LemmaStatement: 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.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Coefficientwise sums and convolution products of finitely supported sequences are finitely supported

Statement

If $a, b ⁣ : N \to R$ have finite support, then their coefficientwise sum and convolution product have finite support.

Facts & Assumptions

Given: A commutative ring $R$ and finitely supported coefficient sequences $a, b \in R [x]$ .

[L1]

A coefficient sequence has finite support when it vanishes beyond some natural-number bound; addition is coefficientwise and multiplication is convolution (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

Proof

technique · direct

1.1givenL1choose

Choose $M, N \in N$ such that $a_{i} = 0$ for $i > M$ and $b_{j} = 0$ for $j > N$ .

2.1step 1.1L1algebra∎

If $k > \max {M, N}$ then $(a + b)_{k} = 0$ , and if $k > M + N$ then every pair $i + j = k$ has $i > M$ or $j > N$ , so every summand $a_{i} b_{j}$ in $(a b)_{k}$ is zero; hence both sequences have finite support.

Depends on

The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution

Dependency tree · 2 levels

Sources

Neil Donaldson, Math 120B Notes, Section 22