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?

  • 2
    $\begingroup$ Well written question, +1 $\endgroup$ – Teresa Lisbon Jan 28 at 14:00
  • 1
    $\begingroup$ 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 $1$. $\endgroup$ – Lubin Jan 28 at 23:28
  • $\begingroup$ Oh wow, thank you! $\endgroup$ – fieke_2000 Jan 30 at 14:08

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.

  • 1
    $\begingroup$ 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? $\endgroup$ – fieke_2000 Jan 28 at 14:04
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    $\begingroup$ Added comment about subfields. $\endgroup$ – Wuestenfux Jan 28 at 14:11
  • $\begingroup$ Ohhhh thank you so much! $\endgroup$ – fieke_2000 Jan 28 at 14:14

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