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Let $R$ be a commutative ring with identity and let $M$ be a module over $R$. We say that $M$ is finitely presented if there exists an exact sequence $$R^{m}\longrightarrow R^{n}\longrightarrow M\longrightarrow0$$ for some nonnegative integers $m$ and $n$.

In the book I am reading, it states that $M$ is finitely presented if and only if $M$ is finitely generated and for every surjective $R$-linear map $\psi:N\rightarrow M$, where $N$ is finitely generated, the kernel $\text{ker}\psi$ is finitely generated.

I know that the $M$ being finitely presented implies the other statement (using the snake lemma), but I don't know how to prove the converse. Any hints would be appreciated.

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  • $\begingroup$ Loosely speaking, your first definition says, that a module is finitely presented if there is one choice of finitely many generators, that has only finitely many relations between them. Your second definition says that a module is finitely presented if any choice of finitely many generators has only finitely many relations beetween them. So you are actually asking for the trivial direction of the equivalence. $\endgroup$ – MooS Nov 13 '17 at 12:57
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Suppose the latter condition holds. Since $M$ is finitely generated, there is an $R$-linear surjection $R^n\to M\to 0$ for some $n$. Now, the kernel is finitely generated so you can get a similar exact sequence for $K=\ker(R^n\to M)$ and, "concatenating" these two sequences, you can produce an exact sequence of the desired form $R^m\to R^n\to M\to0$.

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