Let $R$ be a ring (unital, not necessarily commutative), $M$ a finitely presented left $R$ module. Suppose $m_1,\ldots,m_n\in M$ generate $M$. This determines a surjection $f:R^n\to M$.
Must the kernel of $f$ be finitely generated?
This is true by definition if the generating set $m_1,\ldots,m_n$ is the one coming from the finite presentation of $M$.
We'll prove it under bit more general setting. Let $0\to K\to R^m\to M\to 0$ be an exact sequence where $K$ is finitely generated. Consider now $0\to L\to R^n\to M\to 0$; we wish to prove that $L$ is finitely generated.
By Schanuel's lemma, $L\oplus R^m\cong K\oplus R^n$. Then $L$ is an epimorphic image of $K\oplus R^m$, which is finitely generated.
By the way, this result is a corollary of following more general fact;
If $0\to L\to N\to M\to 0$ is an exact sequence with $M$ a finitely presented and $N$ finitely generated, then $L$ is finitely generated.
Indeed, consider an epimorphism $R^n\to N$ and the commutative diagram $$ \require{AMScd} \begin{CD} 0 @>>> L' @>>> R^n @>>> M @>>> 0\\ @. @VVV @VVV @| \\ 0 @>>> L @>>> N @>>> M @>>>0 \end{CD} $$ where the arrow $R^n\to M$ is the composition of $N\to M$ after $R^n\to N$.
Then $L'$ is finitely generated by the above argument and the map $L'\to L$ is an epimorphism by diagram chasing (or four lemma).
This can also be proved by using Snake lemma instead of Schanuel's lemma (see here).
$L'$, not $L^{'}$: look at the difference between $L'$ and $L^{'}$.
– egreg
Nov 24 '17 at 11:56