## Any field containing the nth roots of unity for odd n also contains the 2nth roots of unity

Let $F$ be a field over which $x^n-1$ splits where $n$ is odd. Show that $x^{2n}-1$ also splits over $F$.

Note that $x^{2n}-1 = (x^n)^2-1 = (x^n+1)(x^n-1)$.

Since $n$ is odd, if $\zeta$ is a root of $x^n-1$, then $(-\zeta)^n+1 = -\zeta^n+1 = 0$. That is, the roots of $x^n+1$ are precisely the negatives of the roots of $x^n-1$ (note that these are all distinct, since the derivative of $x^n-1$ has no nonzero roots). So any field containing the roots of $x^n-1$ also contains the roots of $x^{2n}-1$.

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