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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?

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    $\begingroup$ 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. $\endgroup$
    – Tim Campion
    Commented Feb 3, 2023 at 19:06
  • $\begingroup$ 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 $X$. $\endgroup$
    – Benjamin Steinberg
    Commented Feb 4, 2023 at 16:20

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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.

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    $\begingroup$ 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 $\endgroup$
    – Simon Henry
    Commented Feb 4, 2023 at 17:08
  • $\begingroup$ Great! Could you expand on why $n$ can't be invertible in $Ab(\mathcal E)$ unless $\mathcal E$ is trivial? $\endgroup$
    – Tim Campion
    Commented Feb 4, 2023 at 17:09
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    $\begingroup$ @TimCampion : you can proove constructively that integers are not invertible. $\endgroup$
    – Simon Henry
    Commented Feb 4, 2023 at 17:13
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    $\begingroup$ @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. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Feb 4, 2023 at 17:27
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    $\begingroup$ @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. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Feb 7, 2023 at 10:17

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