1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Module and submodule

  1. Apr 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose M is a D_module and N is a submodule. N is called pure iff for any y [tex]\in[/tex] N and a [tex]\in[/tex] D ax = y is solvable in N iff it is solvable in M. N is a direct summand of M iff there is a submodule K with [tex]M = N \oplus K[/tex]. Prove:
    (1) If N is a direct summand, then N is pure.
    (2) Suppose D is P.I.D and M is a finitely generated torsion module. IF N is pure, then N is a direct summand of M.


    2. Relevant equations

    I am not sure what it means for ax=y is solvable in M iff it is solvable in N

    3. The attempt at a solution
    (1) If M is a direct summand, then there is a submodule K with [tex]M = N \oplus K[/tex]. Let's suppose that ax=y is solvable in M for y [tex]\in[/tex] N and a [tex]\in[/tex] D, then there is a z [tex]\in[/tex] N such that az=y. To prove that N is pure, one needs to prove that z [tex]\in[/tex] N. I do not know if this is what I am supposed to do and if so, I have no idea how to do it.
    (2)Now D is a P.I.D and M is a finitely generated torsion module. Assume that N is pure. Let y [tex]\in[/tex] N and a[tex]\in[/tex] D, then we have z [tex]\in[/tex] N such that az=y implies z [tex]\in[/tex] M. I do not know how to show that there is a submodule K of M such that [tex]M = N \oplus K[/tex].
     
    Last edited: Apr 29, 2008
  2. jcsd
  3. Apr 30, 2008 #2
    when they say solvable in M(or N) they mean "has a solution in".
    So you are to show that for any y in N and a in D, there is x in N s.t. ax = y if and only if there is z in M s.t. az = y.

    So suppose N is a direct summand of M, so M = N + K (here "+" = direct sum) for some submodule K of M.

    One direction is clear, if ax = y is solvable in N then it's solvable in M.

    So suppose the equation ax = y is solvable in M, so there is a z in M s.t. az = y.
    Now use the fact z is in M. (it's pretty straightforward, I had typed up the proof but I really think you can do this!)

    Post again if you are stuck, goodluck!
     
    Last edited: Apr 30, 2008
  4. Apr 30, 2008 #3
    I can see now how to do the first one, the other direction is proved using the fact that the intersection of N and K is trivial. Thanks!
    For part (2): how do I show the existence of K? How does the assumption that D is a P.I.D change the problem?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Module and submodule
  1. Basis for a submodule? (Replies: 1)

  2. Module and submodule (Replies: 1)

Loading...