I am learning about abelian categories for a talk I have to give next week. One of the first questions I had upon learning this definition is "does there exist an additive category that is not abelian?"

The question, Additive category that is not abelian, gives many great answers to the question but I am curious about the details, in particular about the example of finitely generated modules over a non-noetherian ring.

The definition I am using for abelian categories is as followed:

A category $\mathcal{C}$ is abelian if

1) $\mathcal{C}$ is an additive category

2) Every morphism in $\mathcal{C}$ has a kernel and cokernel

3) Every monomorphism is the kernel of a map, and every epimorphism is a cokernel of a map.

Thus my question is:

Given the above definition of abelian categories, why is the category of finitely-generated modules over a non-noetherian ring not abelian?

My guess would be that this could fail because some kernels/cokernels might not be finitely generated, but I am blanking on how to construct an example.

  • 7
    $\begingroup$ Kernels are problematic: Let $R$ be a non-Noetherian ring. Let $I$ be ideal, which is not finitely generated (this exists because $R$ is not Noetherian). $I$ now is the kernel of $R\to R/I$. Also both $R$ and $R/I$ are finitely generated as $R$-modules (by 1 actually). Hence the kernel, in the category of finitely generated $R$-modules, does not always exist. $\endgroup$
    – Hamed
    Jul 12, 2016 at 17:58
  • $\begingroup$ This helps, thank you very much! I am going to write down the extra details I needed to check for myself in case someone else has the same question. This works because a) one of the equivalent definitions of a noetherian ring is that every ideal is finitely generated (therefore in $R$ there exists an $I$ not finitely generated) and b)modules over a ring are finitely generated iff they are a quotient of $R^n$, which $R/I$ clearly is. $\endgroup$
    – CEH
    Jul 12, 2016 at 18:21
  • 4
    $\begingroup$ @Hamed One has to consider the possibility that kernels of f.g modules could a priori differ from those computed in the category of all modules. $\endgroup$
    – Hanno
    Jul 12, 2016 at 18:24
  • $\begingroup$ @Hanno you're absolutely right. I wanted to mention this myself but it was already along comment (Thankfully now you provided the answer). $\endgroup$
    – Hamed
    Jul 12, 2016 at 20:53

1 Answer 1


Claim: The inclusion $R\text{-mod}\hookrightarrow R\text{-Mod}$ preserves kernels.

Once this is known, it follows that a kernel in $R\text{-mod}$, if it exists, must be isomorphic to the corresponding kernel in $R\text{-Mod}$; in particular, the latter is finitely generated.

Specializing this observation to projections $\pi_I: R\to R/I$ for an ideal $I$ in $R$, it follows that $\ker(\pi_I)$ exists in $R\text{-mod}$ if and only if $I$ is finitely generated. Letting $I$ vary, we see that $R\text{-mod}$ is abelian only if all ideals of $R$ are finitely generated.

Proof of claim: Suppose $f: M\to N$ is a morphism of finitely generated $R$-modules and $k: K\to M$ is a kernel of $f$ in $R\text{-mod}$. Further, let $g: T\to M$ be another $R$-module homomorphism with $fg=0$ and $T$ arbitrary. Then, by assumption, for any finitely generated submodule $\iota: S\subseteq T$ the composite $g\iota$ factors uniquely through $k$ via some $t_S: S\to K$. Since any module is the union of its finitely generated submodules, it follows that a factorization of $g$ through $k$ is unique, if it exists. In turn, applying this uniqueness it moreover follows that for any other $\iota^{\prime}: S^{\prime}\subseteq T$, the factorizations $S\to K$ and $S^{\prime}\to K$ of $\iota$ resp. $\iota^{\prime}$ agree on $S\cap S^{\prime}$. Therefore, all $t_S$ glue to a factorization $t: T\to K$ of $g$ through $k$, proving that $k$ is a kernel in $R\text{-Mod}$.

Addendum (independent of the rest): If you like it more technically, you can package the same argument as follows: Consider any category ${\mathscr C}$ (generalizing $R\text{-Mod}$), any diagram $D: I\to {\mathscr C}$ over some index category $I$ (generalizing $\bullet\rightrightarrows\bullet$), and any cone $c: D\to X$ over it, that is, you have $X\in{\mathscr C}$ and for any $i\in I$ you have a morphism $c_i: X\to D(i)$ such that $$X\xrightarrow{c_i} D(i)\xrightarrow{D(\alpha)} D(j) = X\xrightarrow{c_j} D(j)$$ for any arrow $\alpha$ in $I$. In other words, $X\to D$ is a candidate for a limit-cone for $I$, and you might ask:

Question: Which objects of ${\mathscr C}$ indeed 'see' $X$ as the limit of $D$?

Formally, this means that for some $Y\in{\mathscr C}$ you can check whether the natural morphism in $\textsf{Set}$, $${\mathscr C}(Y,X)\to {\lim}_I{\mathscr C}(Y,D(i))$$ is an isomorphism. Call the respective subcategory ${\mathscr C}_D$ for lack of a better name. Now you have two facts:

  1. Since inverse limits commute, ${\mathscr C}_D$ is closed under colimits in ${\mathscr C}$.
  2. If $X$ and the codomain of $I$ are contained in some full subcategory ${\mathscr D}$ in which $X\to D$ is indeed an inverse limit, then ${\mathscr D}\subseteq{\mathscr C}_D$.

Combining both, it follows that ${\mathscr C}_D={\mathscr C}$ if there's a full subcategory ${\mathscr D}\subset{\mathscr C}$ containing $X$ and the codomain of $D$, for which $X\to D$ is a limit, and such that any object of ${\mathscr C}$ is a colimit of a diagram in ${\mathscr D}$.

This applies to $R\text{-mod}\subset R\text{-Mod}$ and shows that the latter embedding preserves all limits, in particular kernels.

  • $\begingroup$ What is the difference between $R-mod$ and $R-Mod$? $\endgroup$
    – gen
    May 29, 2019 at 14:06
  • $\begingroup$ @gen $R\text{-mod}$ is the category of finitely generated $R$-modules, while $R\text{-Mod}$ is the category of all $R$-modules. The latter is always abelian, while the latter is if and only if $R$ is Noetherian. $\endgroup$
    – Hanno
    May 29, 2019 at 19:03
  • $\begingroup$ aha! Thanks a lot for clarifying, I wasn't familiar with this notation $\endgroup$
    – gen
    May 29, 2019 at 19:16
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    $\begingroup$ also I think your second 'latter' was meant to be 'former' $\endgroup$
    – gen
    May 29, 2019 at 19:19

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