Commutative Algebra chapter II

                                         Chapter –II

            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.
                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.
       (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.
 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,
                  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+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.                 
              (ab) (x+N) = a(b(x+N))
                    1.(x+N) = 1.x+N =x+N
    M/N is an R-module
            The natural map of M onto M/N is an R-module homomophism.

            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
                 =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.
            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.
            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)
 (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)
      (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
       i.          I   L  M N are R-modules , then
     ii.          I   M1,M2  are submodules of M, then M1+M2/M1  M2/M1  M2.

           (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) = 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.
          The smallest submodules of M containing N  K is called the submodule generated by N and K.

        The submodule S generated by N and K is a submodule
N+K ={x+y/x Є N, y Є K}
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 .

         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.
         An R-module M is called cyclic if M = Rx  for some x Є M.
     An R-module M is cyclic iff M R/I  for some ideal I in R.
         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
    (a+b)  =  (a) +(b)
    (αa)      =(αa)x
 (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.