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Is there any such thing as a ``global completion functor'' for commutative Noetherian rings?

More specifically, let $R$ be a commutative Noetherian ring. For each maximal ideal $m$ of $R$, let $W_m := R \setminus m$. I want a Noetherian ring $S$ and a ring homomorphism $\phi: R \rightarrow S$ such that for each maximal ideal $m$ of $R$, the localized map $\phi_m: R_m \rightarrow W_m^{-1} S$ is isomorphic to the $mR_m$-adic completion map $R_m \rightarrow \widehat{R_m}$. Does such a thing exist?

I guess from a geometric point of view, I want a scheme $X$ over Spec $R$ whose closed fibers correspond to adic completion.

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    $\begingroup$ Your requirement at the closed point is on the stalks, so it imposes some non-trivial constraints at non-closed points too. In fact, they're so strong that such a ring (noetherian or not) cannot exist for $R=\mathbf{Z}$ already. Indeed, if such an $S$ existed, then for any prime number $p$, the base change along $\mathbf{Z} \to \mathbf{Z}_{(p)} \to \mathbf{Q}$ would have to give $\mathbf{Z}_p[1/p] = \mathbf{Q}_p$. But this is true for all $p$, so we must have $\mathbf{Q}_p = \mathbf{Q}_\ell$ as fields for all primes $p$ and $\ell$, which is not true (e.g., by roots of unity). $\endgroup$
    – Anonymous
    Commented Feb 3, 2021 at 20:40
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    $\begingroup$ Maybe I'm off-base, but I'd have thought that however you precisely formulate the question, the canonical example of a "global completion" should be getting $\widehat{\mathbb Z}$ from $\mathbb Z$, so maybe the issue is in the formalization of the question? $\endgroup$
    – Tim Campion
    Commented Feb 3, 2021 at 21:00
  • $\begingroup$ @Anonymous Wait a minute. Why does it follow that $\mathbb Q_p = \mathbb Q_\ell$? $\endgroup$
    – Neil Epstein
    Commented Feb 4, 2021 at 11:42
  • $\begingroup$ You can base change along $\mathbf{Z} \to \mathbf{Q}$ by going through either $\mathbf{Z}_{(p)}$ or through $\mathbf{Z}_{(\ell)}$. The former gives $\mathbf{Q}_p$ while the latter gives $\mathbf{Q}_\ell$. $\endgroup$
    – Anonymous
    Commented Feb 4, 2021 at 13:20
  • $\begingroup$ Hm. I guess you are using that $\mathbb Z_p \otimes_{\mathbb Z} \mathbb Q = \mathbb Q_p$, right? I don’t quite see why that tensor product has to be a field — just that it has to contain $\mathbb Q_p$ as a subring. $\endgroup$
    – Neil Epstein
    Commented Feb 4, 2021 at 14:14

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