Bijectivity of $f(x) = α^qx + αx^q$ over $\mathbb{F}_{q^3}$

Question

Let $q$ be an odd prime power and $α\in \mathbb{F}_{q^3}$ nonzero. Define the map $$f(x) = α^qx + αx^q.$$ I am supposed to show that $f: \mathbb{F}_{q^3} \to \mathbb{F}_{q^3}$ is bijective.

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My attempt: It is enough to show that $f$ is injective, surjectivity follows from a cardinality argument. Now suppose $f(x) = f(y)$, hence $$0 = α^q(x-y) + α(x^q - y^q) = α(x-y)^q+α^q(x-y)=α(x-y)\left[(x-y)^{q-1}+α^{q-1}\right].$$ All I'd need to show now is that the latter factor $(y-x)^{q-1}+α^{q-1}$ is nonzero for any choice of $x,y\in \mathbb{F}_{q^3}$, which is where I am stuck. To no avail, I tried to write everything as a power of a primitive element (the addition is annoying) and also to express everything in a basis containing $α$, but same problem here.

Edit: I also don't see where to use $q^3$; I noticed that $q-1$ divides $q^3 - 1$, but how is this useful?

Any hints or ideas?

Answer 1

Suppose $x\ne y$ and $f(x)=f(y)$. Then (you have a sign error) $$\alpha^q(x-y)+\alpha(x-y)^q=0,$$ so $$\alpha^{1-q}(x-y)^{q-1}=-1.$$ In other words, you are saying $-1\in(\mathbb F_{q^3}^\times)^{q-1}$. But recall that $\mathbb F_{q^3}^\times$ is a cyclic group of order $q^3-1$, so $(\mathbb F_{q^3}^\times)^{q-1}$ is a cyclic group of order $(q^3-1)/(q-1)=q^2+q+1$. But this does not have an element of order $2$, since $q^2+q+1$ is odd!

Answer 2

This answer doesn't add much to Kenta's solution. Posting it mostly to draw attention to:

• the use of linearity, and
• the role played by the Frobenius automorphism.

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Denote $L=\Bbb{F}_{q^3}$, $K=\Bbb{F}_q$, when $L/K$ is Galois, and the Galois group is generated by the Frobenius map $F:L\to L, x\mapsto x^q$. The first thing that caught my attention is the appearance of $F$. This readily implies that $f$ is a homomorphism of additive groups, a $K$-linear transformation even. Therefore injectivity is equivalent to having a trivial kernel.

When the Frobenius appears one recurring trick is to apply it an appropriate number of times, here thrice. Let's assume that $x\in L$ is in the kernel of $f$. This gives us $$x^q=-\alpha^{q-1}x.\tag{1}$$ Applying $F$ to $(1)$, and using $(1)$ to simplify $x^q$ yields $$ x^{q^2}=-\alpha^{q^2-q)}x^q=\alpha^{q^2-1}x.\tag{2} $$ Repeating the dose leads to $$ x^{q^3}=\alpha^{q^3-q}x^q=-\alpha^{q^3-1}x.\tag{3} $$ We then recall two things: $F^3$ is the identity on $L$, so the left hand side of $(3)$ simplifies to $x$. Also, $\alpha^{q^3}=\alpha$ and therefore $\alpha^{q^3-1}=1$. Hence $(3)$ simplifies to $$ x=-x, $$ implying that $x=0$ (we need $q$ to be odd at this point), when the claim follows.

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Final remarks:

• the claim is false when $q=2^m$ for in that case the kernel of $f$ is the $1$-dimensional subspace (over $K$) spanned by $\alpha$,
• as seen from Kenta's answer, using the cyclicity of the multiplicative group often leads to a simpler argument when we are looking for zeros of a binomial.

Answer 3

Just some generic stuff:

Say we have $L$ field, $\phi\colon L \to L$ morphism, and the additive map

$$x\mapsto \alpha \phi(x) + \beta x$$

Can it have a non-zero $x$ in the kernel? That is equivalent to

$$\frac{\phi(x)}{x} = -\frac{\beta}{\alpha}=\colon \gamma$$

Now, assume that $\phi$ has finite order $d$, that is $\phi^d = \operatorname{Id}$. Then from the equality above we get

$$\gamma \cdot \phi(\gamma) \cdot \cdots \cdot \phi^{d-1}(\gamma) = 1$$

In our case, the expression turns out to be $-1\ne 1$, so such $x$ does not exist.

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Let's also check by hand that if $\gamma$ satisfies the above condition then $x\ne 0$ exists with $\gamma = \frac{\phi(x)}{x}$. (Hilbert's 90)

Indeed, consider the $d$ distinct morphisms of $L$ $\operatorname{Id}$, $\phi$, $\ldots$, $\phi^{d-1}$. They are linearly independent ( Artin's lemma). Let's consider the following non-zero elements of $L$ $$p_k = \prod_{i=k}^{d-1} \phi^{k}(\gamma)$$ with $p_0 = 1$, from the above. Since the linear combination $\sum_{k=0}^{d-1} p_k \phi^k(\cdot)$ is not identically $0$, there exists $y \in L$ such that

$$x \colon = \sum_{k=0}^{d-1} p_k \phi^k(y) \ne 0$$

Now one checks easily that $\phi(x) = \gamma \cdot x$.