Timeline for Is a field that never embeds twice in another field necessarily a prime field?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 28, 2018 at 12:11 | comment | added | Gerald Edgar | Transcendence base, eh? Does this result require the Axiom of Choice, then? | |
| Feb 28, 2018 at 6:38 | vote | accept | Omar Antolín-Camarena | ||
| Feb 28, 2018 at 6:37 | history | edited | Sándor Kovács | CC BY-SA 3.0 |
deleted 44 characters in body
|
| Feb 28, 2018 at 6:35 | comment | added | Sándor Kovács | @OmarAntolín-Camarena: you're right, what I had in mind was to extend that inside $\overline G$, but I did not write it down very well. I'll make the change. Thanks. | |
| Feb 28, 2018 at 6:32 | comment | added | Omar Antolín-Camarena | Oh, I guess that step can be replaced by constructing two different homomorphisms $k \to \bar{G}$ to an algebraic closure of $G$! I think I'm happy with the proof after that change (or after an explanation of why my objection was wrong ;)). | |
| Feb 28, 2018 at 6:04 | comment | added | Omar Antolín-Camarena | (I guess that specific example does extend if $F$ has characteristic 2 because then $(\sqrt{t}+1)^2 = t+1$, but I'm pretty sure $t+1$ is not a square in $F(\sqrt{t})$ for other charactertistics.) | |
| Feb 28, 2018 at 6:01 | comment | added | Omar Antolín-Camarena | How does the step of extending the automorphism of $G$ to one of $k$ work? I don't think it's true in general that you can extend automorphisms to algebraic extensions: the automorphism you mentioned of, say, $F(t)$ given by $t \mapsto t+1$, doesn't extend to $F(\sqrt{t})$, does it? | |
| Feb 28, 2018 at 5:48 | history | edited | Sándor Kovács | CC BY-SA 3.0 |
deleted 79 characters in body
|
| Feb 28, 2018 at 5:42 | history | answered | Sándor Kovács | CC BY-SA 3.0 |