---
tag: 0T233AZ1GE
kind: statement
role: proposition
name: Degree of a product
msc: [13F20]
uses: [0DPF6WRZW6, 09S3PFBEPM, 0P05RJC5XA]
sources:
  - url: https://en.wikipedia.org/wiki/Degree_of_a_polynomial
    locator: Multiplication
---

## Statement

Let $R$ be an [[integral domain|09S3PFBEPM]] and let $f, g \in R[x]$.
Then $\deg(f g) = \deg f + \deg g$, with the [[convention|0DPF6WRZW6]] $\deg 0 = -\infty$.

## Remarks

The proof: if $f$ and $g$ are nonzero with [[leading coefficients|0P05RJC5XA]] $a$ 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.