Factorization with a Primitive Factor of Polynomials

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.

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