TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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Cauchy multiplication makes $R ⟦ x ⟧$ a commutative ring containing $R [x]$ as the finitely supported subring

Statement

For every commutative ring $R$ , the coefficientwise sum and Cauchy product make $R ⟦ x ⟧$ a commutative ring. The coefficientwise inclusion

$j : R [x] ⟶ R ⟦ x ⟧$

is an injective unital ring homomorphism, and its image is exactly the finitely supported formal series.

Facts & Assumptions

Given: The hypotheses and notation of the statement above.

[F1]

Cauchy multiplication is the finite convolution $[x^{n}] (f g) = \sum_{i = 0}^{n} [x^{i}] f [x^{n - i}] g$ (Formal power series over a commutative ring and the coefficient-extraction functional $[x^{n}]$ ).

[F2]

A finite sum over $S \times T$ equals either iterated finite sum, and finite sums are invariant under bijective reindexing (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule).

[F3]

Polynomial coefficientwise addition and convolution make $R [x]$ a commutative ring, and the constant-polynomial map is an injective unital ring homomorphism (Polynomial convolution makes $R [x]$ a commutative ring containing $R$ as its constant subring).

Proof

technique · coefficientwise verification

1.1

Coefficientwise addition inherits associativity, commutativity, zero, and additive inverses from $R$ . For multiplication, the coefficient of both $(f g) h$ and $f (g h)$ at $n$ is the finite sum $\sum_{i + j + k = n} [x^{i}] f [x^{j}] g [x^{k}] h$ by finite Fubini. Reindexing $(i, j)$ as $(j, i)$ gives commutativity, and splitting finite sums gives both distributive laws. The constant series $1$ is a multiplicative identity, since the only nonzero summand involving it occurs at index $0$ .

givenF1F2

2.1

A product of finitely supported series is finitely supported, and its coefficient formula is exactly the published polynomial convolution. Hence $j$ preserves $0, 1, +,$ and multiplication, so it is a unital ring homomorphism; it is injective because equality of coefficient functions is literal equality. Its image consists precisely of the finitely supported functions.

step 1.1givenF3

3.1

Therefore $R ⟦ x ⟧$ is a commutative ring and $j$ identifies $R [x]$ with its finitely supported subring.

step 1.1step 2.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 48 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Benjamin Sambale, An Invitation to Formal Power Series (standard reference, not scraped)
Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics (standard reference, not scraped)

Cauchy multiplication makes R⟦x⟧R[[x]] a commutative ring containing R[x]R[x] as the finitely supported subring