I am trying to figure out the splitting field of $x^4 ＋ x^3 + 1$ over $\mathbb{F}_2$ .
I know $x^4 ＋ x^3 + 1$ over $\mathbb{F}_２$ is irreducible.
Let $α$ be a root of $x^4 ＋ x^3 + 1$, then the spitting field will be the field $\mathbb{F}_2(\alpha)$. So, the spitting field is isomorphic to finite field of $16$ elements.
My question is:
How to formally prove that the spitting field is the field $\mathbb{F}_2(\alpha)$
Let the other roots of $x^4＋x^3＋1$ be $\beta,\gamma,\delta$, I want to show $\mathbb{F}_2(\alpha,\beta,\gamma,\delta)=$$\mathbb{F}_2(\alpha)$.
Why are $α^2, α^4,α^8$ also roots of $x^4＋x^3＋1$? Thank you for your help.
After reading a lot of good comments, could you someone write in answer form, the proof with normality and not using normality? Thank you.
For $a\in \Bbb{F}_{p^k}$ (some field with $p^k$ elements) let $$\prod_{m=0}^{k-1} (x-a^{p^m})=\sum_{n=0}^k c_m x^k$$ Because $b\to b^p$ is an automorphism and $a^{p^k}=a$ we get that $$\sum_{n=0}^k c_m^p x^k=\prod_{m=0}^{k-1} (x-(a^{p^m})^p)=\prod_{m=0}^{k-1} (x-a^{p^m})=\sum_{n=0}^k c_m x^k$$ Whence $c_m=c_m^p$ which implies that $c_m\in \Bbb{F}_p$ so that the minimal polynomial $f$ of $a$ divides $\sum_{n=0}^k c_m x^k$ which means that $f$ splits completely in $\Bbb{F}_p(a)$.