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Module and submodule

  • Thread starter mobe
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  • #1
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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 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].
 
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Answers and Replies

  • #2
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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!
 
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  • #3
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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?
 

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