Es ist

${\displaystyle {}X^{2}+7X+5={\left(6X+3\right)}{\left({\frac {1}{6}}X+{\frac {13}{12}}\right)}+{\frac {7}{4}}\,.}$

Da ${\displaystyle {}X^{2}+7X+5}$ und ${\displaystyle {}{\left(6X+3\right)}{\left({\frac {1}{6}}X+{\frac {13}{12}}\right)}}$ zum Ideal ${\displaystyle {}{\mathfrak {a}}}$ gehören, gehört auch

${\displaystyle {}{\frac {7}{4}}=X^{2}+7X+5-{\left(6X+3\right)}{\left({\frac {1}{6}}X+{\frac {13}{12}}\right)}\,}$

zu ${\displaystyle {}{\mathfrak {a}}}$. Damit gehört auch

${\displaystyle {}1={\frac {4}{7}}\cdot {\frac {7}{4}}\,}$

${\displaystyle {}P=P\cdot 1}$

${\displaystyle {}P}$

${\displaystyle {}{\mathfrak {a}}}$