LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-04

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The Cohen map is surjective modulo every power of the maximal ideal

Statement

Let $(A, m)$ be a complete equicharacteristic Noetherian local ring, let $k \subseteq A$ be a coefficient field, and let $x_{1}, \dots, x_{e} \in m$ lift a $k$ -basis of $m / m^{2}$ . Let $ϕ : k ⟦ X_{1}, \dots, X_{e} ⟧ \to A$ be the continuous $k$ -algebra map with $ϕ (X_{i}) = x_{i}$ . Then for every $n \geq 1$ , the induced map $k ⟦ X_{1}, \dots, X_{e} ⟧ / (X_{1}, \dots, X_{e})^{n} \to A / m^{n}$ is surjective.

Facts & Assumptions

Given: A complete equicharacteristic Noetherian local ring $(A, m)$ , a coefficient field $k \subseteq A$ , and lifts $x_{1}, \dots, x_{e}$ of a basis of $m / m^{2}$ .

[L1]

The continuous substitution map $ϕ$ exists (Formal power-series substitution is the unique continuous k-algebra map).

[L2]

Generators of $m / m^{2}$ generate $m$ (Assuming the Axiom of Choice, generators modulo an ideal in the Jacobson radical lift to generators).

[L3]

In a Noetherian ring, powers of a finitely generated ideal are generated by products of generators (A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member).

Proof

technique · generate $\mathfrak m^r/\mathfrak m^{r+1}$ by degree-$r$ monomials

1.1

By [L2], the elements $x_{1}, \dots, x_{e}$ generate $m$ . Therefore every product of $r$ generators is the image under $ϕ$ of a degree- $r$ monomial, and by [L3] these monomials span $m^{r} / m^{r + 1}$ over $k$ for every $r \geq 1$ .

L2L3givenalgebra

2.1

Modulo $m$ , the map $ϕ$ is already surjective because its image contains the coefficient field $k$ and the quotient $A / m$ equals $k$ . By step 1.1, every class in each successive quotient $m^{r} / m^{r + 1}$ also has a polynomial preimage of total degree exactly $r$ . Summing those representatives for $r = 0, \dots, n - 1$ shows that every class in $A / m^{n}$ has a preimage in the source modulo $(X_{1}, \dots, X_{e})^{n}$ .

L1step 1.1givenalgebra

3.1

Hence the Cohen map is surjective modulo every power of the maximal ideal.

step 2.1∎

Depends on

Used by

The Cohen map is surjective by completeness Lemma

Dependency tree · two levels

16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Melvin Hochster, The structure theory of complete local rings (standard reference, not scraped)
The Stacks Project, Section 10.160: The Cohen structure theorem (standard reference, not scraped)