> [!NOTE] Theorem (Ring of Univariate Polynomial Forms over Integral domain is an Integral domain) > Let $D$ be an [[Integral Domain|integral domain]]. Let $D[x]$ be the [[Ring of Polynomial Forms|ring of polynomial forms]] over $D$ in $x.$ Then $D[x]$ is an integral domain: $\forall f,g\in D[x]: fg =0 \implies f= 0 \lor g=0.$ *Proof*. We prove the contrapositive. Suppose $f\neq 0$ and $g \neq 0.$ Then $\deg(f)\geq 0$ and $\deg(g)\geq 0.$ By [[Degree of Product of Polynomials Over Integral Domain|degree of product]], $\deg(fg)=\deg(f)+\deg(g)\geq 0.$ In particular, $\deg(fg) \neq 0.$