )

s =0

( )

s . In other words,

( )

(

f s

)

ϕ = 0. Since ϕ is a ring isomorphism, f

( )

s = 0, for all sS. Then

( )

ψ

Ker = 0, so ψ is injective.

Furthermore, we will show that ψ is surjective. For all gR2

[[

S( )2

]]

, there exists h1DgR1

[[

S( )1

]]

such that ψ

( )

h =h =ϕDh=ϕDϕ1g

.

=g Then ψ is surjective. So R1

[[

S, ω( )1

]]

~= R2

[[

S( )2

]]

.

(8)

Now we will give the definition of direct sum of two monoid homomorphisms.

Definition 2.4. Let R1 and R2 be rings,

(

S,≤

)

be a strictly ordered monoid, and ω( )1 :SEnd

( )

R1 and ω( )2 :SEnd

( )

R2 be monoid homomorphisms. Then the direct sum of ω( )1 and ω( )2 is defined by

( )1 ⊕ω( )2 :SEnd

(

R1R2

)

, ω

where

(

ω( )1 ⊕ω( )2

) (

s r1, r2

) (

= ω( )s1

( )

r1( )s2

( ))

r2 ,

for all sS and

(

r1, r2

)

R1R2.

Example 2.5. Let monoid S, rings R1 and R2( )1 and ω( )2 be given as in Example 2.2. Then, we can define the direct sum of ω( )1 and ω( )2 by

( )1 ⊕ω( )2 :SEnd

(

R1R2

)

, ω

where

(

ω( )1 ⊕ω( )2

) (

s

(

p, q

) (

, x, y

) ) (

= ω( )s1

( (

p, q

) )

, ω( )s2

( (

x, y

) ))

( ) ( )

(

0, q , x,0

)

,

=

for all sS and

( (

p, q

) (

, x, y

) )

R1R2.

The following lemma shows that the direct sum ω( )1 ⊕ω( )2 that defined in Definition 2.4 is a monoid homomorphism.

Lemma 2.6. Let R1 and R2 be rings,

(

S,≤

)

be a strictly ordered monoid, and ω( )1 :SEnd

( )

R1 and ω( )2 :SEnd

( )

R2 be monoid homomorphisms. Then the direct sum

(9)

( )1 ⊕ω( )2 :SEnd

(

R1R2

)

ω

is a monoid homomorphism.

Proof. For any s,tS and

(

r1, r2

)

R1R2, we have

(

ω( )1 ⊕ω( )2

) (

st r1, r2

) (

= ω( )st1

( )

r1( )st2

( ))

r2

((

ω( ) ( )s1ωt1

) ( ) (

r1 , ω( ) ( )s2ωt2

) ( ))

r2

=

(

ω( )s1

(

ω( )t1

( ))

r1( )s2

(

ω( )t2

( )))

r2

=

(

ω( )1 ⊕ω( )2

) (

s ωt( )1

( )

r1( )t2

( ))

r2

=

((

ω( )1 ⊕ω( )2

) (

s ω( )1 ⊕ω( )2

) )(

t r1r2

)

.

= Hence, we obtain

(

ω( )1 ⊕ω( )2

)( ) (

st = ω( )1 ⊕ω( )2

)( )(

s ω( )1 ⊕ω( )2

)( )

t . So the direct sum ω( )1 ⊕ω( )2 is monoid homomorphism.

Now, based on Definition 2.4 and Lemma 2.6 we get the following proposition.

Proposition 2.7. Let R1 and R2 be rings,

(

S, ≤

)

be a strictly ordered monoid, and ω( )1 :SEnd

( )

R1 and ω( )2 :SEnd

( )

R2 be monoid homomorphisms. Then

(

R1R2

) [[

S, ω( )1 ⊕ω( )2

]]

~= R1

[[

S, ω( )1

]]

R2

[[

S( )2

]]

.

Proof. Let i1: R1R1R2 and i2 :R2R1R2 be natural injections, and let p1:R1R2R1 and p2 :R1R2R2 be natural projections. Then we have

( )s1 = p1

(

ω( )1 ⊕ω( )2

)

si1 ω

(10)

and

( )s2 = p2

(

ω( )1 ⊕ω( )2

)

si2, ω

as seen in the following diagram:

Then we obtain

( )s1 p1 = p1

(

ω( )1 ⊕ω( )2

)

si1p1 ω

(

( )1 ( )2

)

1

1 sidR

p ω ⊕ω

=

(

( ) ( )

)

s p1 ω1 ⊕ω2

= and

( )s2 p2 = p2

(

ω( )1 ⊕ω( )2

)

si2p2 ω

(

( )1 ( )2

)

2

2 sidR

p ω ⊕ω

=

(

( )1 ( )2

)

.

2 s

p ω ⊕ω

=

Now, for any f

(

R1R2

) [[

S, ω( )1 ⊕ω( )2

]]

, we define a map

(

1 2

) [[

, ( )1 ( )2

]]

1

[[

, ( )1

]]

2

[[

, ( )2

]]

: ⊕ ω ⊕ω → ω ⊕ ω

ψ R R S R S R S

by ψ

( ) (

f = f1, f2

)

, where f1 = p1D f and f2 = p2 D f.

For ,i=1,2 we will show piD

(

f +g

) (

= piDf

) (

+ piDg

)

and piD

( )

fg

(

pi D f

) (

pi Dg

)

.

= For any sS, f, g

(

R1R2

) [[

S, ω( )1 ⊕ω( )2

]]

and ,

2 ,

=1

i we have

(11)

( )

(

pi D f + g

) ( )

s = pi

( (

f + g

) ( )

s

) ( ) ( ) (

f s g s

)

pi +

=

( )

(

f s

)

p

(

g

( )

s

)

pi + i

=

(

pi D f

) ( ) (

s + pi Dg

) ( )

s

= and

( )

(

pi D fg

) ( )

s = pi

( ( ) ( )

fg s

)

( ) (

( ) ( )

) ( ( ) )

⎜⎜

⎛ ω ⊕ω

=

=xy s

i f x s g y

p 1 2

( ) (

( ) ( )

) ( ( ) )

=

ω

⊕ ω

=

xy s

i s

if x p g y

p 1 2

( )

( )

( ( ) )

=

ω

=

xy s

i s

if x p g y

p 1

( ) ( )

( )

( ( ) ( ) )

=

ω

=

xy s

i s

i f x p g y

p D 1 D

( ) ( )

(

pi D f pi Dg

) ( )

s .

=

Since for any f, g

(

R1R2

) [[

S( )1 ⊕ω( )2

]]

, we have

(

f + g

) ((

= f + g

) (

1, f + g

) )

2

ψ

( ) ( )

(

p f + g p f + g

)

= 1D , 2D

( ) ( ) ( ) ( )

(

p1D f + p1Dg , p2D f + p2 Dg

)

=

(

f1+ g1, f2 +g2

)

=

(

f1, f2

) (

+ g1, g2

)

=

( )

f

( )

g ψ

=

(12)

and

( ) (( ) ( ) )

fg = fg 1, fg 2 ψ

( ) ( )

(

p1D fg , p2 D fg

)

=

( ) ( ) ( ) ( )

(

p1D f p1Dg , p2 D f p2Dg

)

=

(

f1g1, f2g2

)

=

(

f1, f2

) (

g1, g2

)

=

( ) ( )

f ψ g , ψ

= ψ is a ring homomorphism.

Now, we will show ψ is injective. Let fKer

( )

ψ . Then we will show .

= 0

f Since fKer

( ) ( ) (

ψ , ψ f = 0, 0

)

. So, for any sS and i =1, 2, we have

(

pi D f

)( )

s = 0

( )

s . In other words, pi

(

f

( )

s

)

=0. Since pi is a natural projection, f

( )

s = 0 for all sS. So Ker

( )

ψ =0 or ψ is injective.

Furthermore, we will show ψ is surjective. For all

(

f1, f2

)

R1

[[

S( )1

]]

[[

, ( )2

]]

,

2 ω

R S there exists

[[

( ) ( )

]]

= ∈ ⊕ ω ⊕ω

= 2

1

2 1 2

1 ,

k ik fk R R S

f D

such that ψ

( ) (

f = f1, f2

)

.So, ψ is surjective. Then, ψ is a ring isomorphism.

So R1R2

[[

S, ω( )1 ⊕ω( )2

]]

~= R1

[[

S, ω( )1

]]

R2

[[

S( )2

]]

. References

[1] R. Mazurek and M. Ziembowski, Uniserial rings of skew generalized power series, J. Algebra 318 (2007), 737-764.

[2] P. Ribenboim, Generalized power series rings, Lattice, Semigroups and Universal Algebra, Plenum Press, New York, 1990, pp. 271-277.

[3] G. A. Elliott and P. Ribenboim, Fields of generalized power series, Arch. Math.

54 (1990), 365-371.

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[4] P. Ribenboim, Noetherian rings of generalized power series, J. Pure Appl. Algebra 79 (1992), 293-312.

[5] P. Ribenboim, Rings of generalized power series II: Units and zero-Divisors, J. Algebra 168 (1994), 71-89.

[6] P. Ribenboim, Semisimple rings and von Neumann regular rings of generalized power series, J. Algebra 198 (1997), 327-338.

Ahmad Faisol: [email protected]; [email protected] Budi Surodjo: [email protected]

Sri Wahyuni: [email protected]

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