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May 17, 2017 at 9:54 answer added Doris timeline score: 1
Apr 25, 2017 at 3:20 history edited Myshkin CC BY-SA 3.0
+ top level tag (nt.)
Apr 25, 2017 at 2:50 comment added Tom Goodwillie But you can think of another generator of the cyclotomic field such that its trace is rational. For example, in the $8$th cyclotomic field $\zeta+\zeta^2+\zeta^3$.
Apr 25, 2017 at 2:37 comment added P Vanchinathan Thats my point: Using N, Tr, for relative norm and trace: $N(\zeta)$ and $N(1+\zeta)$ have different degrees as algebraic numbers. But Tr($\zeta$) and Tr($1+\zeta$) have same degrees.
Apr 25, 2017 at 0:51 comment added Tom Goodwillie There is no real asymmetry. It just happens that $\zeta\bar\zeta$ is rational. But use $\alpha=1+\zeta$ instead and you get $\alpha\bar\alpha=2+\zeta+\bar\zeta$.
Apr 25, 2017 at 0:35 history asked P Vanchinathan CC BY-SA 3.0