# Gauss’s lemma II

Let $D$ be a unique factorization domain  and $F$ its field of fractions  .   If a polynomial $P\in D[x]$ is reducible in $F[x]$, then it is reducible in $D[x]$.

Remark. The above statement is often used in its contrapositive form.  For an example of this usage, see this entry (http://planetmath.org/AlternativeProofThatSqrt2IsIrrational).

Proof.  A primitive polynomial in $D[x]$ is by definition divisible by a non invertible  constant polynomial, and therefore reducible in $D[x]$ (unless it is itself constant). There is therefore nothing to prove unless $P$ (which is not constant) is primitive.  By assumption  there exist non-constant $S,\,T\in F[x]$ such that  $P=ST$.  There are elements $a,\,b\in F$ such that $aS$ and $bT$ are in $D[x]$ and are primitive (first multiply by a nonzero element of $D$ to chase any denominators, then divide by the gcd of the resulting coefficients in $D$).  Then $aSbT=abP$ is primitive by Gauss’s lemma I, but $P$ is primitive as well, so $ab$ is a unit of $D$ and  $P=(ab)^{-1}(aS)(bT)$  is a nontrivial decomposition of $P$ in $D[X]$.  This completes      the proof.

From the above proposition and its proof one may infer the

If a primitive polynomial of $D[x]$ is divisible in $F[x]$, then it splits in $D[x]$ into primitive prime factors.  These are uniquely determined up to unit factors of $D$.

 Title Gauss’s lemma II Canonical name GausssLemmaII Date of creation 2013-03-22 13:07:52 Last modified on 2013-03-22 13:07:52 Owner bshanks (153) Last modified by bshanks (153) Numerical id 18 Author bshanks (153) Entry type Theorem Classification msc 12E05 Synonym Gauss’ lemma II Related topic GausssLemmaI Related topic EisensteinCriterion Related topic ProofOfEisensteinCriterion Related topic PrimeFactorsOfXn1 Related topic AlternativeProofThatSqrt2IsIrrational Defines primitive polynomial