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Question: Let $f,g\in\Bbb Q[x]$. Why is it that

$\rm\color{#c00}{(1)}$ if $f$ is monic then $f=\frac{1}{a}f^*$ for some primitive polynomial $f^*\in\Bbb Z[x]$ and $a\in\Bbb Z$ ?

$\rm\color{#c00}{(2)}$ in general $g=\frac{b}{c}g^*$ for some primitive polynomial $g^*\in\Bbb Z[x]$ and $b,c\in\Bbb Z$ such that $\gcd(b,c)=1$ ?

Note: A primitive polynomial is a polynomial such that the $\gcd$ of its coefficients is $1$.

Thoughts: I'm sure there must be an easy argument.

I think we can assume the coefficients are of the form $\frac{x}{y}$ with $\gcd(x,y)=1$.

For $\rm\color{#c00}{(1)}$, for example, I thought about multiplying $f$ or $g$ by the lowest common multiple of the denominators of the coefficients, call it $m$, but I'm unsure about the primitivity of $f^*:=mf$ even though I've found no counter-example.

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As you suggested, let $m$ be the least common multiple of the denominators of the coefficients of $f$, and $f^* = mf$, so that $f^*$ has integer coefficients. Now let $d$ be the greatest common divisor of the coefficients of $f^*$, and set $df'=f^*$ where $f'$ is primitive, so you have $df' = mf$, and $f = \frac{d}{m} f'$. This shows $(2)$.

Now suppose $df'=mf$ as above and $f$ is monic. Then $f'= \frac{m}{d} f$, so $\frac{m}{d}$ is the leading coefficient of $f'$. As that must be an integer, $d | m$, so $m=dd'$ for some integer $d'$. Then $f = \frac{1}{d'}f'$, which shows $(1)$.

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  • $\begingroup$ Where do you use the fact that $f$ is monic when you prove $\rm\color{#c00}{(1)}$ ? $\endgroup$
    – Guest
    Dec 16 '14 at 5:12
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    $\begingroup$ Oh ok when you say $\frac{m}{d}$ is the leading coefficient. Good stuff. $\endgroup$
    – Guest
    Dec 16 '14 at 5:13

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