Let $f(T)=\sum_{i=0}^n{}a_iT^i\in\mathbb{Q}[T]$ be a polynomial of degree $n\geq 1$ with all coefficients $a_i$ integers. Show that if $\frac{x}{y}$, with $x, y$ coprime integers, is a root of $f(T)$, then $x$ divides $a_i$ and $y$ divides $a_n$.
What I've done so far
$f(\frac{x}{y})=0$ with $\gcd(x,y)=1$, then $x/a_1$ and $y/a_n$.
$f(\frac{x}{y})=\sum_{i=0}^{n}a_i(\frac{x}{y})^i=a_i\frac{x^0}{y^0}+a_i\frac{x^1}{y^1}+\dots+a_{n-1}\frac{x^{n-1}}{y^{n-1}}+a_{n}\frac{x^n}{y^n}=0$
With $1=dx+by$, and $d,b \in \mathbb{R}$, $y=\frac{1-dx}{b}$.
Thus, $$f\left(\frac{x}{y}\right)=f\left(\frac{x}{(1-dx)/y}\right)=f\left(\frac{bx}{1-dx}\right)=\sum_{i=0}^{n} a_i\left(\frac{bx}{1-dx}\right)^i=$$ $$=a_i\frac{bx^0}{(1-dx)^0}+a_i\frac{bx^1}{(1-dx)^1}+... +a_{n-1}\frac{bx^{n-1}}{(1-dx)^{n-1}}+a_{n}\frac{bx^n}{(1-dx)^n}=0$$
I'm not sure how to prove that $x$ divides $a_i$ and $y$ divides $a_n$.
Thanks for reading this.
