2

$x^3 + x^2 + 1$ and $x^3 + x + 1$ are both irreducible over $\mathbb{F}_2[x]$, so then we have isomorphism of fields:

$$ \mathbb{F}_2[x]/(x^3 + x^2 + 1) \simeq \mathbb{F}_2[t]/(t^3 + t + 1) \simeq \mathbb{F}_8 $$

Then do we have that $x \mapsto t$ is an isomorphism?

cactus314
  • 24,438
  • 6
    No. You have $t^3 + t + 1 = 0$ but $x^3 + x + 1 \ne 0$. – David Jan 19 '16 at 18:01
  • 1
    On the other hand, notice that $u := [x]^{-1}$ satisfies $0 = [x]^3 + [x]^2 + 1 = [x]^3 (1 + u + u^3)$ and so $1 + u + u^3 = 0$, suggesting a candidate map between the two fields. – Travis Willse Jan 19 '16 at 18:10

0 Answers0