Find the automorphism group of an extension involving polynomial field with coefficients from a finite field.

Question

I want to find the automorphism group of $L/K$ for $L=F_2 (x)$ and $K=F_2(x^2)$. I haven't solved something like this before so I will try a similar approach to extensions like $\mathbb{Q} (\sqrt{2}) / \mathbb{Q}$. So I set $ f(x)=x$ then $$f^2-x^2$$ is the minimal functional equation (not sure if such thing exists). So I will try to find maps that map $x$ to its roots. Is this approach even correct? What should I do?

Answer 1

Your idea is OK. First we have that $L=K(x)$ and also $f(t) = t^2 - x^2 \in K[t]$ is the minimal polynomial of $x$ over $K$. Obviously $f$ is irreducible, as $x \not \in K$. Now we have that $L$ is the splitting field of $f$ over $K$ and also $f(t) = (t-x)^2$ in $L[t]$. Now an automorphism $\tau$ of $L$, fixing $K$ must send $x$ to a root of $f$. So we have that $\tau(x) = x$ and this uniquely determines the automorphism. Hence we must have that the only element of $\text{Gal}(L/K)$ is the identity automorphism and $\text{Gal}(L/K)$ is the trivial group.

Answer 2

Let $a$ be one of those automorphisms. Then $x^2=a(x^2)=a(x)^2$. This equation is equivalent to $(a(x)-x)^2=0$ in $\mathbb{F}_2$. Therefore, $a(x)= x$.

On the other hand, this completely determines the automorphism since any element of your field $p(x)/q(x)$ can be written as $$\frac{xp_1(x^2)+p_0(x^2)}{xq_1(x^2)+q_0(x^2)}$$ for which the action of $a$ results in $$\frac{a(x)p_1(x^2)+p_0(x^2)}{a(x)q_1(x^2)+q_0(x^2)}$$ since $a$ fixes $p_0(x^2),p_1(x^2),q_0(x^2)$, and $q_1(x^2)$.

Therefore, $a$ is the identity.