polynomial ring which is PID

Theorem.  If a polynomial ring $D[X]$ over an integral domain $D$ is a principal ideal domain, then coefficient ring $D$ is a field. (Cf. the corollary 4 in the entry polynomial ring over a field.)

Proof.  Let $a$ be any non-zero element of $D$.  Then the ideal  $(a,X)$  of $D[X]$ is a principal ideal $(f(X))$ with $f(X)$ a non-zero polynomial (http://planetmath.org/ZeroPolynomial2).  Therefore,

$$a=f(X)g(X),X=f(X)h(X)$$

with $g(X)$ and $h(X)$ certain polynomials in $D[X]$.  From these equations one infers that $f(X)$ is a polynomial $c$ and $h(X)$ is a first degree polynomial $b_0+b_{1}X$ ($b_1\ne 0$).  Thus we obtain the equation

$$c{b}_0+c{b}_{1}X=X,$$

which shows that $c{b}_1$ is the unity 1 of $D$.  Thus  $c=f(X)$  is a unit of $D$, whence

$$(a,X)=(f(X))=(1)=D[X].$$

So we can write

$$1=a\cdot u(X)+X\cdot v(X),$$

where  $u(X),v(X)\in D[X]$.  This equation cannot be possible without that $a$ times the constant term of $u(X)$ is the unity.  Accordingly, $a$ has a multiplicative inverse in $D$.  Because $a$ was arbitrary non-zero elenent of the integral domain $D$, $D$ is a field.