Polynomial Forms is PID Implies Coefficient Ring is Field

Theorem
Let $D$ be an integral domain.

Let $D[X]$ be the Ring of Polynomial Forms in $X$ over $D$.

If $D[X]$ is a principal ideal domain then $D$ is a field.

Proof
Let $y \in D$ be non-zero.

Then, using the principle ideal property, for some $f \in D[X]$ we have:


 * $\langle y, X \rangle = \langle f \rangle \subseteq D[X]$

Therefore, for some $p,q \in D[X]$, $y = fp$ and $X = fq$.

By Properties of Degree we conclude that $f = a$ and $q = b + cX$ for some $a,b,c \in D$.

Substituting into the equation $X=fq$ we obtain


 * $X = ab + acX$

which implies that $ac = 1$, i.e. $a \in D^\times$, the group of units of $D$.

Therefore $\langle f \rangle = \langle 1 \rangle = D[X]$.

Therefore there exist $r,s \in D[X]$ such that


 * $r y + s X = 1$

If $d$ is the constant term of $r$, then we have $yd = 1$.

Therefore $y \in D^\times$.

Our choice of $y$ was arbitrary, so this shows that $D^\times \supseteq D\backslash \{0\}$, which says precisely that $D$ is a field.

Also See

 * Polynomial Forms over a Field is Principal Ideal Domain