# Around Gauss' Lemma

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!

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 $$f$$, $$g\in\Bbb Z[x]$$.
• 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? – DesmondMiles Jun 14 at 19:13
• $fg\in\Bbb Z[x]$. @DesmondMiles – Lord Shark the Unknown Jun 14 at 19:14
• Oops, $fg$ is in $Z$, so we are fine, thanks! – DesmondMiles Jun 14 at 19:15