# Minimal Polynomial GF(5)

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

• Do we know the minimal polynomial of $\alpha$? Is it just $p$? – Berci Dec 5 '13 at 2:37
• I believe so. That is correct – user100503 Dec 5 '13 at 2:45
• @Jyrki Lahtonen – user100503 Dec 5 '13 at 16:27

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.
• How the **** would I know? Have I ever watched you calculate over the field of $2^m$ elements? – Gerry Myerson Dec 5 '13 at 3:14
• I read the question. I don't know how you find conjugates, and I don't know how you multiply. If the conjugates of $\alpha$ are $\gamma$ and $\delta$, then the conjugates of $\alpha^2$ are $\gamma^2$ and $\delta^2$. Does that help? – Gerry Myerson Dec 5 '13 at 4:02