Finding an isomorphism between these two finite fields 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? 
 A: You can notice that if $\alpha^3+\alpha+1=0$, then
$$
\alpha^3\left(1+\frac{1}{\alpha^2}+\frac{1}{\alpha^3}\right)=0
$$
and so $\beta=\alpha^{-1}$ is a root of $x^3+x^2+1$. And conversely.
Thus, if you want an isomorphism $\mathbb{Z}_2(\alpha)\to\mathbb{Z}_2(\beta)$ you just send $\alpha$ to $\beta^{-1}$.
A: 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. 
