Answer to Question #320211 in Abstract Algebra for Deepak

Question

Prove or disprove that in Z[x], the ideal <x> + <3> is a principal ideal.

Answer 1

$I=<x>+<3>=\left\{ xp\left( x \right) +3q\left( x \right) ,p,q\in \mathbb{Z} \left[ x \right] \right\} \\Suppose\,\,I=\left\{ p\left( x \right) h\left( x \right) ,p\in \mathbb{Z} \left[ x \right] \right\} \,\,for\,\,some\,\,h\in \mathbb{Z} \left[ x \right] \\Since\,\,3\in I, we\,\,have\,\,3=h\left( x \right) p\left( x \right) \,\,for\,\,some\,\,p, from\,\,which\\h\left( x \right) \in \left\{ \pm 1,\pm 3 \right\} \\If\,\,h\left( x \right) =\pm 1, we\,\,have\,\,\pm 1\cdot h\left( x \right) =1\in I. This\,\,is\,\,false\, because\\xp\left( x \right) +3q\left( x \right) =1\Rightarrow 3q\left( x \right) =1\Rightarrow q\left( x \right) =\frac{1}{3}\notin \mathbb{Z} \left[ x \right] \\If\,\,h\left( x \right) =\pm 3, we\,\,have\,\,I=\left\{ 3p\left( x \right) ,p\in \mathbb{Z} \left[ x \right] \right\} , from\,\,which\,\,x\notin I, \\which\,\,is\,\,false.\\Hence\,\,there\,\,is\,\,no\,\,such\,\,h\,\,that\,\,I=\left\{ p\left( x \right) h\left( x \right) ,p\in \mathbb{Z} \left[ x \right] \right\} ,which\,\,means\,\,I\,\,is\,\,not\,\,a\,\,principal\,\,ideal$