Page 161 - Advanced Linear Algebra
P. 161

Modules Over a Principal Ideal Domain  145





                                rk²4³ ~  rk²ker ² ³³ b  rk²4³
            and so  rk²ker ² ³³ ~  , that is,  ker ² ³ ~¸ ¹ , which implies that   is an  -



                                                                          9
            isomorphism and so   is a basis.…
                            :
            In  general, a basis for a submodule of a free module over a principal ideal
            domain cannot be extended to a basis for the entire module. For example, the set
            ¸ ¹ is a basis for the submodule     of the  -module  , but this set cannot be
                                                {
                                        {
                                                         {
            extended  to a basis for   itself. We  state without proof the following result
                                {
            along these lines.
            Theorem 6.7 Let  4   be a free  -module of rank  , where   is a principal ideal
                                                           9

                                     9
            domain. Let   be a submodule of  4   that is free of rank          . Then there is a
                      5
            basis  8   for  4    that contains a subset  :  ~  ¸  #     Á  Ã  Á  #     ¹   for which
            ¸  # Á ÃÁ  # ¹ is a basis for  , for some nonzero elements    ÁÃÁ     of  9.…
                                    5



            Torsion-Free and Free Modules
            Let us explore the relationship between the concepts of torsion-free and free. It
            is not hard to see that any free module over an integral domain is torsion-free.
            The converse does not hold, unless we strengthen the hypotheses by requiring
            that the module be finitely generated.
            Theorem 6.8 A finitely generated module over a principal ideal domain is free
            if and only if it is torsion-free.
            Proof. We leave proof that a free module over an integral domain is torsion-free
            to the reader. Let . ~ ¸# ÁÃÁ# ¹  be a generating set for 4 . Consider first the


                                          .
            case  ~  , whence . ~¸#¹ . Then   is a basis for 4  since singleton sets are
            linearly independent in a torsion-free module. Hence, 4  is free.
            Now suppose that . ~ ¸"Á#¹  is a generating set with "Á# £   . If   is linearly
                                                                  .
            independent, we are done. If not, then there exist nonzero  Á    9  for which
             " ~ #. It follows that   4 ~ ºº"Á #»» ‹ºº"»» and so   4 is a submodule of a
            free  module  and  is  therefore  free by Theorem 6.5. But the map  ¢4 ¦  4
            defined by  #~ #  is an isomorphism because 4  is torsion-free. Thus 4  is
            also free.
            Now we can do the general case. Write



                                . ~ ¸" ÁÃÁ" Á# ÁÃÁ#      c     ¹
                                                                        (
                                                                     .
            where : ~ ¸" ÁÃÁ" ¹  is a maximal linearly independent subset of  .  Note


            that   is nonempty because singleton sets are linearly independent.)
               :
            For each  , the set  ¸    #  "  Á  Ã  Á  "     Á  #       ¹   is linearly dependent and so there exist
               9 and    ÁÃÁ   9 for which
   156   157   158   159   160   161   162   163   164   165   166