Math, asked by KrishnaKantPathak, 6 months ago

let M be an abelian group. show that there is only one of making G as a left Z module​

Answers

Answered by mohit921
1

Answer:

Modules

Covention: This semester all rings will be assumed to have 1.

Definition. Let R be a ring. A left R-module is a set M with two opera-

tions:

(1) binary operation + on M, that is a map + : M × M → M

(2) an action of R on M, that is, a map R × M → M

(r, m) 7→ rm

satisfying the following axioms:

M1: (M, +) is an abelian group

M2: (r + s)m = rm + sm for all r, s ∈ R and m ∈ M

M3: (rs)m = r(sm) for all r, s ∈ R and m ∈ M

M4: r(m + n) = rm + rn for all r ∈ R and m, n ∈ M

M5: 1 · m = m for all m ∈ M

Elements of R are often called scalars.

Remark: (1) Similarly one defines right R-modules, where the action is

denoted by (r, m) 7→ mr, and axioms M2-M5 are replaced by their “mirror”

images.

(2) If R is commutative, left R-modules = right R-modules

From now on by an R-module we will mean a left R-module.

1.1. Basic examples of modules.

1. Assume that R is a field. Then R-modules = vector spaces over R

2. Let R be any ring and n ∈ N. Let Rn = {(r1, . . . , rn) : ri ∈ R}. Then Rn

is an R-module where

r(r1, . . . , rn) = (rr1, . . . , rrn).

Rn

is called the standard free R-module of rank n.

3. Let R be any ring, S a subring of R with 1. Then R is an S-module with

action = left-multiplication.

In particular, any ring R is a module over itself.

4. Let R be any ring. I an ideal of R. Then R/I is an R-module where

r(a + I) = ra + I.

Modules over some special rings.

Modules over Z

Claim 1.1. Modules over Z = abelian groups.

“Proof ”. If M is a Z-module, then (M, +) is an abelian group by definition.

Conversely, if A is an abelian group, we can turn A into a Z-module by

setting

na =

a + . . . + a

| {z }

n times

if n > 0

0 if n = 0

a + . . . + a

| {z }

−n times

if n < 0

Module axioms trivially holds.

There is no other way to make M a Z-module since for any n ∈ N and a ∈ A

we must have na = (1 + . . . + 1

| {z }

n times

)a = a + . . . + a

| {z }

n times

by M2 and M5; similarly

we must have 0 · a = a and (−n) · a = −(na).

In other works, an action of Z on M is completely determined by addition

on M.

Modules over F[x] where F is a field

Claim 1.2. Modules over F[x] = pairs (V, A) where V is a vector space

over F and A : V → V a linear transformation.

Sketch of the proof. (see [DF,pp. 340-341] for more details) Let V be an

F[x]-module. Then V can also be considered as an F-module = F-vector

space.

Define a mapping A : V → V by A(v) = xv. By module axioms A is a linear

transformation form V to V .

Conversely, given an F-vector space V and a linear transformation A : V →

V , we want to make V into F[x]-module such that xv = A(v) for all v ∈ V .

By module axioms we are forced to set

(x

2

)v = x(xv) = A(xv) = A(A(v)) = A

2

(v).

Similarly, x

nv = Anv for any n ∈ N, and finally for any p(x) ∈ F[x] we must

have p(x)v = (p(A))v, that is,

(cnx

n + . . . + c0)v = (cnA

n + . . . + c0)(v) (∗ ∗ ∗)

Thus, once we decided how x acts on V , the action of any element of F[x]

has to be given by (***). We still have to verify that (***) indeed defines

an F[x]-module structure on V , but this verification is routine.

Submodules, quotient modules and homomorphisms.

Definition. Let M be an R-module. A subset N of M is called an R-submodule

if

(1) N is a subgroup of (M, +)

(2) for any r ∈ R, n ∈ N we have rn ∈ N.

Example: Let R be a ring, M = R (with action by left multiplication). Then

submodules of R = left ideals of R.

Definition. If M is an R-module and N is a submodule of M, we can define

the quotient module M/N. As a set M/N is just the quotient group M/N,

and R-action is given by

r(m + N) = rm + N for all r ∈ R, m ∈ M.

Definition. If M and N are R-modules, a mapping ϕ : M → N is called a

homomorphism of R-modules (alternatively ϕ is an R-linear mapping) if

(1) ϕ is a homomorphism of abelian group

(2) ϕ(rm) = rϕ(m) for all r ∈ R, m ∈ M.

1.4. Modules and group actions.

Definition. Let G be a group. The integral group ring Z[G] is defined as

follows: as a set Z[G] is the collection of formal finite linear combinations of

elements of G with integral coefficients, that is,

Z[G] = X

g∈G

ngg : ng ∈ Z and only finitely many ng are nonzero.

Addition and multiplication on Z[G] are defined by setting

(

P

g∈G ngg) + (P

g∈G mgg) = P

g∈G(ng + mg)g and

(

P

g∈G ngg) · (

P

g∈G mgg) = P

g∈G lgg where lg =

P

h∈G nhmh−1g

.

In other words, multiplication in Z[G] is obtained by first setting g · h to

be the product of g and h in G and then uniquely extending to arbitrary

elements of Z[G] by distributivity.

Theorem (HW#1, Problem 6). Let M be an abelian group. Show that there

is a natural bijection between Z[G]-module structures on M and actions of

G on M by group automorphisms (that is, actions of G on M such that

for any g ∈ G the map m 7→ gm is an automorphism of the abelian group

(M, +)).

I hope this answer will help you in future too .For giving more answers against this topic

please mark this answer brain listed and give a heart and also follow me for getting more interesting answers ☺

Similar questions