What is the Galois group of the extension $\mathbb F_3(x^4)\subset\mathbb F_{3^2}(x)?$ Square brackets $[\;]$ will denote taking the ring of polynomials, and round brackets $(\;)$ will denote taking the field of rational functions.
My homework assignment from about a month ago had the following problem in it.

Find the Galois group of the field extension $\mathbb F_3(x^4)\subset\mathbb F_{3^2}(x).$

I didn't do it then and now I'm trying to do it to prepare myself for the exam.
The problem looks very exotic to me and I'm having trouble starting to do it. I haven't determined any Galois groups by myself yet. I first tried to see what I know about this extension.
First of all, we have $$\mathbb F_3(x^4)\subset \mathbb F_{3}(x)\subset \mathbb F_{3^2}(x).$$
The first extension has degree at most $4$ because $x$ is a zero of $f\in  \mathbb F_{3}(x^4)[y],$ where $$f(y)=y^4-x^4.$$ But $f$ is irreducible by Eisentein's criterion: $x^4$ is a prime element in $\mathbb F_3[x^4]$. Therefore $f$ is the minimal polynomial of $x^4$ and the first extension has degree $4$.
The extension $\mathbb F_3\subset \mathbb F_{3^2}$ has degree $2$ by an element count. Therefore by what's been said here, the second extension has degree $2$. Thus, the extension in the problem is finite and has degree $8$.
Unfortunately, $\mathbb F_3(x^4)$ isn't finite, nor is it of characteristic $0$. If it were one of these things, I could say that the order of the Galois group of the extension is $\leq 8.$ The best I'm able to see is that there is an eight-element $\mathbb F_3(x^4)$-basis of $\mathbb F_{3^2}(x),$ for example $(1,x_1,x_2,\ldots,x_7)$. So $$\mathbb F_{3^2}(x)=\mathbb F_3(x^4)(x_1,x_2,\ldots,x_7).$$
I could try to give an upper bound to the order of the Galois group of the extension by saying that each of $x_i$ must be mapped by any automorphism to some root of its minimal polynomial. But I don't think I can get a satisfactory bound this way.
In general, most of the study of field automorphisms done in classes was in characteristic zero and for finite fields, so most of the theorems don't apply here. I think this must mean that the problem must have an elementary solution, but I don't see it.
 A: Let $\alpha$ be a root of $y^2-2$ in $F_9$; then $F_9 = F_3(\alpha)$, and so $F_9(x) = F_3(x,\alpha)$. Any automorphism is completely determined by what it does to $x$ and to $\alpha$.
Any automorphism of $F_9(x)$ over $F_3(x^4)$ must map $x$ to a root of $y^4-x^4$, because the coefficients of the latter are fixed. So an automorphism of $F_9(x)$ over $F_3(x^4)$ must map $x$ to one of $x$, $-x$, $\alpha x$, or $-\alpha x$, (the four roots of $y^4-x^4$ in $F_9(x)$). Any automorphism of $F_9(x)$ over $F_3(x)$ must map $\alpha$ to either $\alpha$ or $-\alpha$ (the two roots of $y^2-2$, because $y^2-2$ is fixed by the automorphisms; thus, there are at most $8$ automorphisms.
This does not rely on these polynomials being the minimal polynomials. However, it is worth noting that your extension is the splitting field of $(y^4-x^4)(y^2-2)$, and this polynomial is separable; since your extension is a splitting field of a separable polynomial, it follows that it is a Galois extension. Therefore, the order of the Galois group equals the degree of the extension, so we will have the eight automorphisms mentioned above actually occur. 
In general, whether the extension $K/F$ is Galois or not, whether the extension is simple or not, we have that $|\mathrm{Aut}_F(K)|\leq [K:F]$. In fact, we have that the order of the automorphism group is less than or equal to the separable degree of $K$ over $F$. Your extensions here are all separable, so the issue of finite characteristic simply does not show up. In fact, the Primitive Element Theorem holds for any finite separable extensions, and your extension is finite and separable, so the Primitive Element Theorem holds. You will find that the only time that there are potential problems is when the degree of the extension is a multiple of the characteristic, which is not the case here.
