How does one construct the Galois field extension $GF((2^2)^3)$?

Question

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?

Answer 1

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.