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

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 $$\alpha \in \mathbb{F}_{q^3}$$.

• To show that an element $a \in F_{q^3}$, you should show that $a^{q^3}=a$; also $(a+b)^q=a^q+b^q$. Jun 11, 2015 at 17:17
• I tried that but not even closed :( Jun 11, 2015 at 17:18

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 $\alpha^{q^3}=\alpha$.

• This, my friends, is a very fine solution. Nothing wrong with Jürgen's argument (which is also creative), but...(+1 to y'all, of course). Jun 11, 2015 at 18:40

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

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

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.

• This is certainly a valid proof, but it's just a lot of seemingly unmotivated algebraic manipulations. I still find the result very surprising. Where does the $3$ "come from", intuitively? Is there a geometric interpretation? Jun 11, 2015 at 18:08

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.

• A common theme in Jürgen's and your answers is to produce the relative norm map $N:\Bbb{F}_{q^3}\to\Bbb{F}_q, N(x)=x^{q^2+q+1}$ using the given equation. Well done. Jun 16, 2015 at 6:15
• You can simplify the late parts a bit by raising both sides of $a^{q^2+q+1}=1$ to power $q-1$. That gives first $a^{q^3-1}=1$, then $a^{q^3}=a$, meaning that you're done. Jun 16, 2015 at 6:33