PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Degree inequalities for sums and products over a commutative ring

Statement

Let $R$ be a commutative ring and let $f, g \in R [x]$ be nonzero.

If $f + g \neq 0$ , then $\deg (f + g) \leq \max {\deg f, \deg g}$ .
The coefficient of $x^{\deg f + \deg g}$ in $f g$ is $lc (f) lc (g)$ . If $f g \neq 0$ , then $\deg (f g) \leq \deg f + \deg g$ .

Facts & Assumptions

Given: Nonzero polynomials $f = \sum_{i} a_{i} x^{i}$ and $g = \sum_{j} b_{j} x^{j}$ over a commutative ring $R$ .

[L1]

Degree is the greatest index with nonzero coefficient, and the coefficient there is the leading coefficient (Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[L2]

Polynomial addition is coefficientwise and the coefficient of $x^{k}$ in a product is $\sum_{i + j = k} a_{i} b_{j}$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

Proof

technique · direct

1.1

For every $k > \max {\deg f, \deg g}$ both $a_{k}$ and $b_{k}$ vanish, so the coefficient of $x^{k}$ in $f + g$ vanishes; if the sum is nonzero, [L1] gives the stated inequality.

givenL1L2

2.1

Put $m = \deg f$ and $n = \deg g$ . For $k > m + n$ , every pair $i + j = k$ has $i > m$ or $j > n$ , while for $k = m + n$ the only possibly nonzero summand is $a_{m} b_{n}$ ; hence the top displayed coefficient is $lc (f) lc (g)$ and any nonzero product has degree at most $m + n$ .

givenL1L2algebra∎

Depends on

Used by

Dependency tree · two levels

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Sources

Neil Donaldson, Math 120B Notes, Theorem 22.3 (standard reference, not scraped)