The role of coefficients in dividable a polynomial

Question

Let $m$ and $p$, be two positive integer numbers. Consider the following polynomial $$ f(x)=x^p-u_{p-1}\,x^{p-1}-u_{p-2}\, x^{p-2}-\cdots-u_1\, x-u_0 $$ Suppose the coefficient $u_0$, is relatively prime to $m$. Consider $a$ and $b$, be two positive integer numbers such that $a<b$. My question it is, If the polynomial $f(x)$, divides the polynomials $x^a(x^b-1)$ over modulo $m$, why we can conclude that the polynomial $f(x)$, divides the polynomial $(x^b-1)$ over modulo $m$ and not the polynomial $x^a$. In other words, I want to say $$ f(x)\mid x^a(x^b-1) \mod{m} \quad \Rightarrow \quad f(x)\mid (x^b-1) \mod{m} \quad \& \quad f(x)\nmid x^a \mod{m} $$ I would greatly appreciate for any suggestions

Answer 1

This follows immediately by Euclid's Lemma, i.e.

$$ \color{#c00}{(f,x)= 1},\ f\mid x^ag\,\Rightarrow\, f\mid x^a g, fg\,\Rightarrow\, f\mid (x^ag,fg) = (\color{#c00}{x^a,f})g = g$$

In your case $\,(x,f) = (x,u_0) = (1)\,$ since $\,(u_0) = 1\, $ in $ \,\Bbb Z_m[x]\,$

Answer 2

We proof by contradiction or reductio ad absurdum method. Suppose that $f(x)$ dose not divide $(x^b-1)$, which results that $f(x)$ should divides $x^a$ over mod $m$. So, we conclude that there is a polynomial of degree $a-p$, like $v(x)$, in the following form

$$ v(x) \, b(x)=x^a \qquad \mod{m} $$ $$ (x^{a-p}-v_{a-p-1}\,x^{a-p-1}-\cdots-v_1\, x-v_0) (x^p-u_{p-1}\,x^{p-1}-u_{p-2}\, x^{p-2}-\cdots-u_1\, x-u_0)=x^a \, \mod{m} $$ From the above equation, we conclude that $ v_0\, u_0\mid m$. In addition, by assumption $(u_0,m)=1$, which results that $v_0=0$ over modulo $m$. So, we have $$ (x^{a-p-1}-v_{a-p-1}\,x^{a-p-2}-\cdots-v_2\, x-v_1) (x^p-u_{p-1}\,x^{p-1}-u_{p-2}\, x^{p-2}-\cdots-u_1\, x-u_0)=x^{a-1} \, \mod{m} $$ But it is not true that $f(x)$ divides two consecutive power of $x$, except when $f(x)$ be power of $x$ and it is contradiction to assumption that $a_0$ is coprome to modulo $m$. Is it true discussion or not. Tanks again