Isomorphism between two versions of $GF(2^3)$ I have $GF(2^3)$ generated by $\Pi_1(\alpha)=x^3+x+1$ and $GF(2^3)$ generated by $\Pi_2(\alpha)=x^3+x^2+1$.
$\bullet$ $\Pi_1(\alpha)=x^3+x+1$ $000=0, 100=1,010=\alpha,001=\alpha^2,110=\alpha^3, 011=\alpha^4,111=\alpha^5,101=\alpha^6$
$\bullet$ $\Pi_2(\alpha)=x^3+x^2+1$ $000=0, 100=1,010=\lambda,001=\lambda^2,101=\lambda^3, 111=\lambda^4,110=\lambda^5,011=\lambda^6$
The exampla say that $\alpha$ and $\lambda^3$ both have minimal polynomial $\Pi_1$ and thus $\alpha \iff \lambda^3$ form an isomorphism between the two version. 
How can I see in a practical way this fact? Someone can explain me this concept? 

$\Pi_1(\lambda^3) = (\lambda^3)^3 + \lambda^3 + 1 = \lambda^9 + \lambda^3 + 1$
But you know that $\lambda^7=1$. So $\lambda^9=\lambda^2$ and
$\Pi_1(\lambda^3) = \lambda^3 + \lambda^2 + 1 = \Pi_2(\lambda) = 0$
The complete isomorphism is as follows:
$0 \leftrightarrow 0 \\ 1 \leftrightarrow 1 \\ \alpha \leftrightarrow \lambda^3 \\ \alpha^2 \leftrightarrow \lambda^6 \\ \alpha^3 \leftrightarrow \lambda^2 \\ \alpha^4 \leftrightarrow \lambda^5 \\ \alpha^5 \leftrightarrow \lambda \\ \alpha^6 \leftrightarrow \lambda^4$
