Timeline for Existence of a global completion functor
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 5, 2021 at 22:11 | comment | added | Neil Epstein | @TimCampion Well I didn't say what I wanted it for, so there's no way you could know. But yeah, I don't think even $\hat {\mathbb Z}$ does what I want. | |
| Feb 4, 2021 at 21:33 | comment | added | Tim Campion | @NeilEpstein Okay. But I wasn't necessarily suggesting you use $\prod_m \widehat{R_m}$ in general. Only that whatever definition you settle on, it should agree with $\prod_m \widehat{R_m}$ when $R = \mathbb Z$. Perhaps it's wrong even in this case, in which case I have misunderstood your intention. | |
| Feb 4, 2021 at 21:31 | comment | added | Neil Epstein | @TimCampion Thank you. It’s an interesting idea, but for the application I have in mind, I think $\prod_m \widehat{R_m}$ does not do what I want. | |
| Feb 4, 2021 at 21:13 | comment | added | Neil Epstein | Okay. Thank you! Post this as an answer, and I’ll accept it. | |
| Feb 4, 2021 at 16:06 | comment | added | Anonymous | Base changing from $\mathbf{Z} \to \mathbf{Q}$ amounts to inverting all the primes, so it takes $\mathbf{Z}_p$ to $\mathbf{Q}_p$ (as once you invert $p$, nothing needs to be done). | |
| Feb 4, 2021 at 14:14 | comment | added | Neil Epstein | 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. | |
| Feb 4, 2021 at 13:20 | comment | added | Anonymous | 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 $\mathbf{Q}_\ell$. | |
| Feb 4, 2021 at 11:42 | comment | added | Neil Epstein | @Anonymous Wait a minute. Why does it follow that $\mathbb Q_p = \mathbb Q_\ell$? | |
| Feb 3, 2021 at 21:00 | comment | added | Tim Campion | 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? | |
| Feb 3, 2021 at 20:40 | comment | added | Anonymous | 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). | |
| Feb 3, 2021 at 20:04 | history | asked | Neil Epstein | CC BY-SA 4.0 |