How does one construct the Galois field extension $GF((2^2)^3)$? Looking at past exam question, one asks us to construct a Galois field extension $GF((2^2)^3)$ whenever the primitive irreducible polynomial $p(X) = X^3 + \alpha X^2 + \alpha X + \alpha \in GF(2^2)[X]$ is given. How do you answer this?
 A: Ok. It is easy to check that $p(X)$ is irreducible. Let $\beta$ be a zero of $p(X)$. Then
$$
\begin{aligned}
\beta^3&=\alpha\beta^2+\alpha\beta+\alpha,\\
\beta^4&=\beta^2+\beta+\alpha^2,\\
\beta^5&=\alpha^2\beta^2+\beta+\alpha,\\
\beta^6&=\alpha^2\beta+1.
\end{aligned}
$$
Now we can observe that $\beta^6+\beta^5+\beta^3=\beta^2+1$, so the minimal polynomial of $\beta$ over $GF(2)$ is $m(X)=X^6+X^5+X^3+X^2+1$.
At this point it is easy to cheat, fire up Mathematica, and check that $m(X)$ is not a factor of $x^{21}+1$. The only sextic factor of $X^9+1$ is $X^6+X^3+1$, so we now know that $\beta$ is of order 63, i.e. a primitive element.
Let's start squaring (using what we have calculated already) for a change to verify this
$$
\begin{aligned}
\beta^8&=(\beta^4)^2=\beta^4+\beta^2+\alpha^4&=\beta+1,\\
\beta^{16}&=(\beta^8)^2&=\beta^2+1,\\
\beta^{32}&=(\beta^{16})^2=\beta^4+1&=\beta^2+\beta+\alpha,\\
\beta^{64}&=(\beta^{32})^2=\beta^4+\beta^2+\alpha^2&=\beta.
\end{aligned}
$$
Well, well! There's a fair chance that no errors crept in, given that $\beta^{64}$ checks out.
