$\alpha \in \overline{\mathbb{F}}_q$ satisfying $\alpha^{q+1}+\alpha=-1$

Question

Let $\overline{\mathbb{F}}_q$ be the algebraic closure of $\mathbb{F}_q$. Assume that $\alpha \in \overline{\mathbb{F}}_q$ satisfies at $$\alpha^{q+1}+\alpha=-1.$$ Show that $\alpha \in \mathbb{F}_{q^3}$.

Answer 1

For an easier typing let us use $a$ instead of $\alpha$. You are looking for an $s$ such that

$$a^{q^s} = a .$$

Rewrite your equation as $$a^{q+1} = -a -1$$

and raise to the power $q$:

$$a^{q^2+q} = (-1)^q (a^q + 1) = (-1)^q (-\frac 1 a) ,$$

so

$$a^{q^2+q+1} = (-1)^{q+1} = 1$$

(if $q$ is odd this is clear; if $q$ is even, $-1 =1$). But

$$1 = a^{q^s - 1}$$

too, so

$$a ^{q^s - q^2 - q -2} = 1$$

and therefore

$$\text{ord} a | q^s - q^2 - q -2$$

in $\Bbb F _{q^s}$.

But the order of an element divides the order of the group ($q^s - 1$), so you are left with

$$\text{ord} a| q^2 + q +1$$

or equivalently

$$(q-1) \text{ord} a| (q-1) (q^2 + q +1) = q^3 - 1 ,$$

so

$$\text{ord} a | q^3 - 1$$

or equivalently

$$a ^{q^3-1} = 1 ,$$

so

$$a ^{q^3} = a$$

as desired, so $s=3$.

Two final notes: first, the proof is not long, it just seems so because of the typesetting and because I have written down even the most humble details, assuming that the OP is a beginner. Second, @Jürgen Böhm's proof begins by raising to the $q$th power too, but it is different in spirit since it obtains an additive relationship, while from the beginning mine tried to produce a purely multiplicative one in order to be able to use basic concepts from group theory.

Answer 2

As I just did the calculation, I will answer, albeit the hints in the comments nearly told it all.

So start with

$$(1)\quad\alpha^{q+1} + \alpha + 1 = 0$$

Raising to the $q$-th power

$$(2)\quad \alpha^{q^2+q} + \alpha^q + 1 = 0$$

and again

$$(3)\quad\alpha^{q^3+q^2} + \alpha^{q^2} + 1 = 0$$

Multiplying the latter with $\alpha^q$

$$(4)\quad\alpha^{q^3+q^2+q} + \alpha^{q^2 + q} + \alpha^q = 0$$

Substituting for $\alpha^{q^2+q}$ from (2)

$$(5)\quad \alpha^{q^3+q^2+q} - 1 - \alpha^q + \alpha^q = 0$$

Again substituting from (2)

$$(6)\quad \alpha^{q^3} (-\alpha^q - 1) - 1 = 0$$

Multiplying with $\alpha$

$$(7)\quad \alpha^{q^3} (-\alpha^{q+1} - \alpha) - \alpha = 0$$

Substituting from (1) $\alpha^{q+1} + \alpha = -1$ we are done.

Answer 3

The above answer is certainly correct.

To further understand what's going on, observe that the original equation implies $$ \alpha^q = \frac {-\alpha-1}{\alpha}, $$ so the Galois conjugate of $\alpha$ is a Mobius transformation of order $3$ in $\alpha$. If you repeat it three times, you get $\alpha^{q^3}=\alpha$.