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?](https://mathoverflow.net/questions/333955/is-every-grothendieck-category-with-a-generator-a-category-of-sheaves)



Is the category of modules over a commutative ring the category of abelian objects in a topos?
**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](https://ncatlab.org/nlab/show/biproduct#BiproductsImplyEnrichment)). 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.**


