Let $f=X^n-1$ and $g=X^m-1$ be two polynomials. Show that: $$\left(f,g\right)=X^{\left(n,m\right)}-1,$$ where $\left(a,b\right)=$ greatest common divisor of $a$ and $b$.
Asked
Active
Viewed 2,802 times
2
-
And what have you tried? – ajotatxe Feb 17 '18 at 14:21
-
I'm supposed to use the Euclidean Division but I'm getting nowhere. – user528021 Feb 17 '18 at 14:22
-
Try the standard: that is to show that $d=(x,y)$ show that $d$ divides $x$ and $y$ and that every $d'$ that divides $x$ and $y$ also dvides $d$. – ajotatxe Feb 17 '18 at 14:26
-
4Possible duplicate of Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$ and Euclidean algorithm on $(a^n-1,a^m-1)$ – Sil Feb 17 '18 at 14:32
-
Consider their roots of unity. – BAI Feb 17 '18 at 14:46
-
How am I supposed to show that there are $(n,m)$ identic roots? – user528021 Feb 17 '18 at 15:00
-
And will the answer change if we consider GCD over $\mathbb{C} [X] $? – GraduateStudent Dec 10 '19 at 06:07
-
@GraduateStudent The gcd remains same over extension fields, this is in covered in textbooks, just notice that gcd d_L of f, g in larger field has to divide gcd d_S in smaller field as d_S=af+bg. Also, d_L should have degree larger than d_S as d_S is also there in larger field. – Harshavardhan May 01 '21 at 16:51
1 Answers
1
Suppose m and n have a common divisor like k such that $m=m_1k$ and $n=n_1k$ then we can write:
$x^n-1=x^{n_1k}-1=(x^k-1)(x^{n_1k-k}+x^{{n_1k-2k}}+ \ldots+ {x^{n_1k-(n_1-1)k}}+1)$
$x^m-1=x^{m_1k}-1=(x^k-1)(x^{m_1k-k}+x^{{m_1k-2k}}+ \ldots +{x^{m_1k-(m_1-1)k}}+1)$
Which their common divisor is..$(f,g)=(x^k-1)$ or $(f,g)=x^{gcd(m,n)}-1$
Lord Vader
- 961
sirous
- 10,751