Timeline for construct totally real cubic fields
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 8, 2014 at 1:16 | comment | added | Ted Mao | @GerryMyerson I am very sorry for the confusion. As you said I mean to assign $\sigma_i$'s first and want that work for all units. | |
| Dec 8, 2014 at 0:18 | comment | added | Gerry Myerson | Sorry, still don't follow. $u+1$ has norm $-1$, but you are allowing "up to multiplication by $-1$", so look at $-u-1$. Well, that's no good because it's negative, but one of its conjugates is positive, so take that positive conjugate to be $\sigma_1(-u-1)$. Or is it that you want to assign the $\sigma_i$ first, so that one assignment works for all units? | |
| Dec 7, 2014 at 21:18 | comment | added | Ted Mao | @GerryMyerson Thank you. I want all the elements $x$ (up to multiplication by $-1$) in $O_K^\times$ to have that property in my second last paragraph. $u$ is nice but for $x=u+1$ it doesn't hold. | |
| Dec 7, 2014 at 5:59 | comment | added | Gerry Myerson | I don't understand your objection, Ted. That $u$ is a unit, it is positive, and its conjugates are negative. What desired property is it missing? | |
| Dec 6, 2014 at 19:14 | history | edited | Ted Mao | CC BY-SA 3.0 |
deleted 6 characters in body
|
| Dec 6, 2014 at 19:13 | comment | added | Ted Mao | @LiorBary-Soroker Thank you but your $K$ does not have the property I want. If we write $u=\zeta_7+\zeta_7^{-1}$, then $-1, u$ and $u+1$ generate the unit group. | |
| Dec 6, 2014 at 17:15 | comment | added | Lior Bary-Soroker | What about $K=Q(\zeta_7)\cap \mathbb{R}$, does it satisfy what you want? (Here $\zeta_7 ={\rm exp}(2\pi i/7)$ is a primitive a 7th root of unity.) | |
| Dec 6, 2014 at 8:45 | comment | added | Siksek | You should really view the $\sigma_i$ as the embeddings of $K$ into $\mathbb{R}$ and not as elements of the Galois group. | |
| Dec 6, 2014 at 2:34 | history | asked | Ted Mao | CC BY-SA 3.0 |