If $R$ is a PID and $M$ is finitely generated $R$-module, why is $M=\operatorname{coker}[R^m\to R^n]$

Question

I read a book and I don't understand one transition:

$R$ is a PID. $M$ is a finitely generated module. Thus there exists a surjective $\varphi:R^n\to M$. $R$ is PID therefore $R$ is noetherian, thus $R^n$ is noetherian thus $\ker\varphi\subseteq R^n$ is finitely generated. Thus there exists $\psi :R^m\to R^n$ such that $\operatorname{Im} \psi=\ker\varphi$.

The transition that I don't understand is why is $M=\operatorname{coker} \psi$?

I know that $\operatorname{coker}\psi=R^n/\operatorname {Im}\psi=R^n/\ker\varphi$. I don't understand how does $M=\operatorname {coker}\psi$. Thanks.

Answer 1

We have $M = \text{im}(\varphi) \cong R^n/\text{ker}(\varphi) = R^n/\text{im}(\psi) = \text{coker}(\psi)$.

Answer 2

The image of $\varphi$ is isomorphic to $R^n/\ker\varphi$. But $\varphi$ is surjective, so that image is just $M$.

Answer 3

There exists a surjective morphism $\;p:R^m\longrightarrow\ker\varphi$ for some $m$, and $\psi$ is just the composition of $p$ by the canonical injection $\;i:\ker\varphi\hookrightarrow R^n$.