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Let A,B in $\mathbb{Q}[X]$,A and B are monic polynomials. suppose P=AB $\in$ $\mathbb{Z}[X]$. Prove that A, B $\in\mathbb{Z}[X]$.

What I did : let $a,b \in \mathbb{Z}$ such as $a = min\left\{ k \in \mathbb{Z}, aA \in \mathbb{Z}\right\}$ and $b = min\left\{ k \in \mathbb{Z}, bB \in \mathbb{Z}\right\}$ ( or a = LCM of the denominators of the coefficients of A, same for b).

Then : $abP = aAbB$.

Here, if I have to show that $aA$ and $bB$ are "primitives" (that means $cont(aA)= 1$ and $cont(bB)=1$), then I can end my proof, because that involve a=b=1.

How to show that?

(Notations : if $Q=\sum_{k=0}^n{q_k}X^k$, the content of Q is $cont(Q)=GCD(a_1,a2,...,a_n)$)

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    I guess you want $a = \min\left\{ k \in \mathbb{Z}: k > 0, k A \in \mathbb{Z}[X] \right\}$, and similarly for $b$. – 2017-02-14

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If all coefficients of $a A$ are divisible by the prime $p$, say, then $p$ divides $a$, as $A$ is monic, so that $a$ is the leading coefficient of $a A$.

Hence $$\dfrac{a}{p} A = \dfrac{a A}{p}$$ has integers coefficients, contradicting the definition of $a$.

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    So, $a$ isn't divisible by any prime, so $a=1$ ! I think it's right ! Thank you very much ! – 2017-02-14
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    I come back here because I have a problem to understand something in that proof... Where do you use the hypothesis "A and B are monic"? because, without it it's false (for example $\frac{5}{6}X + \frac{5}{6}$).. I know, you used it to say that "a" is the leading coefficient of aA, but why is it contradictory? – 2017-02-14
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    $A$ is monic, so the leading coefficient of $a A$ is $a$, so if $p$ divides all coefficients of $a A$, then $p$ divides $a$. – 2017-02-14
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    You can also rephrase the proof using your function $cont(P)$ as follows : $cont(AB)\in \mathbb{Z}$ but $AB$ is monic so $cont(A)cont(B)=cont(AB)=1$ then one of the two contents is in $\mathbb{Z}$, say $cont(A)$. Now $A$ is monic so $cont(A)=1$ and then $cont(B)=1$ which proves the claim. – 2018-01-02