Polynomial modulo a monomial

Question

in an article I was reading it was mentioned that for some polynomial $p(x)$ in $\mathcal{Z}[x]$, the operation $p(x) \mod (x-r) = p(r)$, that is to compute the remainder of a polynomial modulo some monomial, you simply evaluate the polynomial with the constant term of the monomial. Can someone provide a proof of why this is true?

Answer 1

First of all, notice that, as TheSilverDoe commented, we have an euclidean division on $\mathbb{Z}[x]$, i.e., we for any pair of polyonomials $Q,P\in\mathbb{Z}[x]$, there exist polynomials $A,B\in\mathbb{Z}[x]$ such that $P(x)=A(x)Q(x)+B(x)$, with the degree of $B(x)$ stricly less than the degree of $Q(x)$.

Now, the definition of congruence $mod\,g(x)$ is given by $p(x)\equiv q(x)\,\,mod\,g(x)$ if and only if $p(x)-q(x)=f(x)g(x)$ for some polynomial $f(x)$. In our case, where $g(x)=(x-r)$, we have that $p(x)-q(x)=f(x)(x-r)$. So if we look at the euclidean division we got, dividing $p(x)$ by $(x-r)$, we get $p(x)=f(x)(x-r)+k,\,\,k\in\mathbb{Z}$, as the degree of the reminder is stricly less than 1, i.e., it's a zero degree polynomial, i.e, an integer. So we can see that by definition, $p(x)$ is congruent to an integer $k$, by seeing $p(x)-k=f(x)(x-r)$.

Now let's calculate $k$. Notice that if we plug $x=r$ into the expression $p(x)=f(x)(x-r)+k$, we obtain $p(r)=f(r)\cdot 0+k$. Therefore, $p(r)=k$, and we obtain $p(x)\equiv p(r)\,\, mod\,(x-r)$, as desired.