This question comes from the following proof of the theorem.
Definition: Let $R$ be a P.I.D and $M$ be a left $R$-module.$N$ is a submodule of $M$.
N is a pure submodule if whenever $y\in N$ and $a\in R$ are such that there exists $x\in M$ with $ax=y$, then there exists $z\in N$ with $az=y$.
Theorem:
If $N$ is a pure submodule and $M/N$ is of finite presentation, then $N$ is a direct summand of $M$.
Proof: Consider the sequence:
$$
0 \longrightarrow N \longrightarrow M \overset{\pi}{\longrightarrow} M / N \longrightarrow 0.
$$
Using the P.I.D. structure theorem, $M / N$ is the direct sum of cyclic submodules:
$$M/N = R \bar{x}_1 \oplus \cdots \oplus R \bar{x}_n,$$
where $\bar{x}_i = \pi(x_i)$ for some $x_i \in M$ and $R\bar{x}_i \cong R / (p_i)$ as $R$-modules for some $p_i \in R$.
Because $p_i \bar{x}_i = 0$, so $p_i x_i \in N$.
So there exists $z_i \in N$ such that $p_i x_i = p_i z_i$.
Define:$$f \colon M/N \to M$$
$$\bar{x}_i \mapsto x_i – z_i$$
Thus $\pi f=id_{M/N}$.
This proves the exact sequence splits.
My question is how to proof that $f$ is well-definied?
That is if $\bar{y}_i=\bar{x}_i$, then I can get $y_i-x_i\in N$, $x_i-z_i$ and $y_i-q_i$ where $q_i$ satisfies $p_i y_i=p_i q_i$.
How to proof:$x_i-z_i=y_i-q_i$?
Maybe my thoughts is wrong?
Thanks!
Best Answer
Maybe I have solve the problem.
If there is an error, please point it out!
If $\overline{y}_i$=$\overline{x}_i$, we can get $y_i=x_i+n_i$.
Let $k_i=y_i-(x_i-z_i)=n_i+z_i\in N$,then $p_iy_i=p_i(y_i-(x_i-z_i))$,because $p_i(x_i-z_i)=0$.
So $f(y_i)=y_i-(y_i-(x_i-z_i))=x_i-z_i$.
Thus $f$ is well-defined.