Clearly, if $\zeta_{n}$ is a root of unity for $n$ odd, then $\zeta_{n}^{2}$ is also a primitive $n$th root, since $(n, 2) = 1$. Hence, $\zeta_{n}$ is a $2n$th root of unity. But how do I know all the $2n$th roots are in this field, including the ones that aren't $n$th roots?
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They're not. For example, try $n=3$. – quasi Jul 21 '17 at 00:05
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Really? The exact wording of the question is "Prove that if a field contains the $n$th roots of unity for $n$ odd, then it also contains the $2n$th roots of unity." – BMac Jul 21 '17 at 00:07
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4If $\zeta_n$ is a primitive $n^{th}$ root of unity, then $-\zeta_n$ is a primitive $2n^{th}$ root (when $n$ is odd of course). – lulu Jul 21 '17 at 00:09
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Ah, I stand corrected. I was thinking about the group, not the field. – quasi Jul 21 '17 at 00:10
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I see lulu, so because the field contains all inverses, it's sufficient to prove that result. – BMac Jul 21 '17 at 00:13
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If $n$-th root of unity means the roots of $X^n-1$ then see Robert's answer. If it means primitive $n$-roots of unity, then see lulu's comment, provided the characteristic is not $2$ (in which case primitive $2n$-th roots of unity don't exist). – reuns Jul 21 '17 at 00:14
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$X^{2n}-1 = (X^n - 1)(X^n+1)$, so any $2n$'th root of unity that is not an $n$'th root of unity is a root of $X^n + 1$. But if $n$ is odd, that is $-Y^n + 1$ where $Y = -X$.
Robert Israel
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Oh, right, I see this because we can rewrite it by pulling out the negative: $-((\frac{x}{-1})^{n} - 1)$, but this only works because we're dealing with an odd power of -1. – BMac Jul 21 '17 at 00:19
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This answer really needs to address the failure in characteristic $2$. – Tobias Kildetoft Jul 21 '17 at 07:27
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What failure? In characteristic $2$, this says that any $2n$'th root of unity is an $n$'th root of unity. – Robert Israel Jul 21 '17 at 15:16
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As we know the formula [Q[]:Q]=$\phi(n)$ so with similar argument $[Q[_{2n}]:Q]$=$\phi(2n)$ as (2,n)=1 so $\phi(2n)=\phi(n)$
