Is $\Bbb Q（sin2π/n）$ is Galois over $\Bbb Q$?

Question

I want to know whether$\Bbb Q（sin2π/n）$ is Galois over $Q $or not.

Let ζ be a n-th root of unity. $sin2π/n＝（ζ＋1/ζ）/2i$,

I know

$\Bbb Q$⊂$\Bbb Q（sin2π/n）$⊂ $\Bbb Q（ζ,i）$

and $\Bbb Q（ζ,i）/\Bbb Q $ is Abel extension（Is this right? How to prove it ?）,

$\Bbb Q$（sin2π/n）/$\Bbb Q $is Galois.

Is my attempt right? I would be appreciated if you could help me, thank you.

score 1 · Accepted Answer · answered Jan 30 '20 at 11:45

1

$\Bbb{Q}(\zeta_n,i)=\Bbb{Q}(\zeta_m)$ where $m=lcm(n,4)$. And $\Bbb{Q}(\zeta_m)/\Bbb{Q}$ is Galois and abelian because its Galois group is a subgroup of $\Bbb{Z}/m\Bbb{Z}^\times$ (in fact it is the whole of it) where $a\bmod m$ corresponds to $\zeta_m \mapsto \zeta_m^a$. Yes, the Galois correspondence between each subgroup and its fixed subfield implies that subextensions of an abelian extension are Galois (and abelian).

answered Jan 30 '20 at 11:45

reuns

77,999

Thank you very much.Is my proof right? – Jan 30 '20 at 11:59
And how to prove Q（ζ,i）＝Q（ζ）？ – Jan 30 '20 at 12:10
1

if $n\ne 4$ then $\zeta_m=\zeta_n i$ – reuns Jan 30 '20 at 13:00
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Perhaps a more interesting question would be "where does $\mathbf Q(sin\theta)$ sit in $\mathbf Q(sin\theta,i)$. See e.g. https://math.stackexchange.com/a/2896166/300700 – nguyen quang do Feb 01 '20 at 13:58

Is $\Bbb Q（sin2π/n）$ is Galois over $\Bbb Q$?

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