Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I seem to have come up with a proof of the following statement, but I have not managed to find this statement, either as a proposition or as an exercise in some of the Abstract Algebra books I have looked at. So is this true (?):

Let $R$ be a UFD, $F$ its field of fractions, and $f \in R[x]$ a non-constant irreducible polynomial. If $g \in R[x]$ is a polynomial such that $f|g$ in $F[x]$, then $f|g$ in $R[x]$.

share|improve this question
en.wikipedia.org/wiki/… –  Qiaochu Yuan Oct 21 '11 at 17:59
To find a proof in a book, look for a book that proves the following: If R is a UFD, then R[x] is a UFD. (Lots of abstract algebra books do). The standard proof of this is via Gauss's Lemma, which would be that if $f\in R[x]$ factors over $F[x]$, then it factors over $R[x]$ with polynomials that are constant multiples (over $F$) of the factors in $F[x]$. This implies the result you write: write $g=fh$ in $F[x]$, and apply the Lemma to conclude that there are rational multiples of $f$ and $h$ that lie in $R[x]$, then use the fact that $f$ is already in $R[x]$ to get the result. –  Arturo Magidin Oct 21 '11 at 20:40
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.