Galois Group of $x^4 - 7$ over $\mathbb{F}_5$ I am asked to find the Galois Group of the polynomial $x^4 - 7$ over $\mathbb{F}_5$. I am wondering if the following is correct:
The splitting field of $x^4 - 7 = x^4 - 2$ over $\mathbb{F}_5$ is $\mathbb{F}_5(\sqrt[4]{2},i)$ where $i,\sqrt[2]{2}$ lie in a fixed algebraic closure of $\mathbb{F}_5$ and $i^2 = -1$ and $(\sqrt[4]{2})^4 = 2$. Since $2^2 = 4 = -1 \in \mathbb{F}_5$ we see that $i = 2$. Now, $x^4 - 2$ does not have any roots in $\mathbb{F}_5$ and so $[\mathbb{F}_5(\sqrt[4]{2}): \mathbb{F}_5] = 2$ or $4$ which means the Galois Group is either order $2$ or $4$. If it were $2$, then $\sqrt[4]{2} = a + b\sqrt{2}$ where $a,b \in \mathbb{F}_5$. After squaring we must have that $a^2 + b^2 = 0$ and $2ab = 1$. This is an impossibility and so the degree of this extension (and hence the order of the Galois group) is $4$.
Consider $\sigma: \mathbb{F}_5(\sqrt[4]{2}) \to \mathbb{F}_5(\sqrt[4]{2})$ given by $\sigma(\sqrt[4]{2}) = 2\sqrt[4]{2}$. This is an automorphism of $\mathbb{F}_5(\sqrt[4]{2})$ of order $4$ and so must be the the galois group is $\langle \sigma \rangle$.
 A: I am stating an alternative path.
I will show that $x^4-2$ is irreducible over $\mathbb F_5$ and hence $[\mathbb F_5(\sqrt[4] 2):\mathbb F_5]=4$.
Also $\mathbb F_5$ contains fourth root of unity over $\mathbb F_5$($\because a^{(5-1)}\equiv 1(\mod 5) $ for all $a\in \mathbb F_5^*$
). So $x^4-2$ splits over $\mathbb F_5(\sqrt[4] 2)$.
$$x^4-2=\prod_{a\in \mathbb F_5^*}(x-a\sqrt[4]2)$$
If $g(x)$ is a factor of characteristic $x^4-2$ then $g(-x)$ is also a factor .
$x^4-2$ has no root in $\mathbb F_5$.
Then observe that if $x^4-2$ reducible then only possibility is
$$x^4-2=(x^2+ax+b)(x^2-ax+b)=x^4+(2b-a^2)x^2+b^2x$$
$\therefore 2b=a^2$ and $b^2=2$ . But this is not possible for any $a,b\in \mathbb F_5$.
So $x^4-2$ is irreducible in $\mathbb F_5$. Then you have $[\mathbb F_5(\sqrt[4] 2):\mathbb F_5]=4$.
After this use the fact that every finite extension over a finite field is cyclic. (you can find a proof here)
A: Here’s another way of looking at the problem:
You’re asking for $\sqrt[4]2$ and the extension it generates over $\Bbb F_5$.
Now, $2$ is of (multiplicative) order $4$ in $\Bbb F_5^\times$, so its fourth root will be of order $16$. So you’re looking for the smallest power $5^m$ such that $\Bbb F_{5^m}$ has order ($5^m-1$) divisible by $16$. You see that $25$ and $125$ are no good, but surenough, $16\mid624=5^4-1$. So the degree of the splitting field is four. (Note: all extensions of finite fields are normal, abelian, cyclic.)
