# Polynomial rings, division algorithm

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 $x^n -1|x^m - 1$?

• 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

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.