
  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"

$\require{begingroup} \begingroup$
$\def\coker{\operatorname{Coker}}$
$\def\im{\operatorname{Im}}$
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) that $K'$ is itself finitely generated.
$\endgroup$
