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2:29 PM
@AaronMazel-Gee fwiw $\mathbb{Q}\otimes_{\mathbb{Z}} R = R[p^{-1}]$ for any p-local ring $R$, since all the other primes have already been inverted. (Lots of proofs for this... you can use the universal property of both sides, or use the fact that $\mathbb{Z}[S^{-1}]\otimes \mathbb{Z}[T^{-1}] = \mathbb{Z}[(ST)^{-1}]$ and tensoring with $\mathbb{Z}[q^{-1}]$ has no effect on $R$ when $q$ is already inverted... etc. etc.)
 

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