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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.

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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.

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