Calculate $\mathbb{F}_2(\alpha, \beta)$ where $\alpha^3 + \alpha + 1 = 0, \beta^2 + \beta + 1 = 0$

Question

Calculate $\mathbb{F}_2(\alpha, \beta)$ where $\alpha^3 + \alpha + 1 = 0, \beta^2 + \beta + 1 = 0$ where $\alpha, \beta$ are elements of a field extension $L$ of $\mathbb{F}_2$.

---

I know $\mathbb{F}_2(\alpha) \cong \frac{\mathbb{F}_2[X]}{X^3+X+1}$ and $\mathbb{F}_2(\beta) \cong \frac{\mathbb{F}_2[X]}{X^2+X+1}$.

There are 2 methods that I think could work but I am not sure which one is correct.

1. I know $\mathbb{F}_2(\alpha) \cong \frac{\mathbb{F}_2[X]}{X^3+X+1}$ and $\mathbb{F}_2(\beta) \cong \frac{\mathbb{F}_2[X]}{X^2+X+1}$. Since $X^3+X+1$ is irreducible, it means that $\mathbb{F}_2(\alpha) \cong \mathbb{F}_{2^3}$ since the irreducible polynomial has degree 3. Then since $\mathbb{F}_2(\alpha, \beta) = (\mathbb{F}_2(\alpha))(\beta)$, I thought of maybe doing a similar thing and then getting that it is isomorphic to $\mathbb{F}_{{2^3}^2}$ but I don't know how to check that $X^2 +X +1$ is irreducible in $\mathbb{F}_2(\alpha)$.

2. Since $X^2+X+1$ is in the ideal, I know that $X = -X^2 - X$, by filling this in the other polynomial we get $X^6 + X^4 + X^2 + 1$. This is reducible over $F_2$ since in $\mathbb{Z}_2$ it holds that 1 is a root. But then I'm not sure what I can do, I wanted to divide everything by $(x-1)$ to make it reducible but this doesn't really work since that would be for $\mathbb{Z}_2$.

3. I know that $\mathbb{Z}_2(\alpha) \cong \frac{\mathbb{Z}_2[X]}{X^3+X+1} \cong \mathbb{F}_8$ and $\mathbb{F}_4 \cong \frac{\mathbb{Z}_2[X]}{X^2+X+1} \cong \mathbb{Z}_2(\beta)$. Could I maybe use this in this exercise?

Which method is correct, or is there another method that I could use?

Answer 1

Well, in view of dimensions

$[F_2(\alpha,\beta)] = [F_2(\alpha,\beta):F_2(\beta)] \cdot [F_2(\beta):F_2]$,

where $F_2[\alpha] = GF(8)$ and $F_2[\beta] = GF(4)$ and both extensions are have trivial intersection: $F_2[\alpha] \cap $F_2[\beta] = F_2$.

Trivial intersection since $GF(p^m)$ is a subfield of $GF(p^n)$ iff $m\mid n$.

The last property shows that

$[F_2(\alpha,\beta)] = [F_2(\alpha):F_2] \cdot [F_2(\beta):F_2]$

This should clarify the situation.