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Given some $e_i=0$ or $1$ for $1\le i \le 3$.

I wish to construct a totally real cubic number field $K$ so that $\prod_{i=1}^3 sgn(\sigma_i(u))^{e_i}$ is always $1$ for any $u\in O_K^\times$ where $\sigma_i$'s are real embeddings of $K$.

For quadratic case this question is somehow equivalent to Pell's equation. I am interested in higher degree cases.

Say $e_1=0$, $e_2=e_3=1$. One may want to construct a field $K$ such that all of the elements in $O_K^\times$ is either totally positive or totally negative. I would like to see such an example.

Furthermore, I am interested in whether we can construct $K$ with $\sigma_1(u)>0, \sigma_2(u)\sigma_3(u)>0$ for all $u\in O_K^\times$ up to multiplication by $-1$, but there exists some $u$ with $\sigma_1(u)>0, \sigma_2(u)<0, \sigma_3(u)<0$.

Sorry for being wordy. Thank you in advance!

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  • $\begingroup$ You should really view the $\sigma_i$ as the embeddings of $K$ into $\mathbb{R}$ and not as elements of the Galois group. $\endgroup$
    – Siksek
    Commented Dec 6, 2014 at 8:45
  • $\begingroup$ 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.) $\endgroup$
    – Lior Bary-Soroker
    Commented Dec 6, 2014 at 17:15
  • $\begingroup$ @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. $\endgroup$
    – Ted Mao
    Commented Dec 6, 2014 at 19:13
  • $\begingroup$ 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? $\endgroup$
    – Gerry Myerson
    Commented Dec 7, 2014 at 5:59
  • $\begingroup$ @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. $\endgroup$
    – Ted Mao
    Commented Dec 7, 2014 at 21:18

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