Failure of Fin. Presented and Fin. Generated Modules to be Abelian Categories?  Let R be a ring. I'm trying to understand when the categories of finitely presented R-modules and finitely generated R-modules can fail to be abelian categories. 
Poking around on the internet has lead me to believe that these categories will agree and be abelian if R is a (left?) Noetherian ring, but I'm interested in the more general case. I've found warnings that these categories can fail to be abelian but haven't found much discussion of what goes wrong. I feel like there should be some really good illuminating examples.
I think it's not too hard to show that the cokernel of a map of finitely presented modules is again finitely presented. So one of the things I'm really looking for is an example of a map of between finitely presented modules in which the kernel is not finitely presented. It would be even better if the kernel was not finitely generated. Is this possible?
 A: From Lam's Exercises in Modules and Rings, exercise 4.9, p. 102:
Let R be a ring that is not left coherent, and let I be a left ideal that is finitely generated but not finitely presented, then R→R/I is a surjection of finitely presented modules, whose kernel I is finitely generated but not finitely presented.
Again, if R is not left noetherian, and I is a left ideal that is not finitely generated, then R→R/I is a surjection of finitely generated modules whose kernel is not finitely generated.
These are both just silly rewordings of the definition, but maybe they are useful.
k[x1,x2,...] is coherent.
However k[y,x1,x2,...]/(yx1, yx2, ... ) is not coherent since its ideal (y) is finitely generated but not finitely presented.
A: [Background to the following can be found in Lam's Lectures on Modules and Rings, section 4.G.]
Definition: A finitely generated (f.g.) right module $M_R$ is coherent if every f.g. submodule of $M$ is finitely presented (f.p.).  
Now let $M_R$ be a finitely presented module that is not coherent.  There exists some submodule $N_R\subseteq M$ that is f.g. but not f.p.  Pick a surjection $R^n\twoheadrightarrow N$, and consider the composition $f\colon R^n \twoheadrightarrow N\hookrightarrow M$. Because $N\cong R^n/\ker(f)$ is not f.p., it must be the case that $\ker(f)$ is not f.g.  
So all we need is an explicit example of a f.p. module that is not coherent.  For this, see Jack Schmidt's answer to this question.  Using my approach in his case, instead of looking at a surjection $R\twoheadrightarrow R/I$, we would look at the composition $R^n\twoheadrightarrow I\hookrightarrow R$.
