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

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?

• Well written question, +1 – Teresa Lisbon Jan 28 at 14:00
• One way to check irreducibility of $X^2+X+1$ over $\Bbb F_2(\alpha)\cong\Bbb F_8$ is to check that this polynomial has no root in $\Bbb F_8$. Its roots are cube roots of unity, i.e. of period $3$, whereas the elements of $\Bbb F_8$ all have period $7$ or $1$. – Lubin Jan 28 at 23:28
• Oh wow, thank you! – fieke_2000 Jan 30 at 14:08

$$[F_2(\alpha,\beta)] = [F_2(\alpha,\beta):F_2(\beta)] \cdot [F_2(\beta):F_2]$$,
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. • Thank you! I am still a bit confused. Why is$\mathbb{F}_2$a trivial intersection? And if it is trivial, does that mean that$[\mathbb{F}_2(\alpha,\beta): \mathbb{F}_2] = [\mathbb{F}_2(\alpha): \mathbb{F}_2] \cdot [\mathbb{F}_2(\beta): \mathbb{F}_2]\$ or how does this information help me? – fieke_2000 Jan 28 at 14:04