# 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 $D$.

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

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

Therefore, for some $p, q \in D \left[{X}\right]$, $y = f p$ and $X = f q$.

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

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

- $X = a b + a c X$

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

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

Therefore there exist $r, s \in D \left[{X}\right]$ such that:

- $r y + s X = 1$

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.

$\blacksquare$