> [!NOTE] Theorem (Units of Ring of Polynomial Forms over Integral Domain are The Degree Zero Polynomials) > Let $(R,+,\times)$ be a [[Integral Domain|integral domain]]. Let $R[x]$ be the [[Ring of Polynomial Forms|ring of polynomial forms]] over $R$ in $x.$ Then $f\in R[x]$ is a [[Unit in a Ring|unit]] of $R[x]$ iff $f\in R^*$ . Proof. ($\implies$) Suppose $f\in F[x]$ is a unit. Then there exists $g\in F[x]$ such that $fg=1$ so that $\deg(fg)=0.$ By [[Degree of Product of Polynomials Over Integral Domain|degree of product]], $0=\deg(1)=\deg(f)+\deg(g).$ Therefore $\deg(f)=\deg(g)=0.$ ($\impliedby$) Conversely, suppose $\deg(f)=0.$ Then $f=a_{0}$ where $0\neq a_{0}\in F.$ Since $F$ is a field, there exists $a_{0}^{-1}\in F$ such that $a_{0}^{-1}a_{0}=1.$ Thus $f$ is a unit with inverse $g=a_{0}^{-1}.$