element that is algebraic over a finite field

Question

Let $p$ be a prime. And let $q = p^{2h}$. Suppose I know that an element $\alpha \in \overline{ \mathbb{F}_q }$, satisfies $\alpha^2 + \alpha + 1 = 0$. Does this mean that $\alpha \in \mathbb{F}_{p^2} \subseteq \mathbb{F}_q$? I think the answer is yes, but I can't seem to figure it out... I would appreciate an explanation whether the answer is yes or no. Thanks!

Answer 1

Clearly, $\alpha$ is contained in some quadratic extension of $\mathbb F_p$. But there is only one such extension of $\mathbb F_p$ (in general up to isomorphim, but within a fixed closure it is really unique). So, yes, $\alpha\in\mathbb F_{p^2}\subseteq \mathbb F_q$

Answer 2

If $p=2$ Yes:

$\alpha^{2^2-1}-1=\alpha^3-1=(\alpha-1)(\alpha^2+\alpha+1)=0$. Hence $\alpha\in \Bbb F_{2^2}$.