To calculate minimal polynomials in GF(2^m), I find conjugates and multiply. I am confused about minimal poly in a different GF. p(x)=x^3+x^2+1 over GF(5). It is irreducible over GF(5) and generates GF(5^3). I have proved that it is primitive (124 factors into 2, 4, 31). If alpha is generator, how do I find minimal polynomial of alpha^2. All theorems that I have used so far are exclusively for GF(2^m). How do they generalize? Thanks
Let $\beta=\alpha^2$. Express each of $1,\beta,\beta^2,\beta^3$ in the form $a+b\alpha+c\alpha^2$. Use linear algebra to find a linear dependence relation for $1,\beta,\beta^2,\beta^3$ over the field of 5 elements.
EDIT: Here's a slicker way (though I'm not sure it's similar to whatever method it is you use over the field of $2^m$ elements).
Calculate $$q(x)=p(x)p(-x)=(x^3+x^2+1)(-x^3+x^2+1)=-x^6+x^4+2x^2+1$$ All the exponents are even, so $q$ is a polynomial in $x^2$. Let $$r(x)=x^3-x^2-2x-1$$ so $r(x^2)=-q(x)$. Then $$r(\alpha^2)=-q(\alpha)=-p(\alpha)p(-\alpha)=0$$ so $r(x)$ is the polynomial you are looking for.