Irreducible Polynomials in UFD and corresponding Quotient Field

Hello,

"Let $D$ be a UFD and let $F$ be its quotient field. Further let $f$ be a primitive polynomial of positive degree in $D\left[x\right]$. From this it follows that that $f$ is irreducible in $D\left[x\right]$ if and only if $f$ is irreducible in $F\left[x\right]$."

I've shown the forward direction, i.e. irreducible in UFD $\Rightarrow$ irreducible in quotient field, but am struggling to understand how the converse direction would go.

In particular, it strikes me that it might be productive to try and show that $f$ is prime in the UFD since this is equivalent to irreducibility, but I don't know how to show this. Also, $f$ should have no "denominators" in $F\left[x\right]$ since it's also in $D\left[x\right]$.

Any advice on how to proceed?

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Being irreducible in the quotient field automatically implies being irreducible in the domain. Are you sure you don't mean the other direction? In either case, this is a standard exercise (first show that the product of two primitive polynomials is primitive) called Gauss's lemma. – Qiaochu Yuan Feb 4 '10 at 20:49
Shanest, here's a quick argument for you, if you want irreducible in $F[x]$ implies irreducible in $D[x]$. Let $f\in D[x]$. Assume it factors as $f=gh$ in $D[x]$. Then it factors as $f=gh$ in $F[x]$. This is actually true regardless of primitivity, which you need for the other direction, as Qiaochu mentioned. The proof I'm giving you proves the contrapositive, instead of irred in $F[x]$ implies irred in $D[x]$ it argues that reducible in $D[x]$ implies reducible in $F[x]$ which is equivalent. – Charles Siegel Feb 4 '10 at 21:26
Qiaochu, I meant the easier direction (the forward direction can be adapted from a textbook or the Wikpedia page mentioned below). Charles, thanks. I had been playing around with that kind of idea, but got turned around about what the contrapositive actually was in this case. Too many negatives with IRreducible. – user3795 Feb 4 '10 at 21:53
Yeah, once you actually know what the contrapositive is, this is entirely trivial. – Charles Siegel Feb 4 '10 at 22:37

This is the famous Gauss Lemma (for polynomials) which says $cont(fg) = cont(f) cont(g))$ where $cont(f)$ is the gcd of the coefficients (this must be an ideal for a general UFD).