On Greatest Common Divisors

Question

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.

Answer 1

We can assume $r \geq s \geq 1$. Look at

$$x^r-1 - x^{r-s}(x^s-1)=x^{r-s}-1.$$

Hence $\operatorname{gcd}(x^r-1,x^s-1)=\operatorname{gcd}(x^{r-s}-1,x^s-1)$.

By induction on $r+s$ we can use the assertion for the right hand side and thus deduce

$$\operatorname{gcd}(x^r-1,x^s-1)=\operatorname{gcd}(x^{r-s}-1,x^s-1)=x^{\operatorname{gcd}(r-s,s)}-1=x^{\operatorname{gcd}(r,s)}-1.$$

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Some notes on that proof: Of course the induction makes this look very graceful and you might not see why this works. Then have a look at $$x^r-1 - x^{r-s}(x^s-1)=x^{r-s}-1$$ again. This computation actually shows you that polynomial division of polynomials of the type $x^r-1$ is nothing but the division with remainders applied to the exponents. Since the greatest common divisor is calculated by iterated use of division with remainder, this gives you the result. The induction is just the technial tool to make this explaination rigid.

Answer 2

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 $(X^r-1,X^s-1) = (X^{\gcd(r,s)} - 1)$.