Polynomial Forms is PID Implies Coefficient Ring is Field

Theorem
Let $D$ be an integral domain.

Let $D \left[{X}\right]$ be the ring of polynomial forms in $X$ over $D$.

Let $D \left[{X}\right]$ be 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 $f \in D \left[{X}\right]$ we have:


 * $\left \langle{y, X}\right \rangle = \left \langle {f}\right \rangle \subseteq D \left[{X}\right]$

Therefore:
 * $\exists p, q \in D \left[{X}\right]: y = f p, X = f q$

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

Substituting into the equation $X = f q$ we obtain:


 * $X = a b + a c X$

which implies that:
 * $a c = 1$

That is:
 * $a \in D^\times$

where $D^\times$ denotes the group of units of $D$.

Therefore:
 * $\left\langle{f}\right\rangle = \left\langle{1}\right\rangle = D \left[{X}\right]$

Therefore:
 * $\exists r, s \in D \left[{X}\right]: r y + s X = 1$

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

Therefore $y \in D^\times$.

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

Also see

 * Polynomial Forms over Field form Principal Ideal Domain