### Commutative Algebra chapter II

Chapter –II

MODULES

In this chapter a ring R will always mean a commutative ring with element 1.

Definition :2.1
An R module M is an abelian  group M together with a map
R given  by (a,x) à a.x , satisfying the following conditions.
i.          a .(x+y) = a. x  + a.y    a Є R , x , yЄ M.
ii.           (a+b) .x = a.x + b.x       a,b Є R , x Є M.
iii.          a.(b.x) =(ab) . x              a.b Є  R, x Є M.
iv.          1. x  = x,     x Є M
We denote  a .x by ax.
Examples :2.2
(i)            Any  vector space V over  a field K is a K –module.
(ii)          Any abelian group  G is a  Z –module.
Definition:2.3
A Subset N
M is called a submodule, if N is a subgroup of the abelian group M and a x Є N  for all a Є R and x Є N.
Examples:2.4
(i)  Any subspace W of a vector space V is a submodule.
(ii)          All polynomials of degree atmost n is a submodule of the
R-module R[x].
(iii)        The modules O and M are submodule of M called improper submodules.
Definition :2.5
Let M and N be R-module . A  map : Mà N is called a homomorphism of  R – module  if
(x+y) = (x) + (y)     , x, y Є M.
(ii) (ax) = a  (x) ,  a Є R , x Є M.
Examples :2.6
(I)           For any fixed a Є R ,the  map : MàM given by (x) = ax, is a homomorphism.
(II)         For any submodule  N of M ,the inclusion map  :N àM defined  by  (x)=x , x Є N is a homomorphism.

Definition :2.7
Let N be a submoldule of M. Consider  the quotient abelian group M/N with the scalar multiplication given  by
a.(x  + N) =ax +N , a Є R , x Є M.
Then M/N is an R module called the quotient module M/N.
Definition :2.8
A homomorphism  of modules which is both (1,1) (injective) and onto (surjective) is called an isomorphism.
Definition :2.9
The map p : M à M/N defined by p(x) = x+ N ,x Є M is a homomorphism of modules called the projection.
Result :2.10
Let N be a submodule of M . Then M/N is an R-module.
Proof
Since  M is an abelion group and N is a submodule  of M
M/N is a group
To prove M/N is an abelian group

Let x+N, y+N   M/N x,y,
(x+N)+(y+N)=x+y+N
=y+x+N
=(y+N)+(x+N)
M/N is .an abelian group
Define µ  :A  M/N M/N by  µ (a,x+N) =ax+N =a(x+N)
Let  a,b R,x+N, y+N  M/N.
a((x+N)+(y+N)) =a (x+y+N)
=a(x+y)+N.
=ax+ay+N.
=(ax+N)+(ay+N)
a((x+N)+(y+N) =a(x+N)+a(y+N).
(a+b).(x+N) = (a+b) x +N.
=ax + bx+N.
=ax+N +bx+N.
(a+b.(x+N) =a(x+N) + b(x+N)
(ab) (x+N) = ab.x+N.
=a(bx+N)
(ab) (x+N) = a(b(x+N))
1.(x+N) = 1.x+N =x+N
M/N is an R-module
Result:2.11
The natural map of M onto M/N is an R-module homomophism.

Proof:
Define : M   M/N by  (x) = x+N    x M
To prove  is a R-module homomophism.
Let  x, y M
=x +y+N
= x+N+y+N
(x)+
(ax)=ax+N
=a(x+N)
=a (x).
is an R-module homomophism.
To prove   is onto.
For every x+N  M/N,  x M such  that  (x) =x+N.
: M→ M/N is an onto R-module homomophism.
Theorem:2.12
Let : M → N be a homomophism of M onto N. Then the kernel of  ={x  M / (x) =0}is a submodule K of M and the quotient module M/K isomorphic to N.
Proof:
Kernel of  = K= {x  M / (x) = 0}.
Let  x,y  ker .
Then (x)  =0, (y) =0.
To prove x+y    ker
x+y) = (x) +  (y)
=0+0
(x+y) =0
x+y  ker
To prove –x  ker
(-x) =  (-1.x)
= -1. (x)
= -1.0
(-x) = 0
-x  ker .
Hence  ker  is a subgroup of M.
To prove ax  ker
(ax) =  a f(x)
=a.0
(ax) = 0
ax ker  .
Kernel of  is a submodule of M .
Define the map  M/K→N by (x+K) = (x), x  M.
Claim :  is well – defined.
Let x+K , y+K  M/K.
x+K = y+K
x –y + K = K
x – y  K  =  ker
(x-y)  = 0
(x) – (y) = 0
(x) =  (y)
(x+K) =  (y+K)
is well – defined and 1 – 1.
Claim: is  a homomorphism.
((x+K) + (y+K))  =  (x+y+K)
=  (x+y)
=  (x) + f (y)
=  (x+K) + (y+K).
(a(x+K)) = (ax+K)
=   (ax)
= a   (x+K).
is a homomophism.
Hence    is an is homomorphism.

M/K    N
Theorom:2.13
i.          I   L  M N are R-modules , then
ii.          I   M1,M2  are submodules of M, then M1+M2/M1  M2/M1  M2.

Proof:
(i)Given L M N are R-modules
Define  : L/N à L/M by (x+N) = x+M    x+N L/N.

To prove  is well – defined
Let  x+N, y+N  L/N such that  x+N =y+N
(x-y) +N =N
x –y N  M
x –y  M
x –y +M =M
x + M =y+M
(x+N) =  (y+N)
is well- defined.
is onto
or every x +M  L/M ,   x+ N   L/N  such that
(x+N) = x+M.
is  onto.
To prove   is   an R-module  homomorphism.
Let x+N  y+N  Є L/N , a Є R.

((x+N) + (y+N)) =  ((x+y) +N ))
= x+ y+ M
= (x+M) +(y+M)
((x+N) +(y+ N)) =  (x+N) +  (y+N)

(a(x+ N)) =  (a x +N)
= ax +M
=a(x +M)
= a  (x+N)
is  an R-module  homomorphism.
Hence    is an R-module  homomorphism of L/N onto L/M.
Claim:  ker  = M/N.
Now  x +N Є ker
(x+N) =M
x+M =M
x Є M
x+N Є M/N.
Ker  = M/N.
By fundamental theorem of homomorphism

(iii)         : M2 à M1+M2/M1 defined by  (x) =x+M1 where x Є M2.
Let x, y Є M2 such that   x = y
x-y=0
(x-y) =  (0) = M1
x –y +M1 =M1
x+M1 = y + M1
(x) =  (y)
is well –defined.
To prove  is onto
Let  z +M Є M1 + M2/M1
Then z Є  M1+M2.
z =z1+z2 where z1 Є M1 z2 Є M2.
z2 = z- z1 and
(z2) = z – z1 +M1  = z+M1
z2 Є M2 such that  (z2) =z+M1
is onto .
To prove  is an R-modules homomorphism
Let x, yЄ M2 , a Є R
(x+y) =x+y+M1
=(x+M1 +(y+M1)
(x+y) =  (x)+  (y)
(ax) = ax +M1
=a (x+M1)
=a  (x)

Hence  is an R-module homomorphism  of M2 onto M1+M2/M1.
Claim : ker  = M1 M2.
x Є Ker      (x) = 0, x Є M2
x +M1 = M1, x Є M2.
x Є M1 , X Є M2.
x Є M1  M2.
Ker  = M1  M2
M2/M1 M2  M1+M2/M1
Note :
If N and K are submodules of M then N  K is a submodule of M but N  K is not in general a  submodule of M.
Definition:2.14
The smallest submodules of M containing N  K is called the submodule generated by N and K.

Theorem:2.15
The submodule S generated by N and K is a submodule
N+K ={x+y/x Є N, y Є K}
Proof:
Clearly N + K is a submodule of M since N and K  are  submodules of  M.
Also N N +K and K N+ K so that  S N +K.
Conversely, for any x Є N , y Є K , we have x,y Є S so that x + y Є S.
Thus  N + K S.
Hence  S = N+ K.

Corollary :2.16
If N1, N2,……,NK  are  submodules of  M, the submodule generated by N1,N2,…..NK is equal to
/xi Є Ni } = N1 +N2 +……..NK .

Definition:2.17
Let A be  a subset of M and is denoted by IA.
In particular if I= R and A ={x}, IA is denoted by Rx.
Definition:2.18
An R-module M is called cyclic if M = Rx  for some x Є M.
Theorem:2.19
An R-module M is cyclic iff M R/I  for some ideal I in R.
Proof:
Suppose M is cyclic
M =Rx for some  x Є M.
Define the natural map   : R à M by   (a) = ax .
clearly  is surjective.
To prove    is a homomorphism
Let  a, b Є R
(a+b) =(a+b) x
=ax+bx
(a+b)  =  (a) +(b)
(αa)      =(αa)x
=α(ax)
(αa)         (a) ,α Є R.
is a homomorpohism.
Hence  is a surjective homomorphism.

Let  I =ker .
By fundamental theorem of homomorphism, we have R/I  M.
Conversely suppose M  R/I.
Since R/I is generated by  = 1 +I, R/I is cyclic.
Since R/I  M, M is cyclic.

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