Let $E$ be the splitting field of $x^4+x+1$ over $\mathbb Z_2$. Find $\mathrm{Gal}(E/\mathbb Z_2)$.
I found that $$E=\{ax^3+bx^2+cx+d+\langle x^4+x+1\rangle\mid a,b,c,d\in\mathbb Z_2\}.$$
Let $\alpha=x+\langle x^4+x+1\rangle$. Then I know that $E= \mathbb Z_2(\alpha)$ and $E$ is isomorphic to $\mathrm{GF}(2^4)$.
How can I get $\mathrm{Gal}(E/\mathbb Z_2)$?
Thanks a lot.
A $\Bbb F_2[x]$ polynomial is even in its number of monomial terms iff it vanishes at $\mathbf 0:=0+\frac{\Bbb Z}{2\Bbb Z}$ or $\mathbf 1:=1+\frac{\Bbb Z}{\Bbb Z}$. Consequently, $x^2+x+\mathbf 1$ is the only $\Bbb F_2[x]$ quadratic irreducible and thus $f(x):=x^4+x+\mathbf 1$ is irreducible or else, because $f(x)$ has no $\Bbb F_2$ roots, we'd have the absurdity $$x^4+x+\mathbf 1=f(x)=(x^2+x+\mathbf 1)^2=x^4+x^2+\mathbf 1$$ Fix a field isomorphism $\varphi:\frac{\Bbb F_2[x]}{\langle f(x)\rangle}\to\Bbb F_{2^4}$ and let $\alpha=\varphi(x+\langle f(x)\rangle)$. Now, $\{\alpha,\alpha^2,\alpha^4,\alpha^8\}$ is both the set of four $f(x)$ roots and the permutation orbit of $\alpha$ under the frobenius $\Bbb F_{2^4}$ field automorphism $\varphi:t\mapsto t^2$. $$\therefore\;4\leq\text{ord}(\varphi)\leq|\text{Gal}(\Bbb F_{2^4}/\Bbb F_2)|=4\;\;\;\;\;\therefore\;\text{Gal}(\Bbb F_{2^4}/\Bbb F_2)=\langle\varphi\rangle\approx\Bbb Z_4^+$$