How to construct a finite field modulo $x^5+x^2+1$?

Question

I am trying to construct a finite field modulo $x^5+x^2+1$. As far as I understood its degree is $GF(2^5)$ meaning that I will have only 32 possible equations. But I don't know what to do next. I found that I need to find its roots, but how can I do that?

Answer 1

Construction uses $GF(2^5) = \Bbb Z_2[x]/I = \Bbb Z_2[x]/\langle x^5+x^2+1\rangle$.

$GF(2^5) = \{ax^4+bx^3+cx^2+dx + e\mid a,b,c,d,e\in\Bbb Z_2\}$ where addition is componentwise and multiplication is modulo the ideal/polynomial.

If the polynomial is primitive, take the residue class of $x$ in the quotient ring $\alpha =\bar x = x + I$. Since $\alpha$ is a zero of $x^5+x^2+1$ and so $\alpha^5 + \alpha^2 + 1 =0$. Thus $\alpha^5 = \alpha^2 + 1$ and so you can represent each element of the field as a power $\alpha^k$ of $\alpha$ with $\alpha^{31} = 1$.