6
$\begingroup$

Let $R$ be a nontrivial ring. A right $R$-module $M$ is called coherent if ${\rm Ker} (f)$ is finitely generated for any $R$-module homomorphism $f: L\to M$ with $L$ finitely generated. It is well-known that the full subcategory ${\rm Coh}(R)$ of ${\rm Mod}(R)$ consisting of coherent modules is an abelian category. I wonder that is it possible that ${\rm Coh}(R)$ is the trivial abelian category? In another word, does there exist a ring $R$ over which the zero module is the only coherent right module?

$\endgroup$
1
  • 1
    $\begingroup$ Usually, such modules are called pseudocoherent, while coherent modules are pseudocoherent modules of finite type. $\endgroup$ Mar 9, 2014 at 18:39

1 Answer 1

6
$\begingroup$

Yes, for example the ring $R = k[x_1, x_2, x_3, \ldots]/(x_ix_j)$ where $k$ is a field. Namely, every finite submodule of a coherent module is coherent and so every nonzero coherent module contains a nonzero coherent module of the form $R/I$. Then $I$ has to be finitely generated hence $x_i \not \in I$ for some $i$. Then $R/I$ contains a copy of $k$ (namely, the submodule generated by $x_i \bmod I$) which is not a coherent module over $R$. Hence $R/I$ is not coherent.

$\endgroup$

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.