Find the splitting field $E$ for $x^4+x+1$ over $\mathbb Z_2$.
I proved $x^4+x+1$ is irreducible over $\mathbb Z_2$. So I tried to find the splitting field $E$ using Kronecker's theorem $x+\langle x^4+x+1\rangle$ is a zero of $f(x)$ let $a=x+\langle x^4+x+1\rangle$ and let $F= \mathbb Z_2[X]/\langle x^4+x+1\rangle$ then there is $g(x)$ in $F[x]$ such that $f(x)=(x-a)g(x)$, $\deg g(x)=3$ and I do not know what to do next.
It's easier to try a factorization as two polynomial of degree $2$, i.e $$(ax^2+bx+c)(dx^2+ex+f) = x^4+x+1$$
Since $cd = 1$ and we're in $\mathbb{F}_2$ necessarly we have $c=d=1=-1$. Similarly $a=d=1$. Looking at the coefficient of $x^3$ we have $e+b=0$ but looking to the linear term we have $e+b =1$, contradiction.
Hence having the polynomial no roots and no factorization we have that the splitting field is the $\mathbb{F}_{2^4}$