Injectivity and norm function on finite fields

Question

Let $q$ be an odd prime power. Consider the map $f:\Bbb F_{q^3} \rightarrow \Bbb F_{q^3}$, defined by $$f(x)=\alpha x^q+\alpha^q x$$ for some fixed $\alpha \in \Bbb F_{q^3} \setminus \{ 0 \}$. Show that $f$ is a bijection.

Hint: If $\beta \in \ker(f) \setminus \{ 0 \}$ consider the relative norm map $N_{\Bbb F_{q^3}/{\Bbb F_q}}(\alpha \beta^q)$.

Answer 1

I like Adam's solution a lot. I arrived at the scene late, so I'm just adding this as my best guess as to what the hint means.

The mapping $f$ is linear over the subfield $\Bbb{F}_q$, so it suffices to show that its kernel is trivial. Assume that there exists a $\beta$ such that $f(\beta)=0$. This implies that $$ \alpha\beta^q=-\alpha^q\beta.\qquad(*) $$ Let's appply the Frobenius automorphism to both sides: $$ \alpha^q\beta^{q^2}=-\alpha^{q^2}\beta^q. $$ Repeat the dose remembering that raising to power $q^3$ is the identity mapping: $$ \alpha^{q^2}\beta=-\alpha\beta^{q^2}. $$ Let's multiply these three equations together, and arrive at $$ N(\alpha)N(\beta)=-N(\alpha)N(\beta). $$ Here $N(x)=x\cdot x^q\cdot x^{q^2}$ is the relative norm map. Because $-1\neq1$ (here we need the assumption that $q$ is odd) this implies that $N(\alpha)N(\beta)=0$. But the norm vanishes only at zero, so we can conclude that $\alpha=0$ or $\beta=0$. The former possibility was assumed not to hold, so $\beta=0$. The claim follows.

Using Adam's idea we could deduce (assuming $\alpha\beta\neq0$) from (*) directly that $$ \left(\frac\beta\alpha\right)^{q-1}=-1. $$ This leads to a contradiction by Adam's argument.

Answer 2

Consider

$$g(x) =f(x)-f(a)=\alpha(x-a)^q+\alpha^q(x-a)$$

since $q\equiv 0$ in our field and the binomial theorem holds. Now if $f$ has a double value, say $f(a)$ which is taken on twice, then $g$ has a two zeros. However, if $b\ne a$ is such a pair of zeros, we have

$$\alpha(b-a)^q+\alpha^q(b-a)=0\iff (b-a)^{q-1}=-(\alpha)^{q-1}.$$

Then

$$\left({b-a\over \alpha}\right)^{q-1}=-1$$

So that ${b-a\over\alpha}\not\in\Bbb F_{q}$, as elements of the base field are totally determined by the fact that they are roots of $x^q-x$. However, this implies

$$\left({b-a\over \alpha}\right)^{q^2}={b-a\over\alpha}$$

i.e. ${b-a\over\alpha}\in\Bbb F_{q^2}$ which is impossible since that field is not a sub-extension of $\Bbb F_{q^3}$. Hence no such $b$ exists.