Timeline for Recovering the Zariski topology from the Zariski topology over an extension
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| May 17, 2018 at 21:15 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Apr 17, 2018 at 21:12 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 18, 2018 at 20:33 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Feb 19, 2018 at 11:36 | comment | added | THC | @ nfdc23 (about my focus on $\mathrm{Aut}(\ell/k)$): a very important case for me is the case in which $k$ is a finite field, and $\ell$ is contained in an algebraic closure of $k$. I was hoping that $\mathrm{Aut}(\ell/k)$ would act transitively on the set of prime ideals of $A \otimes_k \ell$ over a given (arbitrary) prime ideal of $A$, and that the action then could be used to elegantly describe closed sets in $A$ ... | |
| Feb 18, 2018 at 22:01 | comment | added | nfdc23 | @anon: I agree that tha case $K/k$ algebraic is easy. But the case of $K/k$ non-algebraic is useful (e.g., $k=\mathbf{Q}$, $K=\mathbf{C}$) and seems to be part of the question, and doesn't seem to reduce to the algebraic case. | |
| Feb 16, 2018 at 19:34 | answer | added | anon | timeline score: 1 | |
| Feb 16, 2018 at 18:51 | comment | added | anon | For the record, there is an easy elementary argument that shows that the map $spec(A\otimes_k K)\to spec(A)$, $K$ the algebraic closure of $k$, is a quotient map (on topological spaces). No need to appeal to EGA IV, 2.3.12,... | |
| Feb 16, 2018 at 17:39 | comment | added | nfdc23 | You are very focused on the action of ${\rm{Aut}}(\ell/k)$, but that is useless when the finite extension is non-trivial but so far from Galois that it has no nontrivial automorphisms at all (e.g., $\mathbf{Q}(m^{1/3})/\mathbf{Q}$ for a non-cube integer $m$). What is your motivation for asking about general extensions and yet trying to shoehorn the (possibly trivial) automorphism group into that case? | |
| Feb 16, 2018 at 17:08 | comment | added | Jason Starr | @THC. A subset of $\text{Spec}(A\otimes_k \ell)$ is the inverse image of a subset of $\text{Spec}(A)$ if and only it has equal inverse images under the two projections $\text{Spec}(A\otimes_k \ell\otimes_k \ell) \to \text{Spec}(A\otimes_k \ell)$. For a finite, Galois extension, equality of these two inverse images is equivalent to be fixed (as a subset) by the Galois group. Anyway, this is a different part of descent: it is not about "openness" or "closedness" of subsets, it is about which subsets are inverse images. | |
| Feb 16, 2018 at 16:44 | comment | added | THC | @ Jason Starr: yep -- unfortunately, I don't see the $\mathrm{Aut}(\ell/k)$-action in this formulation ... :-( | |
| Feb 16, 2018 at 16:38 | comment | added | Jason Starr | @THC. Did you read the reference in the comment by nfdc23? That completely characterizes the open subsets (or, equivalently, the closed subsets). A subset of $\text{Spec}(A)$ is open if and only if the inverse image subset of $\text{Spec}(A\otimes_k \ell)$ is open. | |
| Feb 16, 2018 at 14:24 | comment | added | THC | @ Denis Nardin and Jason Starr: I expanded the statement of the question to make it more precise. Hope it is more clear now. | |
| Feb 16, 2018 at 14:23 | comment | added | THC | @ nfdc23: I changed "$\mathrm{Gal}(\ell/k)$'' in ``$\mathrm{Aut}(\ell/k)$.'' | |
| Feb 16, 2018 at 14:21 | history | edited | THC | CC BY-SA 3.0 |
Changed $\mathrm{Gal}(\ell/k)$ in $\mathrm{Aut}(\ell/k)$, and expanded the question to make it more precise.
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| Feb 16, 2018 at 13:43 | comment | added | nfdc23 | EGA IV$_2$, 2.3.12 (applies with $\ell/k$ an arbitrary extension of fields as in the initial part of the question, whereas "${\rm{Gal}}(\ell/k)$" doesn't make sense in general). | |
| Feb 16, 2018 at 13:16 | comment | added | Jason Starr | @DenisNardin. Only the OP can answer for certain, but I suspect the OP is asking why the morphism $\text{Spec}(A\otimes_k \ell)\to \text{Spec}(A)$ is submersive, i.e., a subset of $\text{Spec}(A)$ is open if and only if the inverse image subset of $\text{Spec}(A\otimes_k \ell)$ is open. | |
| Feb 16, 2018 at 13:15 | comment | added | Denis Nardin | Recover as a space or as a scheme? As a scheme it's just classical Galois descent, I believe. | |
| Feb 16, 2018 at 13:03 | history | asked | THC | CC BY-SA 3.0 |