Page 158 - Advanced Linear Algebra
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142    Advanced Linear Algebra





                                ~    #~   " ~    6   7  "


            Since  ²" ³  and  °         are relatively prime, the order of ² °        ³"     is equal to

             ²" ³ ~                                         ²#³ ~   , which contradicts the equation above. Hence,   .
            It is clear that  ºº" b Ä b " »» ‹ ºº" »» l Ä l ºº" »» . For the  reverse




            inclusion, since         and  °            are relatively prime, there exist  Á          9   for which

                                         b        ~

            Hence

              "~ 6          b    7    "~    "~    ²" b Ä b " ³  ºº" b Ä b " »»








            Similarly, " ºº" b Ä b " »»  for all   and so we get the reverse inclusion.



            Finally, to see that the sum above is direct, note that if
                                     #b Ä b # ~


                       , then each   must be  , for otherwise the order of the sum on the
            where # (           #
            left would be different from  .

                    )                ~°                                   9
            For part 2 , the scalars         are relatively prime and so there exist
            for which
                                       b       Ä  b     ~
            Hence,

                        # ~ ²        bÄb      ³# ~         #bÄb      #


                      #³ ~ °gcd ² Á                  and      are  relatively  prime,
            Since   ²              ³ ~          and since
            we have  ²        #³ ~      . The second statement follows from part 1 . ) …
            Free Modules over a Principal Ideal Domain
            We have seen that a submodule of a free module  need  not  be  free:  The
            submodule  d ¸ ¹  of the module  d  {   over itself is not free. However, if 9
                                         {
                     {
            is a principal ideal domain this cannot happen.
            Theorem 6.5 Let  4   be a free module over a principal ideal domain  . Then
                                                                      9
                                            rk
            any submodule   of 4  is also free and  ²:³  rk ²4  . ³
                        :
            Proof. We will give the proof first for  modules  of  finite  rank  and  then
            generalize to modules of arbitrary rank. Since  4š 9     where    ~ rk ²4³   is
            finite, we may in fact assume that 4 ~9   . For each       , let
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