Show that $\mathbb{F}_{p^2}$ has an 8th root of unity

Question

Let $p$ be an odd prime number. I want to show that $\mathbb{F}_{p^2}$ has a primitive 8th root of unity $\zeta$.

• I know that $\zeta^8 = 1$. So my idea is to define $f = X^8 - 1$ such that $\zeta$ is a root of $f$. But this is for a field extension of degree 8 and $p^2$ is at least 9.

any hints?

Answer 1

Consider the unit group $\mathbb{F}_{p^2}^{\times}$, which has order $p^2-1$.

Finding an eighth root of unity is equivalent to finding an element which has order $(p^2-1)/8$ in this group. Since the unit group of finite field is known to be cyclic, this happens if and only if $p^2-1$ is a multiple of $8$, ie $p^2-1 \equiv 0 \mod{8}.$

Since $p$ is assumed to be odd, this is true since if $p$ is odd then $p \equiv 1,3,5,7 \mod{8}$. This means $p^2 \equiv 1 \mod{8}$ by checking manually and hence we have an eighth root of unity.

Answer 2

We have the cyclic group $\;\Bbb F_{p^2}^*\;$ of order $\;p^2-1=0\pmod8\;$ and thus there exists a (unique) subgroup of order eight there. Any generator of this subgroup will do it.