Units of Ring of Polynomial Forms over Integral Domain
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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 $R$.
$\blacksquare$