I'm reading a paper right now that uses this property but they don't really explain it well. They just write that "it is clear", but it isn't exactly clear to me.
Their proof involves choosing a basis $\{1,\alpha\}$ of $\mathbb{F}_{q^2}$ over $\mathbb{F}_q$ (where $\mathbb{F}_q$ is the finite field of size $q$). What I have so far is:
Take $(m+\alpha n)\in\mathbb{F}_{q^2}$ where $m,n\in\mathbb{F}_q$. Then \begin{align} (m+\alpha n)+(m+\alpha n)^q &= m+\alpha n + m^q + \alpha^q n^q \\ &= 2m + n(\alpha+\alpha^q). \end{align}
I know $2m\in\mathbb{F}_{q}$. What only remains is to show that $(\alpha+\alpha^q)\in\mathbb{F}_q$. Can anyone give me a hint on how to show this?
The Frobenius map $F(x)=x^q$ generates the Galois group of $\mathbb{F}_{q^2}$ over $\mathbb{F}_q$, so an element $x\in\mathbb{F}_{q^2}$ is in $\mathbb{F}_q$ iff $F(x)=x$. Applying this to an element of the form $x+x^q$ we have $$F(x+x^q)=x^q+x^{q^2}=x^q+x$$ (since $x^{q^2}=x$ for all $x\in\mathbb{F}_{q^2}$) so $x+x^q\in\mathbb{F}_q$.