Timeline for Does there exist a scheme $X/{\operatorname{Spec}(\mathbb{Z})}$ such that $\pi_1^\text{ét}(X)=\smash{\hat{\mathbb{Z}}}^2$?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Sep 11 at 14:54 | vote | accept | RJ Acuña | ||
| Sep 11 at 5:19 | answer | added | Will Chen | timeline score: 4 | |
| Sep 10 at 23:55 | comment | added | RJ Acuña | @WillChen but $\pi_1$ is covariant there’s no reason apriori why $\pi_1(X)$ can’t be abelian even if the induced map is into $\pi_1(\text{Spec}(\mathbb{Z}[1/p])$ which isn’t abelian. | |
| Sep 10 at 23:40 | comment | added | RJ Acuña | @WillChen I’m ok with the answer being X doesn’t exist. | |
| Sep 10 at 23:13 | history | edited | LSpice | CC BY-SA 4.0 |
Links to comments; proofreading
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| Sep 10 at 22:45 | comment | added | Will Chen | @RJAcuña that condition doesn’t leave you many options. Connected closed subschemes are just F_p, and open subschemes all include into Z[1/p], whose fundamental group I’m pretty sure is nonabelian… | |
| Sep 10 at 21:59 | history | edited | RJ Acuña | CC BY-SA 4.0 |
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| Sep 10 at 21:56 | comment | added | RJ Acuña | @WillChen yes, I'd like the map to be an embedding. The reason is Spec$(\mathbb{Z})$ has cohomological dimension $3$ (up to $2-$torsion). And well Spec$(\mathbb{F}_p)\hookrightarrow$Spec$(\mathbb{Z})$ is an embedding, and Spec$(\mathbb{F}_p)$ has cohomological dimension 1. I want to know if there is an embedding $X$, with cd$(X)=2$. I'll edit the question. | |
| Sep 10 at 21:12 | comment | added | Cranium Clamp | @WillSawin yes, admittedly, it did sound silly, I’ll just leave it there for now. I’m not sure how accurate this reference is, a link with an example of finite rings math.stackexchange.com/questions/530591/… | |
| Sep 10 at 21:09 | comment | added | Will Chen | You can also take a product of $\mathbb{G}_m$'s over $\mathbb{C}$. | |
| Sep 10 at 21:01 | comment | added | Will Chen | Do you have any requirements on the map $X\rightarrow\text{Spec }\mathbb{Z}$? If not, the fundamental group of an elliptic curve over $\mathbb{C}$ is $\hat{\mathbb{Z}}^2$. | |
| Sep 10 at 20:55 | comment | added | Will Sawin | @CraniumClamp I don't think one normally defines $\pi_1$ for a disconnected space as the products of the $\pi_1$s of the components. Usually it's just taken to be undefined. | |
| Sep 10 at 20:50 | comment | added | Cranium Clamp | Of course, we could take the product of F_p and F_q, which just translates to disjoint union of the Spec but that’s a silly example and I’m not sure if you want that. | |
| Sep 10 at 20:36 | history | edited | RJ Acuña | CC BY-SA 4.0 |
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| Sep 10 at 20:32 | history | edited | LSpice | CC BY-SA 4.0 |
èt -> ét; included title in question
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| S Sep 10 at 20:27 | review | First questions | |||
| Sep 10 at 20:33 | |||||
| S Sep 10 at 20:27 | history | asked | RJ Acuña | CC BY-SA 4.0 |