(March 2008) Let \(g(x)=x^2+3\in\mathbb{F}_7[x]\), where \(\mathbb{F}_7=\{0,1,2,3,4,5,6\}\) is the field of seven elements. (a) Prove that \(g\) is reducible in \(\mathbb{F}_7[x]\). (b) Let \(\langle g\rangle\subseteq\mathbb{F}_7[x]\) denote the principal ideal \(\{gh\mid h\in\mathbb{F}_7[x]\}\). Find an ideal \(I\subseteq\mathbb{F}_7[x]\) such that \(\langle g\rangle\subsetneq I\subsetneq\mathbb{F}_7[x]\).
Solution
Evaluating \(g\) at the elements of \(\FF_7\) gives \(g(2)=7=0\), so \(x-2\) divides \(g\). Long division gives \(g(x)=(x-2)(x+2)\), so \(g\) is reducible.
Let \(I=\langle x-2\rangle\subseteq\FF_7[x]\). Every element of \(\langle g\rangle\) has the form \(hg\) with \(h\in\FF_7[x]\), and \(hg=\bigl((x+2)h\bigr)(x-2)\in I\). Thus \(\langle g\rangle\subseteq I\subseteq\FF_7[x]\). Moreover \(x-2\notin\langle g\rangle\), because a multiple \(hg\) of \(g\) is either \(0\) or has degree \(\deg(hg)\ge\deg g=2\), whereas \(\deg(x-2)=1\). Finally \(1\notin I\), because a multiple \(h\cdot(x-2)\) is either \(0\) or has degree at least \(1\), whereas \(\deg 1=0\). Hence \(\langle g\rangle\subsetneq I\subsetneq\FF_7[x]\).