Let $R$ be a commutative ring and let $\mathrm{Mod}\_R$ be the category of (left) $R$-modules.



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> Question: Is it true that the categories $\mathcal{Z}(\mathrm{Mod}\_R)$ and $\mathrm{Mod}\_R$ are equivalent?
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I have read the previous claim in a couple of places but without any proof or reference, and I cannot really find one. In general if $\mathcal{C}$ is a braided category, then there is a fully faithful functor $\mathcal{C} \hookrightarrow \mathcal{Z}(\mathcal{C})$, but in general this is not an equivalence. Furthermore, it seems to me that the functor $\mathrm{Mod}\_R \hookrightarrow \mathcal{Z}(\mathrm{Mod}\_R)$ has little chance to be an equivalence since $\mathrm{Mod}\_R$ is symmetric monoidal and the center is (highly) braided.



Drinfeld center of $\mathrm{Mod}_R$
Let $(X,\Phi)$ be an object of the Drinfeld center.


We'd like to prove that $(X,\Phi)$ is isomorphic to $(X,$ standard symmetry isomorphism $)$, which would prove the equivalence as you say we already have fully faithfulness.


Compose the isomorphism $\Phi$ with the inverse of the standard symmetry isomorphism, to get a natural map $X\otimes\_R - \to X\otimes\_R-$.


Now, evaluating in $R$ gives an automorphism $f:X\to X$, and, by naturality and the fact that $X\otimes\_R Y$ is generated by pure tensors, it follows that in general, the morphism $X\otimes\_R Y\to X\otimes\_R Y$ is given by $f\otimes\_R \mathrm{id}\_Y$


(here I'm using something specific to $\mathrm{Mod}\_R$, I'm not sure how general it could actually be in the context of symmetric monoidal categories)


In particular, this means that $\Phi $ is actually given by $x\otimes y\mapsto y\otimes f(x)$. Moreover, note that the "associativity" of $\Phi$ imposes $f = f^2$, and because $f$ is an automorphism, it implies $f= \mathrm{id}\_X$.


Therefore $\Phi$ is actually the standard symmetry isomorphism.


