I want to prove that if a rational $y\in \mathbb Q$ is integral over $\mathbb Z$, then it is an integer. We say that $y$ is integral over $\mathbb Z$ if there exists a monic polynomial $F\in \mathbb Z[x]$ such that $F(y)=0$.
I tried to write the integral relation as $$y^n+b_1y^{n-1}+\cdots +b_{n-1}y+b_n=0.$$
Next I write $y=p/q$, where $p$ and $q$ are coprime integers, I factor them into prime factors
\begin{align}
p&=p_1^{e_1}\cdots p_r^{e_r}; & q&=q_1^{f_1}\cdots q_s^{f_s}.
\end{align}
By multiplying the integral relation, I can conclude that $p\mid b_1,\cdots,p^n\mid b_n.$ I have no idea of how to continue.
Best Answer
Since $y=\frac pq$,$$y^n+b_1y^{n-1}+\cdots+b_{n-1}y+b_n=0\iff p^n+b_1qp^{n-1}+\cdots+b_{n-1}q^{n-1}p+b_nq^n=0.$$Therefore, since $q$ divides each of the numbers $b_1qp^{n-1},\ldots,b_{n-1}q^{n-1}p,b_nq^n$, $q$ divides $p^n$. Since $p$ and $q$ are coprime, this means that $q=\pm 1$. Therefore, $y$ is an integer.