# Irreducible Polynomials over Finite Fields

Let f(x), g(x) be irreducible polynomials over the finite field GF(q) with coprime degrees n, m resp. Let $\alpha , \beta$ be roots of f(x), g(x) resp. Then the roots of f(x), g(x), are $\alpha^{q^i}, 0\leq i \leq n-1$, and $\beta^{q^j}, 0\leq j \leq m-1$.

Question: What is the irreducible polynomial over GF(q) of degree nm with roots $\alpha^{q^i}\beta^{q^j}$ where $0\leq i \leq n-1$, and $0\leq j \leq m-1$. Can you define such polynomial explicitly in terms of just f(x) and g(x) without the roots appearing in the formula?

Note: The last sentence/question is what really interests me as the following is the required polynomial (but defined in terms of the roots of f(x))

$$F(x) = \prod_{i=0}^{n-1}\alpha^{mq^i}g\left(\alpha^{-q^i}x\right)$$

Thank you!

Last edited:

Hurkyl
Staff Emeritus
Gold Member
It depends somewhat on just what you mean by "formula" and "appearing in".

I know that F can be expressed as a resultant. The formula you give is presumably comes from one of the methods of computing resultants. There exist other ways to compute resultants, such as as the determinant of a matrix, or a Euclidean algorithm-like method.

You could always find a way to write down the system of equations that literally expresses that $\alpha \beta$ is a root of F(x). Then F would be the solution!

I see. I was expecting some kind of miracle where F(x) could be expressed as some sort of polynomial compositions of f(x), g(x) and their reciprocals, but this matter turns out to be even more complicated than I was hoping. Here's a paper I have just found on this topic.

Theorem 4.2 says it all...

Hurkyl
Staff Emeritus