Question about pure submodules over a polynomial ring

I have had trouble coming up with ideas to solving the following problem on pure modules. We have worked a couple of exercises from Jacobson and Hoffman and Kunze on pure modules but this is a sample test question and I do not see how to apply previous exercises to get the result.

A submodule $N$ of $M$ is pure if whenever $y \in N$ and $a \in D$ are such taht there exists $x\in M$ with $ax = y$, then there exists $z \in N$ with $az =y$

Question: Let $M$ be a finitely generated module over the polynomial ring $F[x]$, where $F$ is a field, and let $N$ be a pure submodule of $M$. Prove that there exists a submodule $L$ of $M$ such taht $N+L = M$ and $N \cap L = 0$

I was thinking that since all the hypothesis of http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain apply to $M$ we can decompose $M$ into a pure and non pure part but I do not know how to proceed from that point.

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In general: If N ≤ M is pure and M/N is finitely presented, then N is a direct summand of M. In particularly over any noetherian ring, the pure submodules of a finitely generated module are exactly its direct summands. This is Lam's LMR Cor 4.91, p. 157. – Jack Schmidt Aug 10 '11 at 13:56
See also Appendix to §7 in Matsumura's Commutative Ring Theory. – Andrea Aug 31 '11 at 22:06

We prove that if $A$ is a PID, $M$ is a finite $A$-module and $N \subseteq M$ a pure submodule, then $N$ is a direct summand of $M$.
If $M$ is free, the proof is simpler. It is a well-known fact (Smith normal form) that there exists a basis $\{ e_1, \dots, e_n \}$ of $M$ and $a_1, \dots, a_s \in A$ such that $\{ a_1 e_1, \dots, a_s e_s \}$ is a basis of $N$. Since $N$ is pure and $M$ is torsion-free, from $a_i e_i \in N$ we have that $e_i \in N$; so $N = \langle e_1, \dots, e_s \rangle$. Picking $L = \langle e_{s+1}, \dots, e_n \rangle$, we have $M = N \oplus L$.
Now, we don't suppose that $M$ is free. We must prove that the sequence $$0 \longrightarrow N \longrightarrow M \overset{\pi}{\longrightarrow} M / N \longrightarrow 0$$ splits. Using the structure theorem, $M / N$ is the direct sum of cyclic submodules: $$M/N = A \bar{x_1} \oplus \cdots \oplus A \bar{x_r},$$ where $\bar{x_i} = \pi(x_i)$ for some $x_i \in M$. $A \bar{x_i} \simeq A / (a_i)$ as $A$-modules for some $a_i \in A$. It is clear that $a_i \bar{x_i} = 0$, so $a_i x_i \in N$ and then, for the purity of $N$, there exists $z_i \in N$ such that $a_i x_i = a_i z_i$. Defining $s \colon M/N \to M$ as $s(\bar{x_i}) = x_i - z_i$ (It is well-defined because $a_i(x_i - z_i) = 0$), we have a section of $\pi$. This prove the exact sequence splits.
Please, what if $A$ is only a UFD? I guess there is a counterexample to your claim? – user237522 Aug 25 '15 at 13:13