# Determine $[K(\zeta_{16}):K]$ when $K=\mathbb{F}_7, \mathbb{F}_9, \mathbb{F}_{17}$

Let $\zeta_{16}$ be a primitive 16-th root of unity over a field $K$. Determine $[K(\zeta_{16}):K]$ when $K=\mathbb{F}_7, \mathbb{F}_9, \mathbb{F}_{17}$.

I know that over $\mathbb{Q}$, the minimal polynomial of $\zeta_{16}$ is $$\Phi_{16}(x)=\frac{x^{16}-1}{x^8-1}=(x^8+1)$$

For first, $$\Phi_{16}(x) = \frac{x^{16}-1}{x^8-1}=x^8+1.$$ Second: the splitting field of $\Phi_n(x)$ over $\mathbb{F}_p$ is isomorphic to $\mathbb{F}_{p^{m}}$, where $m$ is the least integer such that $n$ is a divisor of $|\mathbb{F}_{p^m}^*|=p^m-1$. The least $m$ such that $16\mid 17^m-1$ is $m=1$, hence $x^8+1$ splits completely over $\mathbb{F}_{17}$. The least $m$ such that $16\mid 7^m-1$ is $m=2$, hence $\Phi_{16}(x)$ splits completely over $\mathbb{F}_{7^2}$. The least $m$ such that $3^m-1$ is a multiple of $16$ is $m=4$, hence $x^8+1$ splits completely over $\mathbb{F}_{3^4}$.
• You have all of the computations backwards. For example, $x^8 + 1$ splits completely over $\mathbf{F}_{17}$. You should be looking for the order of $p$ mod $16$, rather than the other way around. – Brandon Carter Jun 3 '15 at 15:49
• @Michael: by definition, the degree of $K(\zeta_m)$ over $K$ is the degree of the minimal polynomial of $\zeta_m$ over $K$. If you know the splitting field of $\zeta_m$, you know such a degree. – Jack D'Aurizio Jun 5 '15 at 13:09
• @JackD'Aurizio I thought $[K(\zeta_m) :K]$ was the degree of the minimal polynomial. If $F$ is the splitting field, all we know is that $[K(\zeta_m):K]$ divides $[F:K]$... – Justine Jun 5 '15 at 13:13