# Is the category of modules over a commutative ring the category of abelian objects in a topos?

The categories of modules over commutative rings are especially notable Abelian categories. Wanting to extend this class a bit, I thought of this question:

Let $$R$$ be a commutative ring with $$1$$. Does there exist a Grothendieck topos $$E$$ such that $$\mathrm{Ab}(E) \simeq R\text{-}\mathrm{Mod}$$ (equivalence of categories)?

See the related question I was able to find: Is every Grothendieck category with a generator a category of sheaves?

• If $E$ is a topos which has a point $x$, then the topos $Set$ is a retract (in the category of topoi) of $E$. It follows that $Ab = Ab(Set)$ is a retract of $Ab(E)$. So for instance $Ab(E)$ can't be $Vect_k$ for $k$ a field. I suspect that the hypothesis of $E$ having a point can be lifted by arguments along the lines here (or maybe the result there can be used directly by passing to hearts of $t$-structures). In short, I'm pretty sure the answer is no in general. Feb 3 at 19:06
• Here are some examples of commutative rings whose module categories are the abelian group objects in a topos. 1. If $M$ is a commutative monoid, then the category of $\mathbb ZM$-modules is equivalent to the category of abelian group objects in the topos of $M$-sets. 2. If $X$ is a compact totally disconnected space and $R$ is the ring of locally constant functions from $X$ to $\mathbb Z$, then the category of $R$-modules is equivalent to the category of sheaves of abelian groups on $X$, which is of course the abelian group objects in the topos of sheaves of sets on $X$. Feb 4 at 16:20

No — in particular, not if $$R$$ has any non-integer rationals.

Briefly: From $$\newcommand{\Z}{\mathbb{Z}}\newcommand{\Ab}{\mathrm{Ab}}\newcommand{\RMod}{{R\text{-}\mathrm{Mod}}}\RMod$$, we can recover $$R$$ as the ring of endomorphisms of the identity functor, $$\newcommand{\End}{\mathrm{End}}\newcommand{\id}{\mathrm{id}}R \cong \End(\id_\RMod)$$. But for a topos $$\newcommand{\E}{\mathcal{E}}\E$$, $$\End(\id_{\Ab(\E)})$$ retracts onto $$\Gamma(\Z_\E)$$ (global sections of the integers of $$\E$$); and this will never have any non-trivial rationals.

Unwinding this more elementarily: Suppose $$n \in \Z$$ is invertible in $$R$$, with $$n \neq \pm 1$$. Then multiplication by $$\frac{1}{n}$$ gives a natural endomorphism $$\mu_{\frac{1}{n}}$$ of the identity functor on $$\RMod$$, satisfying $$n \cdot \mu_{\frac{1}{n}} = 1$$ (where $$1$$ is the identity endomorphism of the identity functor).

But the $$\mathbf{Ab}$$-enrichment of any Abelian category is determined by the category structure (using the biproducts). So an equivalence $$\RMod \cong \Ab(\E)$$ would transfer $$\mu_{\frac{1}{n}}$$ to an endomorphism $$\mu'$$ of $$\id_{\Ab(\E)}$$ with $$n \cdot \mu' = 1$$; then applying this at $$\Z_\E$$, and taking global sections, would give an inverse for $$n$$ in $$\Gamma(\Z_\E)$$. But this can never exist unless $$\E$$ is the trivial topos, since as $$n$$ is non-unital you can prove constructively “$$n$$ is not invertible in $$\Z$$”, so this holds in the internal language of $$\E$$. So then $$\Ab(\E)$$ is also trivial, and the equivalence $$\RMod \cong \Ab(\E)$$ implies $$R \cong 1$$.

So we’ve shown: if an integer $$n \neq \pm 1$$ is invertible in $$R$$, and $$\RMod \cong \Ab(\E)$$, then $$R$$ is the zero ring.

• I think it is not true that the endomorphism of the identity gives the global section of the integer: consider the case of the topos of M-set for M a commutative monoid Feb 4 at 17:08
• Great! Could you expand on why $n$ can't be invertible in $Ab(\mathcal E)$ unless $\mathcal E$ is trivial? Feb 4 at 17:09
• @TimCampion : you can proove constructively that integers are not invertible. Feb 4 at 17:13
• @SimonHenry: Oops, good catch — yes, $\Gamma(\mathbb{Z})(\mathcal{E})$ isn’t generally isomorphic to $\mathrm{End}(\mathrm{id}_{\mathrm{Ab}(\mathcal{E})})$ as I’d claimed. But it’s still a retract, which is enough for the argument to go through. Feb 4 at 17:27
• @JensHemelaer: For the map $\newcommand{\End}{\mathrm{End}}\newcommand{\id}{\mathrm{id}}\newcommand{\Ab}{\mathrm{Ab}}\newcommand{\E}{\mathcal{E}}\newcommand{\Z}{\mathbb{Z}}\End(\id_{\Ab(\E)}) \to \Gamma(\Z_\E)$: given an endomorphism of $\id_{\Ab(\E)}$, apply it at $\Z_\E$ and take global sections to get an ab-gp endomorphism of $\Gamma(\Z_\E)$, then apply it to $1$. Conversely, any $A$ in $\Ab(\E)$ has a natural $\Z_\E$-action $\Z_\E \times A \to A$; so a global section $s : 1 \to \Z_\E$ gives a natural map $s_A : A \to A$. Checking these give a retraction is a follow-your-nose computation. Feb 7 at 10:17