# If a monic polynomial factors over the field of fractions, then the factors must be in the original ring

Let $R$ be a domain with field of fractions $F$. Let $f\in R[x]$ by a monic polynomial and suppose $f = gh$ for $g,h \in F[x]$ monic. I am trying to show that $g$ and $h$ have coefficients in $R$. I tried expanding $f$, $g$, and $h$ as linear combinations of $x^j$ and comparing coefficients, but was not successful. I also tried applying Gauss' lemma, but did not see a way to do so. Any suggestions?

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What is $A{}{}$? – Alex Becker Jan 28 '13 at 23:45
Oh, sorry, yes. $A$ should be $R$. Fixed – user15464 Jan 29 '13 at 0:01
See this proof of Gauss's other lemma. – JSchlather Jan 29 '13 at 1:24

${\bf Counterexample}\,\ \ x^2+x+1 = \left(x- \frac{-1-\sqrt{-3}}2\right) \left(x- \frac{-1+\sqrt{-3}}2\right)\$ with $\: R = \Bbb Z[\sqrt{-3}]$
It is true when $R$ is integrally-closed - look up Dedekind's Prague Theorem or Kronecker's Lemma. This states that the product of any coefficient of $f$ times any coefficient of $g$ is integral over $R$. Therefore if $f$ has a unit coefficient (e.g. is monic) then we deduce that every coefficient of $g$ is integral over $R$ (so $\in R$ if $R$ is integrally closed).