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I'm stuck with a theorem I'm trying to prove, which says that for any $f\in \mathbb{Z}[x]$, and $g,h\in \mathbb{Q}[x]$ s.th. $f=gh$, there is some $\alpha \in \mathbb{Q}$ such that both $\alpha g$ and $\frac{1}{\alpha}h$ are in $\mathbb{Z}[x]$. It seems like just a few lines, and I tried to experiment with some of the functions being primitive, but I am not getting anywhere. What is the trick here?

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Presumably $f = gh$. –  Qiaochu Yuan Nov 21 '11 at 20:33
Google "Gauss's Lemma". –  Bill Dubuque Nov 21 '11 at 20:34
Oh, now it all makes sense. To be clear - it's not true as currently written. For example, taking $g = \frac{1}{5}x$ and $h = \frac{2}{3}x$, you get that both $p > q$ and $q > p$ if $\alpha = \frac{p}{q}$. –  mixedmath Nov 21 '11 at 20:36
yes, sorry, $f=gh$, and this $f\in \mathbb{Z}[x]$ –  Marie. P. Nov 21 '11 at 20:47
well, it is true as currently written. in that example, $gh=\frac{1}{5}x \frac{2}{3}x = \frac{2}{15}x^2\notin \mathbb{Z}[x]$. But thanks for the hint on the name!. –  Marie. P. Nov 22 '11 at 0:05
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