Gauss's lemma II GausssLemmaII 2002-11-04 06:05:53 2007-04-14 02:50:37 Theorem primitive polynomial \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} \PMlinkescapeword{completes} \PMlinkescapeword{name} \PMlinkescapeword{primitive} \PMlinkescapeword{proposition} \textbf{Definition: }\, A polynomial $p(x)=a_nx^n+\cdots+a_0$ over a UFD $R$ is said to be \emph{primitive} if its coefficients are not all divisible by any element of $R$ other than a unit. \textbf{Proposition (Gauss): }\, Let $R$ be a UFD and $F$ its field of fractions. \, If a polynomial $p\in R[x]$ is reducible in $F[x]$, then it is reducible in $R[x]$. \textbf{Remark:}\, The above statement is often used in its contrapositive form. For an example of this usage, see \PMlinkname{this entry}{AlternativeProofThatSqrt2IsIrrational}. \textbf{Proof: }\, We may assume that $p$ is primitive. Suppose $p=qr$ with $q,r\in F[x]$. \,There are unique elements $a,b\in F$ such that $q/a$ and $r/b$ are in $R[x]$ and are primitive. \,But $p/ab=(q/a)(r/b)$. \,Since $p$ is primitive, it follows from Gauss's lemma I that $ab$ is a unit, and therefore so are $a$ and $b$. \,This completes the proof. \textbf{Remark: }\, Another result with the same name is Gauss' lemma on quadratic residues.