Intermediate field of $\mathbb F_9(x)/$ $\mathbb F_3(x^2)$. The title question was a question asked in interview examination in some university.
I want to find all of Intermediate field of $\mathbb F_9(x)/$ $\mathbb F_3(x^2)$.
$\mathbb F_9(x)$＝$\mathbb F_3(x,$$\sqrt{2}$) is OK, but I cannnot proceed from here.
Do I need to rewrite the extension as simple extension?
For the beginning,I cannot specify the extension degree.
Thank you for your help.
 A: Note that $T^2 - x^2 \in \Bbb F_3(x^2)[T]$ and so $[\Bbb F_3(x) : \Bbb F_3(x^2)] = 2$. On the other hand, we have $\Bbb F_9(x) = \Bbb F_3(\sqrt{2})(x)$ and $T^2-2 \in \Bbb F_3[T] \subset \Bbb F_3(x)[T]$, so $[\Bbb F_9(x) : \Bbb F_3(x)] = 2$.
Thus
$$
[\Bbb F_9(x) : \Bbb F_3(x^2)] = [\Bbb F_9(x) : \Bbb F_3(x)] \cdot [\Bbb F_3(x) \colon \Bbb F_3(x^2)] = 4.
$$
All polynomials considered are separable as we are in characteristic $3$, and the extension is normal because $\Bbb F_9(x) = \Bbb F_3(x^2)(x,\sqrt{2})$ is a splitting field, so the original extension is Galois and
$$
|\mathrm{Gal}(\Bbb F_9(x)/\Bbb F_3(x^2))| = 4.
$$
Then $\mathrm{Gal}(\Bbb F_9(x)/\Bbb F_3(x^2))$ is either the Klein four group of the cyclic group of order $4$, but since any automorphism of $\Bbb F_9(x) = \Bbb F_3(x^2)(x,\sqrt{2})$ is determined by the images of $x$ and $\sqrt{2}$ and those must be roots of their respective minimal polynomials, a direct computation shows that
$$
\mathrm{Gal}(\Bbb F_9(x)/\Bbb F_3(x^2)) \simeq \Bbb Z_2 \oplus \Bbb Z_2.
$$
By the Galois corresponence, intermediate fields correspond to subgroups of the Klein group. I'll leave the rest to you.
