---
tag: C49QN234AG
kind: proof
name: Proof of the degree of a product over an integral domain
msc: [13F20]
proves: 0T233AZ1GE
uses: [09S3PFBEPM, 0Q9EZS90GY, 0DPF6WRZW6, 0P05RJC5XA]
sources:
  - url: https://proofwiki.org/wiki/Degree_of_Product_of_Polynomials_over_Integral_Domain
---

## Proof

1. Let $R$ be an [[integral domain|09S3PFBEPM]] and let $f, g$ be elements of the [[polynomial ring|0Q9EZS90GY]] $R[x]$. We use the convention $-\infty + d = d + (-\infty) = -\infty$ for every $d$, including $d = -\infty$.
2. If $f = 0$ or $g = 0$, then $f g = 0$. So both sides are $-\infty$, by the convention for the [[degree|0DPF6WRZW6]] of $0$.
3. Now let $f \neq 0$ and $g \neq 0$. Write $f = \sum_{i=0}^{m} a_i x^i$ and $g = \sum_{j=0}^{n} b_j x^j$ with $m = \deg f$ and $n = \deg g$. Then the [[leading coefficients|0P05RJC5XA]] are $a_m \neq 0$ and $b_n \neq 0$.
4. Put $a_i = 0$ for $i > m$ and $b_j = 0$ for $j > n$. By the multiplication of $R[x]$, $f g = \sum_{k} c_k x^k$ with $c_k = \sum_{i + j = k} a_i b_j$, over pairs of integers $i, j \ge 0$.
5. Let $k > m + n$, and let $i + j = k$. Then $i > m$ or $j > n$, since otherwise $i + j \le m + n$. So $a_i b_j = 0$, and $c_k = 0$.
6. Let $k = m + n$, and let $i + j = k$ with $(i, j) \neq (m, n)$. If $i < m$, then $j = m + n - i > n$. If $i = m$, then $j = n$, which is excluded. So $i > m$ or $j > n$, and $a_i b_j = 0$ for these pairs.
7. Hence $c_{m+n} = a_m b_n$.
8. Since $R$ is an integral domain and $a_m \neq 0$, $b_n \neq 0$, we get $a_m b_n \neq 0$.
9. By steps 5 and 8, $f g \neq 0$, and the largest $k$ with $c_k \neq 0$ is $m + n$. So $\deg(f g) = m + n = \deg f + \deg g$.

## Idea

The highest power that can appear in $f g$ is $x^{\deg f + \deg g}$, and its coefficient is the product of the two leading coefficients. In an integral domain this product is nonzero, so no cancellation of the top term can happen.