Integral rational over $ \mathbb Z$ is an integer

Question

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.

Answer 1

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.

Answer 2

Gauss’s lemma for primitive polynomials states that if two polynomials are primitive (which means that the GCD of all coefficients is $1$) then the product is primitive too. Suppose $f$ is an integral polynomial and $p/q$ a rational root. Then $f$ is divisible by $x-p/q$ or $qx - p$. Assuming $(p,q)=1$ this is a primitive polynomial. Thus $$ f = (qx-p) g $$ If $g$ is not integral then there is a $c\in\mathbb Z$ so that $cg$ is integral and primitive. In fact if one $k$ divides all coefficients of $cg$ then it also divides all coefficients of $cf$, so it must divide $c$. So we can choose $c$ minimal so that $cg\in\mathbb Z[X]$. But then $cf$ must be primitive. As $f\in\mathbb Z[X]$ this implies $c=\pm1$. But this means $g\in\mathbb Z[X]$.

But $LC(f) = qLC(g)$ this implies that $f$ is monic exactly if $q=1$ and $LC(g) = 1$ (assuming $q>0$).