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
Because an integral domain is reduced, $D$ is reduced.

By the corollary to Units of Ring of Polynomial Forms over Commutative Ring, this implies that the units of $D \left[{X}\right]$ are precisely the units of $D$.

This is the required result.