Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-04
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The Cohen map is surjective by completeness

Statement

Let (A,m) be a complete equicharacteristic Noetherian local ring, let k⊆A be a coefficient field, let x1,…,xe∈m lift a k-basis of m/m2, and let ϕ:k⟦X1,…,Xe⟧→A be the continuous k-algebra homomorphism with ϕ(Xi)=xi. Then ϕ is surjective.

Facts & Assumptions

Given: A complete equicharacteristic Noetherian local ring (A,m), a coefficient field k⊆A, lifts x1,…,xe∈m of a k-basis of m/m2, and the resulting continuous k-algebra map ϕ.

[L1]

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

[L2]

For every n≥1, each class in mn/mn+1 has a homogeneous degree-n polynomial preimage under the Cohen map (The Cohen map is surjective modulo every power of the maximal ideal).

Proof

technique · build a compatible inverse system of polynomial approximations
1.1L2givenchooseinduction

Let a∈A. Because k→A/m is an isomorphism, choose F0∈k with ϕ(F0)≡a(modm). Inductively, if Fn−1 is chosen with ϕ(Fn−1)≡a(modmn), then [L2] applied to the error class in mn/mn+1 gives a homogeneous polynomial correction Hn of degree n such that ϕ(Fn)≡a(modmn+1),Fn:=Fn−1+Hn.

2.1L1step 1.1algebra

The formal sum F=∑n≥0Hn defines an element of k⟦X1,…,Xe⟧. By [L1], the series ϕ(F)=∑n≥0ϕ(Hn) converges in A, and step 1.1 says its partial sums are congruent to a modulo arbitrarily high powers of m. Since A is separated, the limit must equal a.

3.1step 2.1∎

Therefore every a∈A lies in the image of ϕ, so ϕ is surjective.

Depends on

Used by

Dependency tree · two levels

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Sources