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