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Let $f, g$ be monic polynomials in $\mathbb{Q}[x]$ whose product is a (monic) polynomial in $\mathbb{Z}[x]$. Is it true that both $f,g$ are in $\mathbb{Z}[x]$ and why?

It seems to me that is the case, but I have a trouble proving it. Tried to write contents in $\mathbb{Q}$ but for now I can only show (with a corollary of Gauss' lemma about primitive polynomials) that the product of contents of $f$ and $g$ is 1, i.e. $f = cP$, $g = dQ$ for some rational numbers $c,d$ with $cd = 1$ and primitive (but not necessarily monic) polynomials $P,Q \in \mathbb{Z}[x]$.

Any help appreciated!

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Yes.

Write $c(h)$ for the content of a polynomial in $\Bbb Z[x]$.

Let $m$ and $n$ be positive integers with $mf$, $ng\in\Bbb Z[x]$. I claim that $c(mnfg)=mn$. Certainly, $mn$ divides all coefficients of $(mn)(fg)$ but its leading coefficient is $mn$. Then $mn=c(mf)c(ng)$ (Gauss's lemma). But $c(mf)\mid m$ as its leading coefficient is $m$, and $c(ng)\mid n$. Therefore $c(mf)=m$ and $c(ng)=n$. So $f$, $g\in\Bbb Z[x]$.

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  • $\begingroup$ Why does $mn$ divide all coefficients of $mnfg$? After all, $f,g$ are in $\mathbb{Q}[x]$ so it might be that $mnfg$ has "killed the denominators of coefficients of f,g" but that it has coefficients which are no more divisible by $m$ or $n$, I think? $\endgroup$ – DesmondMiles Jun 14 at 19:13
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    $\begingroup$ $fg\in\Bbb Z[x]$. @DesmondMiles $\endgroup$ – Lord Shark the Unknown Jun 14 at 19:14
  • $\begingroup$ Oops, $fg$ is in $Z$, so we are fine, thanks! $\endgroup$ – DesmondMiles Jun 14 at 19:15

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