Let $m,n$ be non-negative integers and $m>n$. Find polynomials $g(x),r(x)$ from the ring $R[x]$ such that $x^m -1 =q(x)(x^n-1) + r(x)$ , $r(x)=0$ or $\deg(r(x))<n$. In which case $x^n -1|x^m - 1$?
Asked
Active
Viewed 276 times
1
-
as an answer to this question i suggested a reference to "Division $Algorithm$. though i think Bill Dubuque has a complete answer(+1), but maybe a reference can help you too. – user 1 Jan 25 '15 at 17:27
1 Answers
1
Hint $\rm\ mod\,\ x^{\Large n}\!-1\!:\ x^{\Large n}\equiv 1,\ \ so\ \ x^{\Large m}\!-1 \equiv x^{\Large m\ mod\ n}\!-1 \equiv 0 \!\iff\! m\ mod\ n = 0 \!\iff\! n\mid m$
Remark $\ $ One can go further. The polynomial sequence $\rm\ f_n = (x^n-1)/(x-1),\, $ jut like the Fibonacci sequence, is a strong divisibility sequence, i.e. $\rm\: (f_m,f_n)\: =\: f_{\:(m,n)}.\,$ The proof is simple - essentially the same as the proof of the Bezout identity for integers - see my post here. We can view the polynomial Bezout identity as a q-analog of the integer Bezout identity, e.g. let's compare the Bezout identity for the gcd $\rm\ \color{#c00}3 = (\color{#0a0}{15},\color{blue}{21})\ $ in polynomial and integer form:
$$\rm\displaystyle \color{#c00}{\frac{x^3-1}{x-1}}\ =\ (x^{15} + x^9 + 1)\ \color{#0a0}{\frac{x^{15}-1}{x-1}}\ -\ (x^9+x^3)\ \color{blue}{\frac{x^{21}-1}{x-1}}$$
for $\rm\ x = 1\ $ this specializes to $\ \color{#c00}3\ =\ (3)\ \color{#0a0}{15}\ -\ (2)\ \color{blue}{21},\, $ the integer Bezout identity for the gcd.
It is well-worth studying these binomial divisibility properties since they occur quite frequently in number theoretical applications. Moreover, they provide excellent motivation for the more general study of divisibility theory, $ $ esp. in divisor theory form. For an introduction see Borovich and Shafarevich: Number Theory.
Bill Dubuque
- 272,048