
> 
> Precisely, if an R-module M *has* a finite presentation, and Rk → M is some unrelated **surjection** (k finite), is the kernel necessarily also finitely generated? 
> 
> 
> 


Basically I want to believe I can choose generators for M however I please, and still get a finite presentation. I have reasons from algebraic geometry to believe this, but it seems like a very basic result, so I would like to understand it directly in terms of the commutative algebra, which I just can't seem to figure out...


(Here R is an arbitrary commutative ring, with no other hypotheses.)


**Edit**: All maps here are maps of R-modules. Also, the reason this is not the same as "does finite presentation imply coherent?" is that I am only asking for finite type kernels of **surjections** Rk → M. That the hypotheses assume surjectivity is a common misreading of the general definition of "coherent".


If the answer to the above is "yes", then coherent will mean "finite type, and all finite type submodules are finite presentation"



Does "finitely presented" mean "always finitely presented"?  (Answered: Yes!)
$\require{begingroup} \begingroup$
$\def\coker{\operatorname{Coker}}$
$\def\im{\operatorname{Im}}$


Suppose that we have a short exact sequence $0 \to K \to R^m \to M \to 0$ with $K$ finitely generated over $R$ and that $0 \to K' \to R^n \to M \to 0$ is another short exact sequence. Your question is: is $K'$ necessarily finitely generated?


The answer is yes and we can see this as follows:


First, we argue for the existence of a commutative diagram


$$
\require{AMScd}
\begin{CD}
0 @>>> K @>>> R^m @>>> M @>>> 0 \\
@. @VV{\tilde{f}}V @VV{f}V @| \\
0 @>>> K' @>>> R^n @>>> M @>>> 0 \\
\end{CD}
$$


Using the fact that free modules are projective we can lift the identity map $M = M$ to an $f\colon R^m\to R^n$ which makes the right hand square commute. Restricting $f$ to a map $\tilde{f}\colon K → K'$ fills in the last square and so we have the diagram as claimed.


Now using Snake's lemma we find that there is an isomorphism $\coker{\tilde{f}} \cong \coker{f}$. Thus We have a short exact sequence;
$$
0\to \im{\tilde{f}}\to K'\to \coker{f}\to 0.
$$
Since $K'$ is squeezed between two finitely generated $R$ modules, it follows (by a [well-known-fact](https://math.stackexchange.com/questions/234753/finitely-generated-modules-in-exact-sequence/3490591#3490591)) that $K'$ is itself finitely generated.


$\endgroup$


