$\gcd(X^n - 1, nX^{n-1}) = 1$ in a field $K$ of characteristic $0$ or characteristic $p \nmid n$.

Question

Why is $\gcd(X^n - 1, nX^{n-1}) = 1$ in a field $K$ of characteristic $0$ or characteristic $p \nmid n$.

Most book on abstract algebra don't prove explicitly this, but I wonder what technique I should use to prove this to be formal. Perhaps induction?

Answer 1

Recall that the polynomial ring over a field is a principal ideal domain, hence it has unique factorization.

If $n=1$ the statement is trivial, so suppose $n>1$.

In a field of characteristic zero or $p\nmid n$ the polynomial $nX^{n-1}$ has degree $n-1$ and the only irreducible polynomial that divides $nX^{n-1}$ is $X$. Since $X$ doesn't divide $X^n-1$, we're done.

A different way is to observe that $n$ can be removed, because it's a nonzero constant. Since $$ X^n-1=X\cdot X^{n-1}-1 $$ you're done using the Euclidean algorithm.

By the way, if the characteristic $p$ divides $n$, then $nX^{n-1}=0$ and the greatest common divisor is $X^{n-1}$.

Answer 2

The gcd of polynomials $f$ and $g$ is $1$ if and only if there are polynomials $A$ and $B$ with $Af+Bg=1$. Here, take $A=-1$ and $B=X/n$.