Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-04
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Formal power-series substitution is the unique continuous k-algebra map

Statement

Let (A,m) be a complete local ring, let k→A be a ring map, and let x1,…,xn∈m. Then there is a unique continuous k-algebra homomorphism ϕ:k⟦X1,…,Xn⟧→A such that ϕ(Xi)=xi for every i.

Facts & Assumptions

Given: A complete local ring (A,m), a ring map k→A, and elements x1,…,xn∈m.

[L1]

Degreewise substitution converges for every formal series (Formal power-series substitution converges in a complete local algebra).

Proof

technique · define the map by convergent substitution and use density of polynomials
1.1L1givenconstruct

By [L1], every series F=∑αaαXα has a convergent substituted sum ϕ(F):=∑αaαxα∈A. Finite truncations show that ϕ respects addition and multiplication, and by construction ϕ is a k-algebra map with ϕ(Xi)=xi.

2.1step 1.1givenalgebra

The map is continuous for the (X1,…,Xn)-adic topology on the source and the m-adic topology on the target, because every series all of whose monomials have total degree at least N maps into mN.

3.1step 1.1step 2.1given

If ψ is another continuous k-algebra map with ψ(Xi)=xi, then ψ agrees with ϕ on the polynomial subring k[X1,…,Xn]. Every formal series is the limit of its polynomial truncations, and both maps are continuous, so they agree on all of k⟦X1,…,Xn⟧. Therefore ϕ is unique.

4.1step 3.1∎

Thus formal substitution is the unique continuous k-algebra map sending each indeterminate to the chosen maximal-ideal element.

Depends on

Used by

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Sources