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

by J. Doe   Last Updated September 11, 2019 19:20 PM

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.



Answers 2


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

Eric Wofsey
Eric Wofsey
September 11, 2019 19:15 PM

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

ThorWittich
ThorWittich
September 11, 2019 19:17 PM

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