Timeline for Are there primitive quartic CM fields whose norms of units give all totally positive units of the real quadratic subfield?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 7, 2017 at 11:45 | vote | accept | Bernie | ||
| Jul 2, 2017 at 9:01 | answer | added | Franz Lemmermeyer | timeline score: 2 | |
| Jun 29, 2017 at 8:57 | comment | added | Bernie | @Noam D. Elkies : Thanks! I am no expert in this, but what do you mean by: "there is no criterion known for the existence of a negative unit"? I was reading the article "On real quadratic fields containing units with norm -1" by Yokoi and it says $\mathbb{Q}(\sqrt{D})$ has a unit of negative norm if and only if $D-4$ is a square. Or did I misunderstand something? What do you mean by "even $\mathbb{Q}(\sqrt{t^2+1})$"? Does it mean all fields I get using $\sqrt{t^2+4}$ I can get using $\sqrt{t_0^2+1}$ for some $t_0\neq t$? | |
| Jun 28, 2017 at 15:50 | comment | added | Noam D. Elkies | a) Yes, or even ${\bf Q}(\sqrt{t^2+1})$ (cubing the fundamental unit if necessary); but that's begging the question if you're given only the discriminant of $K_0$ (e.g. for ${\bf Q}(\sqrt{193})$ the first $t$ is $1761432$). I think there's still no full criterion known for the existence of a negative unit, only some partial results (most easily: yes if the discriminant is prime, no if it has a factor $\equiv 3 \bmod 4$). | |
| Jun 28, 2017 at 13:34 | history | asked | Bernie | CC BY-SA 3.0 |