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A complete equicharacteristic Noetherian local ring is a power-series quotient

Statement

Assume the Axiom of Choice.

Let $(A, m)$ be a complete equicharacteristic Noetherian local ring, let $k = A / m$ , and let $e = \dim_{k} (m / m^{2}) .$ Then there is a surjective $k$ -algebra homomorphism $k ⟦ X_{1}, \dots, X_{e} ⟧ ↠ A .$

Facts & Assumptions

Given: A complete equicharacteristic Noetherian local ring $(A, m)$ and the Axiom of Choice.

[L1]

The ring $A$ contains a coefficient field mapping isomorphically to its residue field (Complete equicharacteristic local rings have coefficient fields).

[L2]

Once the coefficient field and lifts of a basis of $m / m^{2}$ are chosen, the associated formal-series map is surjective (The Cohen map is surjective by completeness).

Proof

technique · choose a coefficient field and generators of $\mathfrak m/\mathfrak m^2$

1.1

By [L1], choose a coefficient field $k \subseteq A$ . Choose elements $x_{1}, \dots, x_{e} \in m$ lifting a $k$ -basis of $m / m^{2}$ .

L1givenchoose

2.1

The universal substitution construction gives a continuous $k$ -algebra map $ϕ : k ⟦ X_{1}, \dots, X_{e} ⟧ \to A, X_{i} \mapsto x_{i} .$ By [L2], this map is surjective.

L2step 1.1

3.1

Therefore $A$ is a quotient of the formal power-series ring in $e$ variables over its residue field.

step 2.1∎

Depends on

Used by

Dependency tree · two levels

5 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

Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, 13th ed., Theorem 22.33 (standard reference, not scraped)
The Stacks Project, Section 10.160: The Cohen structure theorem (standard reference, not scraped)