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,……,N_K  are  submodules of  M, the submodule generated by N₁,N₂,…..N_K is equal to

/x_i Є N_i } = N₁ +N₂ +……..N_K .

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)       =α  (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.