Determine the Galois group of the splitting field of $f(x) = x^4+x+t \in F_2(t)[x]$ I used Gauss's lemma to show that the polynomial is irreducible since it is irreducible in $F_2[t,x]$, used the derivative GCD test to conclude that the polynomial is separable, and finally started by taking the quotient $F_2(t)[x]/(f(x))$ which gives a degree $4$ extension that contains at least $2$ roots: $\bar{x}, \bar{x}+1$ (using $a^2+b^2= (a+b)^2$). However, from here, I can't seem to find other roots or clearly show the non-existence of any other roots to have to take another quotient by a degree $2$ factor to find that the Galois group is $D_4$ (thought of as a subgroup of $S_4$). It seems that all subgroups of $S_4$ of order $2^n$ are transitive so that doesn't help.
Let me know if you can think of any ways to finalize the computation of the Galois group.

Let $\alpha$ be a root of $f=X^4+X+t$ in an algebraic closure of $k=\mathbb{F}_2(t)$. If $j$ is a primitive third root of $1$ (so $j^2+j+1=0$), the roots of $f$ are $\alpha,\alpha+1,\alpha+j,\alpha+j+1$ and the splitting field of $f$ is $L=k(\alpha,j)$.
Let us prove that $j\notin k(\alpha)$. Otherwise, $j=a+b\alpha+c\alpha^2+d\alpha^3$.
Then $j^2=a^2+b^2\alpha^2+c^2(\alpha+t)+d^2(\alpha^3+t\alpha^2)$ since $\alpha^4=\alpha+t$ and $\alpha^6=\alpha^2\alpha^4=\alpha^2(\alpha+t)$.
Thus
$0=j^2+j+1=(a^2+a+tc^2+1)+(b+c^2)\alpha+(b^2+c+td^2)\alpha^2+(d+d^2)\alpha^3$.
In particular, $b=c^2$, so $0=(a^2+a+tc^2+1)+(c^4+c+td^2)\alpha^2+(d+d^2)\alpha^3$.
Now $d^2+d=0$, so $d=0$ or $1$. If $d=1$, we would get $c^4+c+t=0$, which is not possible since $f$ is irreducible. So $d=0$ and $c^4+c=0=c(c+1)(c^2+c+1)$. Since $X^2+X+1$
is irreducible over $k$ (easy), we get $c=0$ or $1$.
If $c=0$, we get $a^2+a+1=0$, which is not possible since $X^2+X+1$ is irreducible over $k$.
Hence $c=1$, and $a^2+a+t+1=0$. But $X^2+X+(t+1)$ is irreducible over $k$ , so we get a contradiction.
Finally, $j\notin k(\alpha)$, $[k(\alpha)(j):k(\alpha)]=2$ and $[L:k]=8$. Now, the subextension $k(\alpha)/k$ is non Galois (since the splitting field of $f$ has degree $8$)  of degree $4$, so the Galois group has a non normal subgroup of order $2$. The only group of degree $8$ satisfying this property is the dihedral group $D_4$.
