Theorem. If $N$ is a pure submodule and $M/N$ is of finite presentation, then $N$ is a direct summand of $M$.
Proof: 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 $x_i \in M$ and $A\bar{x}_i \simeq A / (a_i)$ 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, from the purity of $N$, there exists $z_i \in N$ such that $a_i x_i = a_i z_i$. Defining $s \colon M/N \to M$ by $s(\bar{x}_i) = x_i - z_i$, we have a section of $\pi$. This proves the exact sequence splits.
Definition. A submodule $N$ of $M$ is pure if whenever $y∈N$ and $a∈A$ are such that there exists $x∈M$ with $ax=y$, then there exists $z∈N$ with $az=y$.
It is also related to a theorem (Thm 7.14) at Appendix to §7 in Matsumura's Commutative Ring Theory.
My question might be very simple, but it is not very obvious to me. I just need an explanation to understand it well. Is the map $s$ well-defined? We choose $x_i$ as pre-images of $\pi(m+N)$ since $\pi$ is surjective. I thought that we may have different $x_i$'s here, then how is the map $s$ can be well-defined? Could you explain why this is well defined? Thank you in advance.