# 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?

• 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. Feb 4, 2010 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. Feb 4, 2010 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. Feb 4, 2010 at 21:53
• Yeah, once you actually know what the contrapositive is, this is entirely trivial. Feb 4, 2010 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).