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# Gauss’s lemma II

Definition. A polynomial $P=a_{n}x^{n}+\cdots+a_{0}$ over an integral domain
$D$ is said to be *primitive* if its coefficients are not all divisible
by any element of $D$ other than a unit.

Proposition (Gauss). 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.

*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.

Remark. Another result with the same name is Gauss’ lemma on quadratic residues.

From the above proposition and its proof one may infer the

Theorem. 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 $D$.

## Mathematics Subject Classification

12E05*no label found*

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