Let $\alpha$ be a root of $x^3 + x + 1 \in \mathbb{Z}_2[x]$ and $\beta$ a root of $x^3 + x^2 + 1 \in \mathbb{Z}_2[x]$. Then we know that $$\mathbb{Z}_2(\alpha) \simeq \frac{\mathbb{Z}_2[x]}{(x^3 + x + 1)} \simeq \mathbb{F}_8 \simeq \frac{\mathbb{Z}_2[x]}{(x^3 + x^2 + 1)} \simeq \mathbb{Z}_2(\beta). $$
I need to find an explicit isomorphism $\mathbb{Z}_2(\alpha) \to \mathbb{Z}_2(\beta). $
I was thinking of finding a basis for $\mathbb{Z}_2(\alpha)$ and $\mathbb{Z}_2(\beta)$ over $\mathbb{Z}_2$.
I let $\left\{1, \alpha, \alpha^2\right\}$ and $\left\{1, \beta, \beta^2\right\}$ be these two bases. Now suppose I have a field morphism $$ \phi: \mathbb{Z}_2(\alpha) \to \mathbb{Z}_2(\beta) $$ which maps $1$ to $1$. how can I show that the image of $\alpha$, i.e. $\phi(\alpha)$, completely determines this map?
An important thing to notice in how we send the base is to notice that $$\alpha^3+\alpha=1$$ and $$\beta^3+\beta^2=1$$ and as such we have to have that the image of the former will equal the latter. Using binomial theorem we get that $$(\beta+1)^3=\beta^3+3\beta^2+3\beta+1=\beta^3+\beta^2+\beta+1$$ and then we get that $$(\beta+1)^3+(\beta+1)=\beta^3+\beta^2=1$$ so we see that $\alpha\mapsto\beta+1$ will sate our requirements. This is important as the multiplication ins both basis is slightly different by the letters and such but fundamentally follow the same pattern. A quick check will show this mapping gives us the entire basis $$\alpha\mapsto\beta+1$$ $$\alpha^2\mapsto(\beta+1)^2=\beta^2+1$$ $$\alpha^3\mapsto(\beta+1)^3=\beta^3+\beta^2+\beta+1$$ from this we see we can acquire the entire basis of $\Bbb Z_2[\beta]$ through these mappings.