Question about a proof in Cohn's Algebraic Numbers and Algebraic Functions

Question

On page 85 of Paul Cohn's Algebraic Numbers and Algebraic Functions, Cohn is showing that the minimal polynomial of an algebraic number has integer coefficients.

He says, let $\alpha$ have minimal polynomial $g$ and suppose $\alpha$ satisfies $f=0$, where $f$ is monic with integer coefficients. Since $g$ is the minimal polynomial for $\alpha$, $f=gh$ for some $h$. Choose integers $a$ and $b$ such that $ag$, $bh$ are primitive with integer coefficients...and so on.

My question: How exactly does one know that such $a$ and $b$ exist? I think you could take $a$ and $b$ to be the least common multiples of all the denominators in $g$ and $h$, respectively, so $ag$ and $bh$ would have integer coefficients, but I don't feel sure they are primitive.

Answer 1

I think this is on the right track.

Let $a$ be the least common multiple of the denominators of the coefficients of $g$. If $p$ is a prime which is not a divisor of any denominator, then $p\nmid a$. Then $p$ does not divide the leading coefficient of $ag$, since $g$ is monic. If $p$ divides some denominator, then let $p^j$ be the greatest power of $p$ in any of the denominators. Thus the power of $p$ in $a$ is $j$. However, upon multiplying through by $a$, the term with $p^j$ in the original denominator now has no factor of $p$ in its coefficient. Thus $p$ does not divide that coefficient of $ag$. So $ag$ has integer coefficients, and $ag$ is primitive since no prime divides all the coefficients.

Since $f$ is monic, and $g$ is monic, $h$ must be monic as well, so the same argument applies to $h$. So such a $b$ also exists.