Prove: Module & Submodule Homework

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The discussion focuses on proving two key points about submodules in the context of D-modules. First, it asserts that if N is a direct summand of M, then N is pure, which involves demonstrating that solvability of the equation ax = y in M implies solvability in N. Second, it states that if D is a principal ideal domain and M is a finitely generated torsion module, then a pure submodule N must also be a direct summand of M. Participants express uncertainty about the definitions and the steps needed to prove these properties, particularly regarding the solvability conditions and the existence of a complementary submodule K. The thread highlights the challenges in understanding the implications of purity and direct summands in module theory.
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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 \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.

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 M = N \oplus K. Let's suppose that ax=y is solvable in M for y \in N and \in, then there is a \in 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 \inD, 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.
 
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