Splitting field of $x^3+x+1$ over $\mathbb{F}_2$ and the Galois group

Question

I'm trying to find the splitting field $\Sigma$ of $f(x)=x^3+x+1$ over $\mathbb{F}_2$ and determine $\text{Gal}(\Sigma/\mathbb{F}_2)$. Please check if my reasoning is correct.

Since $f$ is an irreducible polynomial of degree 3 over $\mathbb{F}_2$, $\Sigma$ is isomorphic to $\mathbb{F}_{2^3}$. Since $\text{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_{p})$ is cyclic of order $n$, $\text{Gal}(\Sigma/\mathbb{F}_{2})$ is cyclic of order $3$, and we're done?

And I'm not sure how to express all three roots of $f$ in terms of one of its roots, $\alpha$. $f$ is separable over $\mathbb{F}_2$, as $f$ has a term $x$ whose exponent is not a multiple of 2. So $f$ must have 3 distinct roots.

If $\alpha$ is a root, then dividing $x^3+x+1$ by $x-\alpha$ gives $x^2+\alpha x + (1+\alpha^2)$, but I'm not sure how to find the other two roots from this.

Answer 1

HINT: Do you know the generator $\sigma$ of the Galois group? If so, what are $\sigma(\alpha)$ and $\sigma^2(\alpha)$?

Answer 2

observe that $0 = (\alpha^3 + \alpha + 1)^2 = (\alpha^2)^3 + \alpha^2 + 1$, so if $\alpha$ is a root of the polynomial f, then so is $\alpha^2$.

In addition, you may now show that $\alpha^2+\alpha$ is the third required root, so all 3 roots lie in $\mathbb{F}_2(\alpha)=\mathbb F_2[x]/(f(x))$.

It might be worth mentioning that in general in a field of characteristic p, if $\alpha$ is a root of $f(x)$, then so is $\alpha^{p}$, and so forth (here $\alpha^4=\alpha^2+\alpha$ is a root of f over the field with two elements with no need to actually factorise f).