Exercise A0.V.2.12

Proposition

Let $R[x]$ be a PID. Then $R$ is a field.

Proof

Let $I = (x)$ and let $J$ be an ideal such that $I \subsetneq J.$ Since $R[x]$ is a PID, there exists $j \in R[x]$ such that $J = (j).$ Inclusion of ideals forces that $x = rj$ for some $r \in R.$ But note that

$$1 = \deg(x) = \deg(rj) = \deg(r) + \deg(j).$$

Suppose $\deg(j) = 0.$ In this case, $r = ax + b$ such that $ja = 1$ and $jb = 0.$ Clearly $j \neq 0$ so $b = 0$ and $j$ is a unit with $j^{-1} = a.$ Thus $J = R[x].$

Otherwise, suppose that $\deg(r) = 0.$ In this case, $j = ax + b$ such that $ra = 1$ and $rb = 0.$ Similarly, this forces $b$ to be $0$ and $r$ to be a unit. Thus $j$ is associate to $x$ and hence $(x) = (j),$ but this contradicts the inclusion being proper.

This proves that $I$ is maximal in $R[x],$ hence $R[x]/I \cong R$ is a field.

$\blacksquare$