Show that \(R[x]\) a PID \(\iff R\) is a field.
Hint
Take \(r\in R\), then \(\gens{r, x} = \gens{f}\) for some \(f\).
Solution
Let \(R\) be a commutative ring with identity.
1. If \(R\) is a field, then \(R[x]\) is a PID.
Division with remainder in \(R[x]\) shows that a nonzero ideal is generated by any of its nonzero elements of least degree.
2. If \(R[x]\) is a PID, then every \(0\neq r\in R\) is a unit.
\(R[x]\) is a domain, hence so is \(R\subseteq R[x]\). Write \(\gens{r,x}=\gens{f}\), so \(r=fp\) and \(x=fq\) for some \(p,q\in R[x]\). In the domain \(R[x]\), degrees add, so \(\deg f=0\) from \(r=fp\); write \(f=c\in R\). Then \(x=cq\) forces \(q=ax+b\) with \(ca=1\), so \(c\) is a unit and \(\gens{r,x}=R[x]\). Hence \(1=rg(x)+xh(x)\) for some \(g,h\in R[x]\), and evaluating at \(x=0\) gives \(1=rg(0)\).
3. Q.E.D.
Steps 1 and 2, together with \(1\neq0\) in the domain \(R\).