Minimal Polynomial of $\alpha^2$

Question

Having already proved that $p(x)=x^5 + x^2 + 1$ is primitive in $GF(2)$ and assuming that $\alpha$ is a primitive element representing a root of $p(x)$, I am trying to minimal polynomial of $\alpha^2$ and $\alpha^3$. I do not want anyone to solve this for me. Can someone point me to an authoritative source where I can read about this and then solve the problem myself?

Answer 1

One way of finding the minimal polynomial is the following. You can write all the elements of $GF(32)=GF(2)[\alpha]$ as (at most) quartic polynomials in $\alpha$, and try to use that. Let $\beta=\alpha^3$. Then $$ \begin{aligned} \beta^2&=\alpha^6=\\ &=\alpha^3+\alpha,\\ \beta^3&=\alpha^3\beta^2=\alpha^6+\alpha^4=\\ &=\alpha^4+\alpha^3+\alpha,\\ \beta^4&=\beta(\alpha^4+\alpha^3+\alpha)=\alpha^7+\alpha^6+\alpha^4=\\ &=\alpha^3+\alpha^2+\alpha,\\ \beta^5&=\beta(\alpha^3+\alpha^2+\alpha)=\alpha^6+\alpha^5+\alpha^4=\\ &=\alpha^4+\alpha^3+\alpha^2+\alpha+1. \end{aligned} $$ Now finding a polynomial $p(x)=x^5+b_4x^4+b_3x^3+b_2x^2+b_1x+b_0$ with the property $p(\beta)=0$ is equivalent to solving the unknown coefficients $b_0,b_1,b_2,b_3,b_4\in GF(2)$ from the linear system of equations that you get by equating like power of $\alpha$ from the two sides of $$ (\beta^5=)\alpha^4+\alpha^3+\alpha^2+\alpha+1=b_4\beta^4+b_3\beta^3+b_2\beta^2+b_1\beta+b_0 $$ with the aid of the above table of power of $\beta$.