> [!NOTE] Theorem (Degree of Product of Polynomial Forms Over Integral Domain) > Let $(D,+,\times)$ be an [[Integral Domain|integral domain]]. Let $D[x]$ be the [[Ring of Polynomial Forms|ring of polynomials]] over $D$ in the indeterminate $x.$ For $f\in D[x],$ let $\deg(f)$ denote the [[Degree of a Polynomial|degree]] of $f.$ Then the degree of their [[Polynomial Multiplication|product]] is given by $\forall f,g \in D[x]: \deg(fg)= \deg(f) + \deg(g).$ *Proof*. WLOG suppose $f=0.$ Then $\deg(f)=-\infty .$ Then $fg=0$ and so $\deg(fg)=-\infty=\deg(f)+\deg(g)$by conventions adopted about addition involving $-\infty.$ Now suppose $f,g$ are both non-zero. Then $m:=\deg(f)\geq 0$ and $n:=\deg(g)\geq 0.$ Let $a_{n}$ and $b_{m}$ be the leading coefficients of $f$ and $g$ (the coefficients of $x^{m}$ in $f$ and $x^{n}$ in $g$ which are non-zero, by definition of degree) respectively then by definition of polynomial multiplication $fg=a_{n}b_{m} x^{n+m} +\dots+a_{0}b_{0}$Since neither $a_{m}$ nor $b_{n}$ equal $0_{D}$ and $D$ has no [[Proper Zero Divisor|proper zero divisors]] $a_{m}b_{n}\neq 0_{D}.$ It follows that $\deg(f+g)=m+n.$ # Applications If $f\neq 0 \land g \neq 0,$ then $\deg(fg)=\deg(f)+\deg(g) \geq 0$ so [[Ring of Polynomial Forms over Integral Domain is Integral Domain]]. A polynomial is a [[Units of Ring of Polynomial Forms over Integral Domain|unit in the ring of polynomial forms over a field]] if and only if its degree is zero.