Show that if $D$ is integral domain and $b(x) = b_0 + b_1x + ... + b_mx^m$ and $b_m \ne 0$ satisfies for all $a(x) \in D[x]$, $q(x)$ and $r(x)$ $\in D[x]$ exist such that $a(x) = b(x)q(x) + r(x)$ with $r(x) = 0$ or $\deg(r(x)) < \deg(b(x))$; $\Rightarrow b_m$ is unit.
I'm not sure how to begin, in case $r(x)=0$ then $b(x)\mid a(x)$ but I think this path doesn´t help me to prove $b_m$ is a unit. Is there any theorem I can use? Can you provide me with tips?
Let $a(x) = x^m$. If $q(x)$ has degree greater than $0$, $q(x)b(x)$ will have degree greater than $m$ (because the degree of a product is the sum of the degrees of the factors when working over an integral domain), so since $r(x)$ has degree less than $m$, our equation cannot hold. Therefore we must have $q(x) = c$ for some $c \in D$. Then, since $r(x)$ has degree less than $m$, the leading term of $q(x)b(x) + r(x)$ will be $cb_m x^m$. Since this must actually equal $a(x)$, we get $cb_m = 1$.