11
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Dec 18 '18 at 12:05
5
$\begingroup$

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

$\endgroup$
0
7
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$ Apr 29 '10 at 17:37
  • 2
    $\begingroup$ kernels of derivations between polynomial rings sometimes have infinitely generated kernels: See ams.org/mathscinet-getitem?mr=1971045 $\endgroup$
    – J.C. Ottem
    Apr 29 '10 at 18:09
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 29 '10 at 20:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.