planetmath / 12_field_theory_and_polynomials Goto Github PK

OLD:
\emph{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.\\
NEW:
\emph{Proof.}\, A non-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.\\