Let $F$ be a field of characteristic 2. Consider the field $F(x^2)$ that consists of the rational functions in $x^2$ over $F$. $F(x^2) \subset F(x)$. How to prove that $[F(x):F(x^2)]=2$? Furthermore, why $F(x)$ is not a Galois extension of $F(x^2)$?
This is an exercise from Emil Artin's lecture notes "Algebra with Galois Theory" (Exercise 12, pp.75).
My thought is to find a basis for $F(x)$ such that any element of $F(x)$ can be expressed by this basis with coefficients from $F(x^2)$. I started with a polynomial in $F[x]$. Note that $\forall p(x) = a_0 + a_1x + ... + a_nx^n\in F[x]$, $$p(x) = (a_1 + a_3x^2 + ... + a_{2\lfloor\frac{n-1}{2}\rfloor+1}x^{2\lfloor\frac{n-1}{2}\rfloor})x + a_0 + a_2x^2 + ... + a_{2\lfloor\frac{n}{2}\rfloor}x^{2\lfloor\frac{n}{2}\rfloor}.$$In other words, $p(x)\in F[x]$ can be uniquely written as $p(x) = ax + b$, where $a,b\in F[x^2]$. Therefore, $\forall f(x)=\frac{p(x)}{q(x)}\in F(x)$, $f(x)$ can be written as $f(x)=\frac{ax+b}{cx+d}$, where $a,b,c,d\in F[x^2]$. If $c\neq 0$, then $f(x)=\frac{a}{c}+\frac{bc-ad}{c^2}\frac{1}{x+\frac{d}{c}}$, and thus $f(x)$ can be expressed by a linear combination of $1$ and $\frac{1}{x+\frac{d}{c}}$ with coefficients from $F(x^2)$. However, since the latter relies on $f(x)$, they do not form a basis for $F(x)$. I've gotten this far and now I'm not sure what to do next. I haven't used the condition that $\text{char}F = 2$ yet, but I have no ieal how to utilize it. I only find out the square of any element in $F(x)$ belongs to $F(x^2)$, but this conclusion seems useless. Maybe there's another better way to avoid directly constructing a basis for $F(x)$?