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Show $$ \gcd(x^n-1,x^m-1)=x^d-1$$

Trying Euclidian Algorithm where $n=mq+r$ $$ x^n-1=x^m-1 q(x)+x^{r}-1$$

Now

$$\begin{aligned} \gcd(x^n-1,x^m-1)&=gcd(x^n-1,x^r-1) \\&= \vdots \\&= x^d-1 \end{aligned}$$

Trying out with basic defitions

GCD polynomials $\in F[x]$

$\forall f,h \in F[x]$ ,where ($f\neq 0 \vee g\neq 0$) $\exists $(unique) $d\in F[x]$

s.t.

  1. $d|f\wedge d|h$

  2. $d$ is monic

  3. $d$ has maximal degree and $d|f\wedge d|h$

Def Euclidean Domain $gcd$

Let $R$ be a Euclidean domain

$\forall a,b \in R$ where at leats one is not zero ,$\exists d \in R$ s.t

  1. $d|a \wedge d |b$
  2. $c|a,c|b \Rightarrow \delta(c) \leq \delta (d)$

Guessing $\delta$ is somekind of norm

Def gcd Integers

$\forall a,b \in Z$ both not zero, $\exists d\in Z$ s.t

  1. $d|a \wedge d|b$
  2. $c|a\wedge c|b \Rightarrow c\leq d$

Using basic tools

does $d|n$ so $n=k_1 d $ ,$\exists k_1 \in Z$ s.t

$$\begin{aligned} x^n-1&=(x^d-1)(x^{k_1}+1)=x^{d+k_1}-1 \end{aligned} $$

More tools

Thm 1.2 Hungerford $\forall a,b \in Z$ (not both 0) \exists u,v\in Z (not necassary unique)$

s.t. $$ d=nu+mv$$

Thm for polynomials

$\forall f(x),h(x) \in F[x] , \exists u(x),v(x) \in F[x]$

s.t. $$ d(x)=f(x)u(x)+g(x)v(x)$$

Trying an example to see somekind of a pattern and then try to work an argument in genal

considering $x^6-1$ and $x^2-1$

$$ x^6-1=(x^2-1)(x^4+x^2+x)+0$$

the $gcd(6,2)=2$ so the gcd in polynomials is

$ x^6-1=(x^2-1)(x^4+x^2+1)+0$

Tiger Blood
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2 Answers2

2

You can use induction on $\max\{m,n\}$. Let $m\ge n$. If $m=1$ the requested is obvious. Let's assume that it is true for values $<m$. The requested is also obvious when $m=n$, so we can assume that $m>n$. Then $m=nq+r$ with $q,r\in\mathbb{Z_{>0}}$ and $0\leq r<n$. Then we know that gcd$(n,r)$=gcd$(m,n)=d$. Now we have: $$x^m-1=x^{nq+r}-1=x^rx^{nq}-x^r+x^r-1=x^r(x^{nq}-1)+x^r-1=x^rq(x)(x^n-1)+x^r-1.$$ So $\gcd(x^m-1,x^n-1)=\gcd(x^n-1,x^r-1)=x^{\gcd(n,r)}-1=x^d-1$.

user26857
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1123581321
  • 5,108
1

We have: $$x^m-1=\prod_{k=1}^{m}\left(x-e^{\frac{\pi i k}m}\right)$$ $$x^n-1=\prod_{k=1}^{n}\left(x-e^{\frac{\pi i k}n}\right)$$ Since all those factors are prime in $\mathbb{C}[X]$ and $\mathbb{C}[X]$ is a UFD, $\gcd(x^m-1,x^n-1)$ is equal to the product of the common factors, aka factors $$\left(x-e^{\frac{i\pi k}m}\right)$$ for which there is an $1\le l\le n$ with: $$e^{\frac{\pi i l}n}=e^{\frac{\pi i k}m}\implies \frac ln=\frac km$$ looking at those fractions in their lowest terms, both must have equal denominators and that denominator must divide both $m$ and $n$. So if we let $d=\gcd(n,m)$, then that fraction must be $j/d$ for some $j$. Therefore: $$\gcd(x^m-1,x^n-1)\mid\prod_{k=1}^{d}\left(x-e^{\frac{\pi i k}d}\right)=x^d-1$$ Now, write $n=dn'$ and $m=dm'$. Since $$x^n-1=(x^d-1)\sum_{k=0}^{n'}x^k$$ $$x^m-1=(x^d-1)\sum_{k=0}^{m'}x^k$$ it follows that $x^d-1\mid \gcd(x^n-1,x^m-1)$ and we conclude: $$\gcd(x^n-1,x^m-1)=x^d-1$$

Mastrem
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