Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11
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Polynomial convolution makes R[x] a commutative ring containing R as its constant subring

Statement

For every commutative ring R, the coefficientwise addition and convolution multiplication of The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution make R[x] a commutative ring. The constant-polynomial map c ⁣:R→R[x] is an injective unital ring homomorphism (Ring homomorphism: additive, multiplicative, and required to send 1 to 1).

Facts & Assumptions

Given: A commutative ring R and the operations on R[x] defined by coefficientwise addition and finite convolution.

[L1]

The set R[x] consists of finitely supported coefficient sequences, with (ab)n=∑i+j=naibj (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

[L2]

Coefficientwise sums and convolution products of finitely supported sequences are finitely supported (Coefficientwise sums and convolution products of finitely supported sequences are finitely supported).

[L3]

Finite sums in a commutative monoid are invariant under bijective reindexing, split over disjoint unions, and may be summed in either order over a finite product (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule).

[L4]

A ring homomorphism preserves addition, multiplication, and the multiplicative identity (Ring homomorphism: additive, multiplicative, and required to send 1 to 1).

Proof

technique · direct
1.1givenL1L2

Closure follows from [L2]; the additive group laws, including the zero sequence and coefficientwise negatives, follow coefficient by coefficient from the additive group laws in R.

2.1step 1.1L1L3L4algebra∎

Distributivity follows by splitting each finite convolution sum, commutativity follows by reindexing (i,j) as (j,i) and using commutativity in R, and associativity follows by [L3] from the equality of the two finite sums ∑i+j+k=naibjck; the sequence 1 is a multiplicative identity because only the index-0 coefficient contributes. Finally c(r+s)=c(r)+c(s), c(rs)=c(r)c(s), and c(1R)=1, so [L4] makes c a unital ring homomorphism, while equality of constant sequences forces equality of their index-0 coefficients and makes c injective.

Depends on

Dependency tree · 2 levels

Sources