Galois Group of $x^4 - 7$ over $\mathbb{F}_5$

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)

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.)