# 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?

• Just a remark: Additionally to the category of finitely generated and to the category of finitely presented modules, we have the category of coherent modules. This category is always abelian. If $R$ is coherent over itself, then an $R$-module is coherent iff it is finitely presented. This happens, for instance, if $R$ is Noetherian. The basic properties of coherent modules are nicely led out in Ravi Vakil's lecture notes on algebraic geometry, as a fun series of exercises. Dec 18 '18 at 12:05

[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 $R^n\twoheadrightarrow I\hookrightarrow R$.

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.

• Ahh, now we are getting there. Okay, is there an example of a map of finitely presented modules for which the kernel is not even finitely generated? Apr 29 '10 at 17:37
• kernels of derivations between polynomial rings sometimes have infinitely generated kernels: See ams.org/mathscinet-getitem?mr=1971045 Apr 29 '10 at 18:09
• I believe Manny Reyes has now given a nice example for this; his finitely presented modules are as nice as they come too! His map is not surjective of course, but again it is focussing on the key feature of not-being-coherent: submodules of f.p. ned not be f.p. Apr 29 '10 at 20:45