Degree of a product

Statement

Let $R$ be an integral domain and let $f, g \in R[x]$. Then $\deg(f g) = \deg f + \deg g$, with the convention $\deg 0 = -\infty$.

Remarks

The proof: if $f$ and $g$ are nonzero with leading coefficients $a$ and $b$, then the coefficient of $x^{\deg f + \deg g}$ in $f g$ is $a b \neq 0$. Over every ring, $\deg(f g) \le \deg f + \deg g$. The equality fails in $(\mathbb{Z}/4\mathbb{Z})[x]$: $(2x + 1)(2x + 1) = 4x^2 + 4x + 1 = 1$.

Explanations

L1 human

When you multiply two polynomials, the highest terms multiply to give the highest term of the product. For example, $(x^2 + 1)(x^3 - x)$ has degree $2 + 3 = 5$. This needs the product of the leading coefficients to be nonzero.