International Journal of Mathematics and Computer Applications Research (IJMCAR) ISSN(P): 2249-6955; ISSN(E): 2249-8060 Vol. 4, Issue 2, Apr 2014, 45-54 © TJPRC Pvt. Ltd.

SOME ASPECTS OF MODULE THEORETIC HOMOMORPHISMS MAITRAYEE CHOWDHURY Department of Mathematics Gauhati University, Guwahati, Assam, India

ABSTRACT Here we try to highlight some aspects of module homomorphisms when the attached rings are different and hence, the already known cases of module homomorphisms come up as in some sense as corollaries to what we have proposed. This very approach may open up new windows so far the module homomorphisms are concerned (at least in some sense -the so called categorical approach is concerned). We would like to explore some interesting phenomena of module homomorphisms in some broader aspects. Moreover, it would be interesting to note how already available module theoretic results with same ring may have more than one version.

KEYWORDS: Homomorphisms, Module Theoretic, Injective Modules 1. INTRODUCTION This paper is an attempt for some sort of generalization of modules and its homorphisms [1, 3, and 4] with different rings –best suited for some aspects of categorical study of rings and homomorphisms. As it appears as a first step towards the expected theory, we would like to present here some very familiar aspects of what we are attempting for. And in a natural approach we, here present, as most beginning to it the main aspect of the theory, viz., the homomorphism theorems etc with examples. Here we would like to comprise some torsion character of modules with different rings, with some direct sum related problems. We see that homomorphic image of such a left module epimorphism (in broader aspect) appears with invariant character in the corresponding torsion part of the codomain module [4]. Moreover, some necessary and sufficient condition type property is carried over in case of integral domain makeup of the codomain module. It is also observed that simple character is also preserved in such type of left module morphisms. We here skillfully show the isomorphism character in case of quotient structure with well-behaved nature. If two left module morphisms agree on the generating set of a submodule of the codomain module, then the two left module morphisms agree also on the codomain submodule [2,3]. Finally we try to highlight some categorical property so far the kernel is concerned in such type of left module morphisms.

2. PRELIMINARIES We begin with some definitions: 2.1 Definition A left ring module is a pair ( R, M ) , where R is a ring; M is an additive abelian group with •

r(x+y)=rx+ry

•

(r1+r2)x=r1x+r2x

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Maitrayee Chowdhury

•

(r1r2)x=r1(r2x), where r,r1,r2 ∈R and x,y∈M. Let ( R1 , M 1 ) and ( R2 , M 2 ) be two left ring modules. Then

2.2 A Left Module-Morphism

(α , f ) is a system α : R1 → R2

(α , f )

( R1 , M 1 ) → ( R2 , M 2 ) where f : M 1 → M 2 is an additive group homomorphism and

is a ring homomorphism with

f(r1m1)= α(r1 )f(m1), with r1 ∈R1, m1∈M1 When the ring R1 is considered as a left R1- module, and a ring homomorphism

α : R1 → R2

f : R1 → R2 is such that for

with α( r )= f(r ), then we have f(rs)=f( r )f(s)= α(r ) f (s) . Thus f(rs)= α(r ) f (s) and this ring

homomorphism is a left module-morphism ( f ,f )

( R1 , R1 ) → ( R2 , R2 ) too. The left module morphism 2.3

“

f : M1 → M 2 ”

(α , f ) under consideration may also be called is

an

(α , R1 − R2 ) homomorphism.

[so

the

left

module

morphism

( f ,f )

( R1 , R1 ) → ( R2 , R2 ) is an (f-R1-R2) homomorphism (i, f )

Note: If R1 = R2 =R then we get the left module morphism,

( R, M 1 ) → ( R, M 2 ) - the (i, R-R) homomorphism

as our well known module homomorphism viz., R-homomorphism. For here,

f : M 1 → M 2 , where i : R → R , with

i(r)=r f(a1+a2)=f(a1)+f(a2) and i(ra1)= i(r) f(a1)= rf(a1) , for r, ∈R and a1,a2∈M (α , f )

2.4 A left module morphism ( R1 , M 1 ) if

→ (R , M 2

2

) is an α- monomorphism, (α-epimorphism, α- isomorphism)

f : M 1 → M 2 is injective(, surjective, bijective). If (α , f ) : ( R1 , M 1 ) :→ ( R2 , M 2 ) is an α- isomorphism, we denote

it by

( R1 , M 1 ) ≅α ( R2 , M 2 ) f

α ( S1 )

An R2-submodule N2 of M2 is an

(α , R2 ) -submodule of M2 [denoted: N 2 ≤ M ] if N2 is an α ( S1 ) -submodule 2

of M2 for some subring S1 of R1. If α is onto, then R2= α(R1) thus, M2 is an 2.5

(α , R2 ) -submodule of M2 ;this M2 , is then an

(α , R2 ) - module.

2.6 Lemma •

R1

α ( R1 )

If N1 ≤ M then f ( N1 ) ≤ M , and 1 2

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Some Aspects of Module Theoretic Homomorphisms

•

R2

For A2 ≤ M we have f −1 ( A2 ) ≤ α 2

R1

R1

M 1 [ it is to be noted the difference between N ≤ M1 and f 1

−1

( A2 ) ≤

R1

α

M1 ]

Proof f(N1)={f(n1)|n1 ∈ N1}. Now, α(r1) ∈ α(R1), f(n1) ∈ f(N1) give α(r1) f(n1)= f(r1n1)=f(t1) ∈M2 , [for, t1=r1m1∈M1] •

α ( R1 )

Thus f ( N1 ) ≤ M

2

//

And f −1 ( A2 ) = {m1 ∈ M 1 | f (m1 ) ∈ A2 } . Now, suppose, r1 ∈ R1, m1,m2∈ f-1(A2). Then f(m1), f(n1) ∈ A

2

and

f(m1)-f(n1)=f(m1-n1) =f(t1) ∈ A 2 , as A2 is an R2-submodule of A2. And, f(r1m1)= α(r1)f(m1) ∈ A 2 gives, r1m1∈ f-1(A2) and hence, f −1 ( A2 ) ≤

α

Now, in case of if a,b∈ kerα

M 1 Here we know that our familiar kernel, kerf or, f-1(02) is an R1 -submodule of M1

(α , f ) : ( R1, M 1 ) :→ ( R2 , M 2 )

f , then f(a1)-f(b1)= 02⇒f(a1-b1)= 02 [ a1-b1∈M1]

so a1-b1∈ kerα r1∈R1, m1∈ kerα

R1

f And, if r1∈R1, m1∈ kerα f , then, f(r1m1)=α(r1)f(m1)= 02⇒f(r1m1)= 02⇒r1m1∈ kerα f . Thus,

f ⇒r1m1∈ kerα f , Hence, f-1(02) or kerα f or,

f −1 (0 2 ) ≤α

i.e. f (02) is an (α,R1)-submodule of M1.[thus, f (0 2 ) ≤ α -1

morphism, i.e.,

−1

R1

M1 (α , f )

M 1 ]. ( R1 , M 1 ) → ( R2 , M 2 ) is

R1

a left module

f : M 1 → M 2 is an (α , R1 − R2 ) homomorphism. So f : M 1 → M 2 and α : R1 → R2 are as

described. Now, 2.7 Definition α-kernel of f [or, kernel

(α , f ) ] or, Kα =Kα(f)= {m1∈M1|f(m1)=02}= f-1(02)

When R1=R2=R and i, the identity homomorphism then, α-kernel of f is kernel of f, our familiar kernel of f viz., kerα

f = {m1 ∈ M 1 | f (m1 ) = 0 2 } . If (A,+) is a sub group of (M2,+), then f-1(A) = { m1∈M1| f(m1) ∈A} is an

(α,R1)-submodule of M1 2.8 Definition Let C be a class for each pair A,B ∈C we have morc(A,B) , a set . Elements of morc(A,B) are “arrows”

f : A → B . A is called its domain and B its codomain. If A,B, C ∈C then we have a function : morc ( B, C ) × morc ( A, B ) → morc ( A, C ) such that for g : B → C ∈ morc(B,C) and f : A → B ∈ morc(A,B) we have

g

f :A→C∈

morc(A,C).

If

the

system

triple h : C → D , g : B → C , f : A → B , h ( g www.tjprc.org

C=

(C,morc, )

is

such

that

C1.

For

every

f ) = (h g ) f C2. For each A∈ C, there is a unique editor@tjprc.org

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Maitrayee Chowdhury

1A ∈ morc ( A, A) such that if f : A → B , g : C → A , then f 1A = f and 1A g = g , then C is a category. Each arrow is a morphism. 2.9 Definition α

f : A → B or A → B has a kernel if there is a morphism M → A we have f α = 0 f

A morphism

g

C → A with f g = 0 - the zero morphism, we get a unique morphism

the zero morphism, and for any morphism such that α

h

C→ M

h=g.

3. MAIN RESULTS 3.1 Theorem Fundamental Theorem of α-Homomorphism (α , f )

(R1,M1) and (R2,M2) are two left ring modules and an α-R1-R2 homomorphism from M1 to M2, then

( R1 ,

M1

( R1 , M 1 ) → ( R2 , M 2 ) is an left module- morphism [or, f is ) ≅α ( R2 , f (M 1 )) with ф: φ

kerα f

M1

kerα f

→ f (M 1 ) such

that ф(a1+kerαf)= f(a1) [a1∈M1] Proof: Here we have two mappings, viz., We define a mapping ф:

M1

kerα f

α : R1 → R2 , with r1 → α (r1 )

and

f : M 1 → M 2 , with m1→f(m1).

→ f (M 1 ) by ф(a1+kerαf)= f(a1) [a1∈M1]

•

Ф is well defined. For, a1+kerαf=b1+kerαf ⇒ a1-b1∈ kerαf⇒ f(a1-b1)=0 ⇒ f(a1)= f(b1) [f is additive group homo]

•

Ф is one-one , φ ( a1 ) = φ (b1 ) [ i.e. ф(a1+kerαf)= ф(b1+kerαf)] ⇒ f(a1)= f(b1)⇒ f(a1)- f(b1)=0 ⇒ f(a1- b1)=0⇒ a1- b1 ∈ kerαf ⇒a1+kerαf= b1+kerα ⇒ a1

•

= b1

Ф is onto-easy! Now we see that Ф is an (R1,R2) –α homomorphism . Here, Ф ( a1 + b1 ) = Ф (a1

ф (a1 ) + ф (b1 ) and Ф(r1 a1 )= Ф( r1a1 ) = f(r1a1)= α(r1) f(a1)= α(r1) (α-R1,R2)

then

isomorphism,

i.e.,

( R1 ,

f ( M 1 ) = M 2 And then ( R1 , M 1

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M1

) ≅α ( R2 , f (M 1 )) φ

kerα f

φ (a1 ) . If

Hence, ф:

+ b1 ) = f(a1+b1) =f(a1)+f(b1) = M1

f : M1 → M 2

kerα f is

→ f (M 1 ) is an an

epimorphism,

φ

kerα f

) ≅α ( R2 , M 2 ) //

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Some Aspects of Module Theoretic Homomorphisms

3.2 Theorem Correspondence Theorem for Left Module Morphism (α , f )

(R1,M1) and (R2,M2) are two left ring modules. If

( R1 , M 1 ) → ( R2 , M 2 ) is a left module epimorphism, then

there is a one-one correspondence between the set of all R2-submodules A2 of M2 and those R1-submodules of A1 such that kerαf⊆ A1 and; where, in other words the correspondence is of type:

(kerα f ⊆) A1 = f −1 ( A2 ) ↔ A2

(α , f )

Proof: Here

( R1 , M 1 ) → ( R2 , M 2 ) is a left-module epimorphism [or, (α-(R1-R2) epimorphism. i.e.,

f : M 1 → M 2 is an additive group epimorphism with the attached homomorphism α : R1 → R2 as r1 → α (r1 ) . And R1

A2 is an R2-submodule of M2. We take f-1(A2)=A1 Here kerα f ⊆ f −1 ( A2 ) ≤ M

1

For x1,y1∈f-1 (A2)⇒f(x1), f(y1) ∈ A2⇒

f(x1-y1)∈ A2⇒ x1-y1∈ f-1(A2). For x1∈ f-1 (A2), r1∈R1, we have f(x1) ∈ A2,α(r1) ∈R2 ⇒ α(r1) f(x1) ∈ A2⇒ f(r1x1) ∈ A2⇒ r1x1∈ f-1 (A2) Therefore, f-1(A2)=A1 is an R1-submodule of M1 And clearly the correspondence is one-one- for if I1 and J1 are two R1-submodules of M1 with kerαf⊆ I1,J1 with f(I1)=f(J1), then I1= f-1(f(I1)) and J1= f-1(f(J1)).// Consider the map from the set of R1-submodules of M1 to the set R2-submodules of f(M1)=M2, i.e.,

R2

R1

R2

β : A2 | A2 ≤ M 2 → f −1 ( A2 )(≤ M 1 ) | A2 ≤ M 2 ,[ β is one-one]

3.3 Example

{

}

Z [ x] = a0 + a1 x + a2 x 2 + ... + an x n | an ∈ Z , n = 0,1,2.. , Z = {o,±1,±2,±3,...}

{

}

{ }

Z 2 [ x] = 0, x, x 2 , x + x 2 , x 3 , x + x 3 , x + x 2 + x 3 ,..... Z 2 = 0,1 Here

Z [x] is a left Z -module. Similarly Z 2 [ x] is a left Z 2 -module w.r.t. the maps .

Z × Z [ x] → Z [ x] :

(n, (a0 + a1 x + a2 x 2 )) → (na0 ) + (na1 ) x + (na2 ) x 2

Z 2 × Z 2 [ x] → Z 2 [ x] : (n, (a0 + a1 x + a2 x 2 )) → (na0 ) + (na1 ) x + (na2 ) x 2 [In Z2 n is either 0 or 1 ] We i.e.,

consider

the

maps:

f : Z [ x] → Z 2 [ x] :

a0 + a1 x + a2 x 2 → a0 + a1 x + a2 x 2

f (a0 + a1 x + a2 x 2 ) = a0 + a1 x + a2 x 2 and α : Z → Z 2 , n → n i.e. α (n) = n

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Now

f (n(a0 + a1 x + a2 x 2 )) = f ((na0 ) + (na1 ) x + (na2 ) x 2 )

= na0 + na1 x + na2 x 2 = na0 + na1 x + n2 a2 x 2 = n(a0 + a1 x + a2 x 2 ) = α (n) f (a0 + a1 x + a2 x 2 ) f (n(a0 + a1 x + a2 x 2 )) = α (n) f (a0 + a1 x + a2 x 2 ) Hence, f is an α , Z − Z 2 morphism from,

Thus

(α , f )

Z Z [x ] to

( Z , Z [ x]) → ( Z 2 , Z 2 [ x]) is a left module morphism with

Z 2 Z 2 [ x]

f : Z [ x] → Z 2 [ x] and

α : Z → Z 2 , as defined above. 3.4 Direct Sum of Modules over Different Rings M1 is a left R1-module and R2 is a left R2-module. Now

M 1 ⊕ M 2 = {(m1 , m2 ) | mi ∈ M i }and

R1 ⊕ R2 = {(r1 , r2 ) | ri ∈ Ri } It is easy to see that R = R1 ⊕ R2 is a ring and M = M 1 ⊕ M 2 is a left R-module it is not difficult to see that

M = M 1 ⊕ M 2 is an additive abelian group. Now we define the map R×M→M as (r,m)→rm ,

where, m=(m1,m2) and r=(r1,r2), ring

rm = (r1 , r2 )(m1 , m2 ) = (r1m1 , r2 m2 ) and it appears that M is a left R-module .And as

R2 ≅ 0 ⊕ R2 , R1 ≅ R1 ⊕ 0 . Thus in this sense, as ring we can consider each of R1 and R2 as a subring of

R = R1 ⊕ R2 and so M = M 1 ⊕ M 2 is also an R1-module as well as a left R2-module. Here for m=(m1,m2) ∈M and r1∈R1 gives r= (r1,0) ∈R1⊕ R2 and so

rm = (r1 ,0)(m1 , m2 ) = (r1m1 ,0) It is clear that M is a left R1 ⊕ 0 –module i.e., M is a left

R1-module. Now M is a left R-module and M1 is a left R1-module. We define f: M →M1 such that f(m)=f(m1,m2)=m1 Clearly, the map is onto and we see the map ф:R →R1 where ф(r)=ф(r1,r2)=r1 is also an onto map. Now we see that f is an R-R1 homomorphism. For 1) f(x+y)=f(( x1,x2)+ (y1,y2)) = f(x1+y1,x2+y2)=x1+y1=f(x1,x2)+f(y1,y2)=f(x)+f(y) For r=(r1,r2) ∈R, f(rx)=f[(r1,r2) (x1,x2)]= f(r1x1,r2x2)= f(r1x1) =r1.x1=ф(r1,r2).f(x1,x2)=ф (r) f(x) Thus f(rx)= ф (r) f(x) Clearly f is onto and one-one. And kerфf= {m∈M| f(m)=01} ={(m1,m2)| f(m1,m2)=01}= {(0,m2)|m2∈M2}( ≅M2). So by fundamental theorem of homomorphism, we get: 3.5 Theorem

M 1 ≅φ M f

⊕M2)/M2. Similarly,

kerφ

f =

M 1 ≅ψ f

(M 1 ⊕ M 2 )

(M1 ⊕ M 2 )

appears as a Quotient modules: Let

kerφ f

M2

≅

(M 1 ⊕ M 2 )

M2

i.e.

M 1 ≅φ

f

(M1 ⊕ M 2 )

M2

M1 ≅

ф

(M1

, with ψ(r)=f(r1,r2)=r2 We now note when a Ring modulo a one-sided ideal

( R1 , M 1 ) and ( R2 , M 2 ) be two left ring modules. In other words, M1 is a left

R1 –module and R2 is a left R2-module. Define A1=A(M1)={r1∈R1|r1m1=0, m1∈M1}, then A1 is an R1-submodule of R1 (i.e. it is a left ideal of R1). For r,s∈A1⇒rm1=0, sm1=0⇒(r-s)m1=0⇒r-s∈A1 for all m1∈M1. And for r∈R1 and r1∈A1,

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Some Aspects of Module Theoretic Homomorphisms

(rr1)m1=r(r1m1)=r.0=0⇒rr1 ∈A1 , for all m1∈M1 Similarly, A2=A(M2)={r2∈R2|r2m2=0, m2∈M2} is an R2-submodule of R2 Thus, R1/A1 is an R1-module similarly R2/A2 is an R2-module. Then 3.6 Theorem

R1 φ R R1 , ≅α R2 , 2 , for some epi α:R1 → R2. Define: ф: R1→ R2/A2 with ф (r1)=α(r1)+A2 then for r1 , r,s∈R, A1 A2 Ф(r+s)=α(r+s)+A2 = (α(r)+ α(s))+A2= (α(r) +A2 )+ (α(s)+A2)=ф(r)+ф (s) and, ф(rs)=α(rs)+A2=α(r)α(s)+A2=α ( r) (α(s) +A2)=α(r) ф(s). ф is an epi, for r2+A2 ∈ R2/A2,r2 ∈R2 and α being epi, we have r1∈A1with α(r1)=r2 thus r2+A2= α(r1)+A2= ф (r1), i.e.

r2 = φ (r1 ) Therefore, ф is an α-R1-R2 epimorphism, or an α-epimomorphism. Now, by fundamental theorem,

R1 φ R R1 , ≅α R2 , 2 // A1 A2 3.7 Some More Results (α ,φ )

( R1 , M ) and ( R2 , N ) are two left ring modules. ( R1 , M ) → ( R2 , N ) is a left module morphism. [φ : M

→ N , α : R1 → R2 ] Then ( φ (Tor ( R1 M )) ⊆ Tor ( R2 N ) with Proof: Let

α : R1 → R2

Tor ( R1 M ) = {m∈M|r1m=0, for some r1∈R1}

be an epimorphism. Now we have,

Tor ( R1 M ) = {m∈M|r1m=0, for some r1∈R1}

φ (Tor ( R M )) = {φ (m) | m ∈ Tor ( M )} = {φ (m) | φ (r1m) = 0, for r1 ∈ R1} 1

= {φ ( m) | α ( r1 )φ ( m) = 0, Thus,

for α (r )1 ∈ R2 } ⇒ф(m)[Є Tor ( R2 N ) ]⊆ Tor ( R2 N ) [ as, φ ( R1 M ) ⊆ R2 N

φ (Tor ( R M )) ⊆ Tor ( R N ) .// 1

2

Note:

For ( R1 , M ) and ( R2 , N ) two

(α ,φ )

if ( R1 , M )

→ ( R , N ) is a left module morphism, with φ : M → N , α : R 2

1

left

ring

modules,

→ R2 .Then Ф(r1x)+y)=α(r1) ф(x)+ф(y),

for all x,y∈M and for all r1∈R1 Now 3.8 Theorem We consider each of R1 and R2 is with unity. If R2 is an integral domain with unity, then for a (α ,φ )

φ : M → N , α : R1 → R2 , ( R1 , M ) → ( R2 , N ) is a left module morphism if Ф (r1x) +y) =α (r1) ф(x) +ф(y). Proof: Taking r=1, we get Ф(x+y) = ф(x) +ф(y) [only when R2 is an integral domain] And, for y=0, Ф (r1x) =α (r1) ф(x). Therefore, ф is an α -R1-R2 homomorphism. i.e.,

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Maitrayee Chowdhury

3.9 Example A left module morphism, α-epi[

f : M 1 → M 2 , α : R1 → R2 If M1 is simple, then f(M1) is a simple

R2-submodule of N and if f(M1)≠0 then f is one-one. Proof: Here, clearly f(M1) is an α(R1) ( if it is not epi) R2-submodule of N-as α-epi. Now let T2 be an R2-submodule of f(M1) Claim: For

f −1 (T2 ) is an R1-submodule of M1. Now, f −1 (T2 ) = {m1 ∈ M 1 | f (m1 ) ∈ T2 } .

m1 , n1 ∈ f −1 (T2 ) ⇒ f (m1 ), f (n1 ) ∈T 2 ⇒ f (m1 − n1 ) ∈ T2 ⇒ m1 − n1 ∈ f −1 (T2 ) And for m1∈M1, r1∈R1,

f(r1m1)= α(r1)f(m1) ∈T2 ⇒ r1m1 ∈f-1(T2)⇒ f-1(T2) is an R1-submodule of M1. Since M1 is simple, so either f-1(T2) is 0 or M1.Thus T2=f(M1) or, T2=f(0)=0 ⇒ f(M1) is simple [R2-submodule of M2]. If f(M1) ≠0 then T2≠0. We have, f-1(0)⊆ f1

(T2)[=M1 or (0)] ⇒ f is a monomorphism.

3.10 Example An R1-submodule <S> of M1 generated by a non-empty subset S1 of M1 [ S1 that <

⊆ M 1 ] such

(α , f )

S >= {∑i ri xi | ri ∈ R, xi ∈ S } . Consider two modules M 1 , N1 with morphism ( R1 , M 1 ) → ( R2 , M 2 ) And (α , g )

( R1 , M 1 ) → ( R2 , M 2 ) If f(x)=g(x) for all x∈S, then f and g agree on the R1-submodule < S > . To show that f and g agree on

< S > i.e., f(s)=g(s) for all s∈S For s∈S , we have r1 x1 + r2 x2 + .....rt xt , t=1,2,… Now,

f ( s ) = f (r1 x1 + r2 x2 + .....rt xt ) = f (r1 x1 ) + f (r2 x2 ) + ..... + f (rt xt ) = α ( r1 ) f ( x1 ) + α ( r 2 ) f ( x2 ) + ..... + α ( rt ) f ( xt ) = α ( r1 ) g ( x1 ) + α (r 2 ) g ( x2 ) + ..... + α ( rt ) g ( xt ) = g ( r1 x1 ) + g ( r2 x2 ) + ..... + g ( rt xt ) = g (r1 x1 + r2 x2 + .....rt xt ) =g(s) Therefore, f and g agree on the R1-submodule < S > // 3.11 Theorem Let A1 and A2 be R1 and R2 submodules of M1 and M2 respectively. Then,

M × M2 R1 × R2 , 1 ≅α ×α A1 × A2 1 2 (ν 2 , i 2 )

and ( R2 , M 2 )

M2

→ (R , A 2

(ν 1 , i1 ) M M M R1 × R2 , 1 × 2 , where, ( R1 , M 1 ) → ( R1 , 1 ) A1 A2 A1

) are left module morphisms with ν 1 : M 1 →

2

M1 M2 and ν 2 : M 2 → as natural additive A1 A2

group epimorphisms and i1, i2 are respective identity homomorphisms with

α1 : R1 → R1 and α 2 : R2 → R2 be any ring homomorphisms.,

Impact Factor (JCC): 4.2949

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53

Some Aspects of Module Theoretic Homomorphisms

M 1 × M 2 and

Proof: Here, the corresponding homomorphisms are as follows:

M1 M 2 × are both A1 A2

R1 × R2 (= R) -modules with rm= r (m1 ,m2 ) = (r1 , r2 )(m1 , m2 ) = (r1m1 , r2 m2 ) and =r

(m , m ) = (r , r )(m , m ) = (r m , r m ) 1

2

1

Define

2

1

2

1

1

2

2

(ν =)ν 1 ×ν 2 : M 1 × M 2 →

M1 M 2 × such that A1 A2

ν (m ) = (ν 1 ×ν 2 )(m1 , m2 ) = (ν 1 (m1 ),ν 2 (m2 ))[= (m1 , m2 ) = (m1 + A1 , m2 + A2 )] . [where m=m1 ×m2] Clearly,ν [: R1 × R2

= ν 1 ×ν 2 is well defined as well as onto. Moreover it is α1 × α 2

→ R1 × R2 ] homomorphism. For

ν (rm ) = (ν 1 ×ν 2 )[(r1 , r2 )](m1 , m2 ) = (ν 1 ×ν 2 )(r1m1 , r2 m2 )

= (ν 1 (r1m1 ),ν 2 (r2 m2 )) = (α 1 (r1 )ν 1 (m1 ),α 2 (r2 )ν 2 (m2 ))

(

)

Here, [= m1 , m2 = (m1 + A1 , m2 + A2 )] (α1 (r1 )ν 1 (m1 ), α 2 (r2 )ν 2 (m2 )) = (α1 × α 2 )(r1 × r2 )(ν 1 × ν 2 )(m1 , m2 ) = α (r )ν (m )

{

=

{(m , m ) | (m , m ) = (0 , 0 )}= {(m , m ) | m ∈ A , m ∈ A }= A × A 1

2

1

2

1

2

}

= (m1 , m2 ) | ν (m1 , m2 ) = (01 , 02 ) = ( A1 , A2 )

Thus,ν (rm ) = α (r )ν (m ) Now kerα ν

1

2

1

1

2

2

1

2

And by fundamental theorem

M × M 2 (ν ×ν 2 ) M M R1 × R2 , 1 ≅(α ×α ) R1 × R2 , 1 × 2 or 1 2 A1 × A2 A1 A2 1`

M × M2 ν M M R1 × R2 , 1 ≅α R1 × R2 , 1 × 2 A1 × A2 A1 A2 4. SOME CATEGORICAL OBSERVATION ( f ,α )

4.1 Every left morphism has a kernel in categorical sense [1]. . morphism, where

( R1 , M 1 ) → ( R2 , M 2 ) is a left module

f : M 1 → M 2 is group homomorphism and α : R1 → R2 is ring homomorphism. We write ( f ,α )

(i , j )

M ′ = Kerf and R′ = Kerα Then we have ( R′, M ′) → ( R1 , M 1 ) → ( R2 , M 2 ) , i : M ′ → M and j : R′ → R1 are natural

inclusion ( g ,b )

maps

such

that

( f , a ) (i, j ) = ( f i, a j ) = (0,0)

We

assume

that

( f ,α )

( R3 , M 3 ) → ( R1 , M 1 ) → ( R2 , M 2 ) with ( f , a ) ( g , b) = (0,0) . Then ( h, c )

( f , a ) ( g , b) = (0,0) ⇒ ( f

g , a b) = (0,0) ⇒ f

( R3 , M 3 ) → ( R′, M ′) with h(m3)=g(m3) and c(r3)=b(r3) As

www.tjprc.org

g = 0,a b = 0

f ( g (m3 )) = ( f

------

(i)

We

consider

g )(m3 ) = 0 ⇒ g (m3 ) ∈ M ′ and

editor@tjprc.org

54

Maitrayee Chowdhury

a (b(r3 )) = (a b)(r3 ) = 0 ⇒ b(r3 ) ∈ R′ . Thus (h,c) is well defined. Now g(m3)=h(m3)=i(h(m3))= (i h)(m3 ) for all m3 ∈M3. Thus

i h = g . Similarly, j c = b . Therefore, ( g , b) = ((i h, j c) ⇒ ( g , b) = (i, j ) (h, c) . Also (h,c) is

unique. Hence (f,a) has a kernel in categorical sense.// Note: One may look into the possibility of some approach for another interesting isomorphism character as described below. However this is our suggestion only. If we have a left module morphism of the type (α , f )

( R1 , M 1 ) → ( R2 , M 2 ) , where f : M 1 → M 2 is an additive group homomorphism and α : R1 → R2 is a ring homomorphism with f(r1m1)= α(r1 )f(m1), with r1 ∈R1, m1∈M1 Then we may give the following definitions: The left (α , f )

module morphism

( R1 , M 1 ) → ( R2 , M 2 ) is an f-isomorphism if α : R1 → R2 is bijective. We denote it by

( R1 , M 1 ) ≅ f ( R2 , M 2 ) ] ACKNOWLEDGEMENTS We would like to acknowledge Professor K.C. Chowdhury of Department of Mathematics, Gauhati University for his new suggestion, guidance and encouragement towards the finishing of the paper.

REFERENCES 1.

Behrens, E.A.

: Ring theory, New York, Academic press;1972

2.

Bourbaki, N.

: Algebra, Paris : Hermann & Cie, 1958

3.

Faith, C.

: Lectures on injective modules and quotients rings., Lecture Notes in Mathematics, Springer

verlag-1967 4.

Fuller, K.R.

: Rings and category of modules Springer-Verlag Newyork1973 Anderson, F.W.

5.

Lembak, J

: Lectures on Rings and modules: Waltham-Toronto-London: Blaisdell,1969

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6 maths some aspects of module maitrayee chowdhury1

Published on May 16, 2014

Here we try to highlight some aspects of module homomorphisms when the attached rings are different and hence, the already known cases of mo...

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