# Finite fields of order 8 and isomorphism [duplicate]

Given the following finite fields, $$F_1=\frac{\mathbb{Z}_2[x]}{\langle x^3+x+1\rangle}$$ and $$F_2=\frac{\mathbb{Z}_2[x]}{\langle x^3+x^2+1\rangle},$$ I know that they are isomorphic because their orders are both equal $8$. How can I find an isomorphism between them?

We can write $$F_1=\left\{c_0+c_1 \alpha + c_2 \alpha^2 : \alpha^3=-\alpha-1,\; c_0,c_1,c_2 \in \mathbb{Z}_2 \right\}$$ and $$F_2=\left\{c_0+c_1 \alpha + c_2 \alpha^2 : \alpha^3=-\alpha^2-1,\; c_0,c_1,c_2 \in \mathbb{Z}_2 \right\},$$ still I have no idea how to go ahead. Thanks for your time!

• Proper notation is $\displaystyle \frac{\mathbb{Z}_2[x]}{\langle x^3+x+1\rangle},$ not $\displaystyle \frac{\mathbb{Z}_2[x]}{<x^3+x+1>}.$ I edited accordingly. If you don't know things like this, then google "latex symbols" and you'll find it. – Michael Hardy Jan 17 '18 at 18:39
• Okay, thank you. The ideal generated by $x^3+x+1$ is often also written as $(x^3+x+1)$. Is that correct? – Nicola M. Jan 17 '18 at 19:05
• I believe I've seen that notation. $\qquad$ – Michael Hardy Jan 17 '18 at 19:15

If $f(x) = \sum_{i=0}^d f_i x^i$ is irreducible over a field $F$, then its reversal $g(x) = \sum_{i=0}^d f_{d-i} x^i$ is also irreducible and defines an isomorphic field.

In your case, $x^3+x+1$ and $x^3+x^2+1$ are reversals.

To see this, let $\alpha$ be a root of $f$, so $f(\alpha)=0$. Let $\beta = \tfrac1\alpha$, then $$0=f(\alpha)=f(\tfrac 1 \beta) = \sum_{i=0}^d f_i \beta^{-i} = \beta^{-d} \sum_{i=0}^d f_i \beta^{d-i} = \beta^{-d} g(\beta)$$ and so $\beta$ is a root of $g$.

• But note that this takes advantage of the special shape of your polynomials. If they didn't have this property, you would have to show there is an isomorphism more directly. – Doris Jan 17 '18 at 18:28
• Right you are, @Doris. In my experience, it may require special techniques or arguments to find an isomorphism between two fields of the same cardinality (but experts may have techniques that I don’t know). – Lubin Jan 17 '18 at 19:16

The zeros of $x^3+x^2+1$ are the reciprocals of the zeros of $x^3+x+1$.

Better: write $F_2=\{c_0+c_1\beta+c^2\beta^2:\cdots\}$. We can map $\alpha\in F_1$ to $1/\beta=\beta^2+\beta\in F_2$.

• Wow! Really thanks. It works. Does it work in general for every finite group? – Nicola M. Jan 17 '18 at 18:22