Prove that $\gcd(f(x),g(x)) = 1$.

Question

Let $F$ be a field, let $p , q$ be two coprime natural numbers and we consider the two polynomials $$ f(X) = \sum_{i = 0}^{p - 1} X^{k i} \qquad \mbox{ and } \qquad g(X) = \sum_{j = 0}^{q - 1} X^{k j} $$ in $F[X]$, where $k \in \mathbb{N}$. If $x$ is an arbitrary element in $F$, how can I show that $\gcd(f(x) , g(x)) = 1$?

Answer 1

Hint (for the polynomial $\gcd$): $\;X^{kp} = \left(X^{k}-1\right)f(X)+1\,$ and $\,X^{kq} = \left(X^{k}-1\right)g(X)+1\,$.

Since $\,p,q\,$ are coprime, there exist integers $\,a,b\,$ such that $\,ap-bq=1\,$, and it can be assumed WLOG that they are positive i.e. $\,a,b \in \Bbb N\,$. Then $\,X^{kap}=X^{k(bq+1)} = X^k \cdot X^{kbq}\,$, and therefore:

$$ \left(\left(X^{k}-1\right)f(X)+1\right)^{a} - X^k \cdot \left( \left(X^{k}-1\right)g(X)+1\right)^b = 0 $$

Expanding the binomials, $\,h(X) = \gcd\left(f(X),g(X)\right)\,$ would have to divide $\,1-X^k\,$, but both $\,f(X)\,$ and $\,g(X)\,$ are coprime with $\,1-X^k\,$.

Answer 2

Perhaps this will help:

$$ 1+x+x^2+...+x^{n-1} = {x^n-1\over x-1}$$

and a fact that $$\gcd(x^n-1,x^m-1) = x^{\gcd(m,n)}-1$$