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I think this is a really basic question, but it had been a little while since I dealt with this material and I was hoping to get a bit of assistance here. Let $q = p ^{2M}$ for some prime $p$ and $M \in \mathbb{N}$. Let $\mathbb{F}_q$ be the finite field of $q$ elements. Let $1,-1 \in \mathbb{F}_q$ and $\alpha \in \overline{\mathbb{F}_q}$, the algebraic closure. Suppose I know that $\alpha$ satisfies $$ \alpha^2 - \alpha - 1 = 0. $$

How does it follow that $\alpha \in \mathbb{F}_q$? Thanks!

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  • $\begingroup$ For information. You can easily use the quadratic formula to show that for $p\neq 2$ the equation is reducible if $5$ is a square modulo $p$. Modulo $5$ for example, it is equivalent to $(x+2)^2$ and modulo $11$ you get $(x-4)(x+3)$. $\endgroup$ – Mark Bennet Jan 28 '15 at 16:42
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$\newcommand{\F}{\mathbb{F}}$If the polynomial $f(x) = x^{2} - X - 1$ is reducible in $\F_{p}[x]$, then its root $\alpha$ is in $\F_{p} \subseteq \F_{q}$.

If $f(x)$ is irreducible in $\F_{p}[x]$, then $\F_{p}(\alpha)$ is a field with $p^{2}$ elements, which is contained in the field $\F_{q}$, as $q = p^{2 M}$.

Here you use the fact that the subfields of $\F_{p^{t}}$ are all $\F_{p^{s}}$, as $s$ ranges over the divisors of $t$.

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Hint: $\alpha$ is a root of a quadratic polynomial over $\mathbb{F}_p$, so what can you say about the possible degree of the field extension $\mathbb{F}_p(\alpha)/\mathbb{F}_p$? Can you use this to determine whether $\mathbb{F}_p(\alpha) \subseteq \mathbb{F}_q$?

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