Prove: Module & Submodule Homework

  • Thread starter Thread starter mobe
  • Start date Start date
  • Tags Tags
    module
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
mobe
Messages
19
Reaction score
0

Homework Statement


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.

Homework Equations



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

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 [tex]\in[/tex], then there is a [tex]\in[/tex] 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].[/tex][/tex]
 
Last edited:
Physics news on Phys.org
I could not fix my post and so I posted it again. Can someone delete one post for me? Thanks!