Timeline for Elements whose conjugates are of the same absolute value in cyclotomic fields
Current License: CC BY-SA 3.0
6 events
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| Sep 25, 2012 at 4:00 | comment | added | Binzhou Xia | Oh,I think my second question is to find the so called $\sqrt{kp+1}$-Weil numbers in cyclotomic fieds. In general, $m$-Weil numbers are not easy to classify. | |
| Sep 24, 2012 at 16:34 | history | edited | Binzhou Xia | CC BY-SA 3.0 |
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| Sep 24, 2012 at 16:25 | history | edited | Binzhou Xia | CC BY-SA 3.0 |
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| Sep 23, 2012 at 11:22 | comment | added | François Brunault | Also, note that $|z|^2=z\bar{z} = kp+1$ implies $|\sigma^r(z)|^2= kp+1$ for all $r$. If there exists $z_0$ such that $|z_0|^2 = kp+1$, then all other $z$ are of the form $z=z_0 \cdot u/\bar{u}$ with $u \neq 0$. I don't see any reason why such $z_0$ should exist in general. For example if $p=3$ and $k=1$ then $z_0=2$ works, but if $p=3$ and $k=3$ then the equation $|z|^2=10$ has no solution in $\mathbf{Q}(\zeta_3)$. | |
| Sep 23, 2012 at 10:49 | comment | added | François Brunault | Note that the condition $|\sigma(z)|=|z|$ can be written $\sigma(z) \sigma(\bar{z}) = z \bar{z}$, which implies $\sigma^r(z) \sigma^r(\bar{z}) = z \bar{z}$ for all $r$. So you're asking what are the elements $z \in \mathbf{Q}(\zeta_p)^\times$ such that $\sigma(z)/z$ has modulus 1. Note that the map $z \mapsto \sigma(z)/z$ is surjective onto the elements of norm 1 (Hilbert 90), thus it attains all elements of modulus 1. | |
| Sep 22, 2012 at 11:34 | history | asked | Binzhou Xia | CC BY-SA 3.0 |