Number of solutions in a finite field corresponding to one-dimensional isotropic subspaces Let $q$ be a prime power and define the bilinear form
$$((x_1, x_2, x_3),(y_1, y_2, y_3)) \mapsto x_1 y_3^q + x_2 y_2^q +x_3 y_1^q$$ on the three-dimensional vector space $V$ over the finite field $F_{q^2}$ of $q^2$ elements.
Direct computation shows that the set of 1-dimensional isotropic subspaces of $V$ is
$$\Omega = \{ \langle(1,0,0) \rangle \} \cup \{ \langle(\alpha,\beta,1) \rangle : \alpha+\alpha^q + \beta \beta^q=0\}.$$
It is known that $|\Omega|=q^3+1$ but I do not see how this is so.
Clearly, the problem reduces to showing that there are $q^3$ solutions $(\alpha,\beta)$ to $$\alpha+\alpha^q+\beta\beta^q =0$$ but I am unsure how to proceed.
For context, this claim is from the Unitary groups section of the seventh chapter in the book `Permutation groups' by Dixon and Mortimer. It provided no proof that the size of $\Omega$ is $q^3+1$ and the citations therein for further information on the unitary groups either do not have the proof or are not available to me.
 A: Basic properties of the relative trace
$$T:\Bbb{F}_{q^2}\to\Bbb{F}_q, T(x)=x+x^q$$ and the relative norm
$$N:\Bbb{F}_{q^2}\to\Bbb{F}_q, N(x)=x\cdot x^q$$
give this quickly.

*

*Because $T(x)$ is a sum of $G$-conjugates, $G=Gal(\Bbb{F}_{q^2}/\Bbb{F}_q)=\{id=F^2, F\}$, $F$ the Frobenius automorphism $x\mapsto x^q$, it follows that $T(x)\in\mathrm{Inv}(G)=\Bbb{F}_q$ for all $x\in\Bbb{F}_{q^2}$.

*Because the elements of $G$ are $\Bbb{F}_q$-linear, so is $T$.

*Because $T(x)$ is a polynomial of degree $q$, its kernel cannot have more than $q$ elements, so $T$ is not the zero map.

*Because it is a linear transformation from a 2-dimensional space to a 1-dimensional space, we can conclude that $T$ is surjective, and that its kernel has exactly $q$ elements.

*So to each $z\in\Bbb{F}_q$ there are exactly $q$ solutions $x\in\Bbb{F}_{q^2}$ to the equation $T(x)=z$.

*Because $N(y)=y\cdot y^q$ is the product of $G$-conjugates we also have $N(y)\in\Bbb{F}_q$ for all $y\in\Bbb{F}_{q^2}$.

*So to each $\beta\in\Bbb{F}_{q^2}$ there are exactly $q$ elements $\alpha\in\Bbb{F}_{q^2}$ such that $T(\alpha)=-N(\beta)$. Hence $T(\alpha)=-N(\beta)$ has $q^3$ solutions altogether.


We can equally well use the fact $N(y)=0$ if and only if $y=0$, and otherwise $N$ is a
$(q+1)$-to-$1$ mapping onto $\Bbb{F}_q^*$. Therefore for $q$ choices of $\alpha$ (namely those in the kernel of $T$) there is a single solution $\beta=0$ such that $N(\beta)=0=T(\alpha)$. On the other hand for any of the other $q^2-q$ choices of $\alpha$ there are $q+1$ solutions $y$ to $N(y)=-T(\alpha)$. Hence a total of
$$q+(q^2-q)(q+1)=q^3$$
solutions.
