Simple exercise on Galois theory

Question

Find the splitting field of $x^6-2x^4-8x^2+16$ over $\mathbb {F}_3$ and list the intermediate fields between the base camp and the splitting field.

Answer 1

Hint: Over $\mathbb{F}_3$ we have $$ x^6-2x^4-8x^2+16=(x^2 + 2x + 2)(x^2 + x + 2)(x^2 + 1). $$

Answer 2

$x^6-2x^4-8x^2+16 = x^6+x^4+x^2+1$ in $\mathbb {F}_3[x]$.

$x^6+x^4+x^2+1 = \dfrac{x^8-1}{x^2-1} = (x^2 + 1) (x^4 + 1)$

Therefore, the splitting field of $x^6-2x^4-8x^2+16$ is the same as the splitting field of $x^4 + 1=(x^2 + x + 2) (x^2 + 2 x + 2)$, which is $\mathbb {F}_9$.

Answer 3

My factorization was incorrect, I have understood my error. Now to calculate the Galois group of the extension (which is actually a Galois extension, as $\mathbb {F}_9$ is the splitting field of a separable polynomial, having first derivative different from zero, with coefficients in $\mathbb {F}_3$) I just note that the degree of the extension is $[\mathbb {F}_9:\mathbb {F}_3]=2$, so the group has 2 elements and it is isomorphic to $\mathbb {Z}_2$. Is it right?