Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17
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Universal property of a polynomial ring on an arbitrary family of indeterminates

Statement

Let R,S be commutative rings, let φ:RS be a ring homomorphism, and let (si)iI be a family in S. There is a unique ring homomorphism

Φ:R[xi:iI]S

whose restriction to R is φ and which satisfies Φ(xi)=si for every iI.

Facts & Assumptions

Given: Commutative rings R,S, a ring homomorphism φ:RS, and a family (si)iI in S.

[L1]

The finite convolution construction is a commutative ring containing R (Finite convolution makes R[xi:iI] a commutative ring containing R).

[L2]

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

Proof

technique · direct
1.1

For aM(I) define sa:=isupp(a)siai, and for c=acaxa define Φ(c):=aφ(ca)sa; both expressions are finite and the empty product is 1S.

L1construct
2.1

Pointwise addition gives Φ(c+d)=Φ(c)+Φ(d), while sa+b=sasb and finite reindexing give Φ(cd)=Φ(c)Φ(d).

step 1.1L3algebra
3.1

The zero monomial gives Φ(1)=1, constants give Φ(r)=φ(r), and the one-supported exponent family gives Φ(xi)=si; hence Φ is the required ring homomorphism by [L2].

step 1.1step 2.1L2
4.1

Any ring homomorphism with these values must send caxa to φ(ca)sa and therefore, by finite additivity, must equal the formula in step 1.1.

step 1.1L2

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 48 results over 12 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