# Module and submodule

1. Apr 29, 2008

### mobe

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 $$\in$$ N and a $$\in$$ 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 $$M = N \oplus K$$. 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 $$M = N \oplus K$$. Let's suppose that ax=y is solvable in M for y $$\in$$ N and a $$\in$$ D, then there is a z $$\in$$ N such that az=y. To prove that N is pure, one needs to prove that z $$\in$$ 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 $$\in$$ N and a$$\in$$ D, then we have z $$\in$$ N such that az=y implies z $$\in$$ M. I do not know how to show that there is a submodule K of M such that $$M = N \oplus K$$.

Last edited: Apr 29, 2008
2. Apr 30, 2008

### ircdan

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
3. Apr 30, 2008

### mobe

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?