TITLE: On Greatest Common Divisors
QUESTION [8 upvotes]: Let $F$ be a field and $r$ and $s$ positive integers. Prove that, in $F[x]$,
$\gcd(x^r-1,x^s-1)=x^{\gcd(r,s)}-1$.
If $r$ and $s$ were known numbers, I could be able to attempt the problem, but I don't know how to do about this.

REPLY [2 votes]: Lemma. Let $I$ be any ideal of $F[X]$, and let $H$ be the set of $n \in \mathbb{N}$ such that $X^n - 1 \in I$. Then $H$ is the intersection with $\mathbb{N}$ of a subgroup of $\mathbb{Z}$.
Proof of Lemma. Let $A = F[X]/I$, and let $a = \bar{X} \in A$. If $a$ is not invertible in $A$, then $H = \{0\}$. Otherwise, $H$ is the intersection with $\mathbb{N}$ of the kernel of the morphism $n \mapsto a^n$ from $\mathbb{Z}$ to $A^{*}$.
Applying the lemma to the ideal $(X^r-1,X^s-1)$, we find (by Bezout's lemma) that $X^{\gcd(r,s)} - 1 \in (X^r-1,X^s-1)$. Conversely, applying the lemma to $(X^{\gcd(r,s)} - 1)$, we see that $X^r - 1, X^s - 1 \in (X^{\gcd(r,s)} - 1)$. Thus $(X^r-1,X^s-1) = (X^{\gcd(r,s)} - 1)$.