Units of Ring of Polynomial Forms over Integral Domain

Theorem
Let $\left({D, +, \circ}\right)$ be an integral domain.

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

Then the group of units of $D \left[{X}\right]$ is precisely the group of elements of $D \left[{X}\right]$ of degree zero that are units of $D$.

Proof
It is immediate that a unit of $D$ is also a unit of $D \left[{X}\right]$.

Let $P$ be a unit of $D \left[{X}\right]$.

Then there exists $Q \in D \left[{X}\right]$ such that $PQ = 1$.

By Degree of Product of Polynomials over Ring we have:
 * $0 = \deg\left({ 1 }\right) = \deg\left({ P }\right) + \deg\left({ Q }\right)$

Therefore $\deg\left({ P }\right) = \deg\left({ Q }\right) = 0$.

That is, $P \in R$ and $Q \in R$.

Moreover $PQ = 1$ in $R$, so it follows that $P$ is a unit of $R$.