# Units of Ring of Polynomial Forms over Integral Domain

## Theorem

Let $\struct {D, +, \circ}$ be an integral domain.

Let $D \sqbrk X$ be the ring of polynomial forms in an indeterminate $X$ over $D$.

Then the group of units of $D \sqbrk X$ is precisely the group of elements of $D \sqbrk X$ of degree zero that are units of $D$.

## Proof

It is immediate that a unit of $D$ is also a unit of $D \sqbrk X$.

Let $P$ be a unit of $D \sqbrk X$.

Then there exists $Q \in D \sqbrk X$ such that $P Q = 1$.

By Corollary 2 to Degree of Product of Polynomials over Ring we have:

- $0 = \map \deg 1 = \map \deg P + \map \deg Q$

Therefore:

- $\map \deg P = \map \deg Q = 0$

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

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

$\blacksquare$