Galois group of $x^5 + x^2 + 1$ over the field $\mathbb{F}_{2}$

Question

I'm uncertain if I am doing this problem correctly. This is an old algebra prelim problem regarding Galois theory. We have to find the Galois group of the irreducible polynomial (the problem already assumes so) $f(x)=x^5 + x^2 + 1$ over the field of two elements, i.e. $\mathbb{F}_{2}$.

My attempt at a solution:
Let $F$ be a splitting field of $f(x)= x^5 + x^2 + 1$ over $\mathbb{F}_{2}$. Now let $\alpha$ be a root of $f(x)$. Since any finite extension of $\mathbb{F}_{p}$ is Galois for $p$ prime, in this case we have $F= \mathbb{F}_{2}(\alpha)$. In the particular case of this problem, then the Galois extension $F/\mathbb{F}_{2}$ has degree $[F:\mathbb{F}_{2}]=5$ since $f(x)$ is irreducible of degree $5$ over $\mathbb{F}_{2}$. I want to say that the Galois group of $f(x)$ over $\mathbb{F}_{2}$ is cyclic of order $5$, but I don't think that's correct.

My reasoning:
If I ended up with the result that $\textit{Gal}(F/\mathbb{F}_{2}) \cong \mathbb{Z}_{5}$, that would have meant that $F$ was isomorphic to the splitting field of $x^{2^5}-x$, i.e. $\mathbb{F}_{2^5}$. Because of this, it would follow that $$ \textit{Gal}(F/\mathbb{F}_{2})= \langle \sigma_{2} \rangle \cong \mathbb{Z}_{5},$$ where $\sigma_{2}$ is the Frobenius automorphism for $p=2$. So my question is, is it true that $F= \mathbb{F}_{2^5}$, or is it isomorphic to it? Am I on the right track?

Answer 1

Yes. Your argument is correct.

• As the comments from KCd and Libertron explained, every finite extension of a finite field is Galois with a cyclic Galois group.
• There is no mystery attached to the fact that the polynomials $x^5+x^2+1$ and $x^{32}-x$ share the same splitting field over $\Bbb{F}_2$. For more details about this specific instance you can read my answer here, where it is explained (when read with a suitable goal in mind) that actually all the zeros of $(x^{32}-x)/(x^2-x)$ can be gotten from those of $x^5+x^2+1$ by some sequence of functions $a\mapsto a+1$, $a\mapsto 1/a$, $a\mapsto a^2$. Thar is very specific to this field though.