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

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Transcendental residue elements adjoin across a maximal subfield

Statement

Let $(A, m)$ be a local ring, let $K \subseteq A$ be a residue-injective subfield, and let $\overline{u} \in A / m$ be transcendental over the residue image $ρ (K)$ . Then there exists a larger residue-injective subfield $K^{'} \subseteq A$ whose residue image contains $\overline{u}$ .

Facts & Assumptions

Given: A local ring $(A, m)$ , a residue-injective subfield $K \subseteq A$ , and a residue element $\overline{u}$ transcendental over $ρ (K)$ .

[L1]

A coefficient-field argument enlarges a residue-injective subfield by adjoining new residue elements when injectivity is preserved (Maximal residue-injective subfields exist).

[L2]

The residue image of a subfield is a field inside the residue field (Equicharacteristic local rings and coefficient fields).

Proof

technique · evaluate rational functions at a lift of the transcendental residue element

1.1L2givenchoose

Choose any lift $u \in A$ of $\overline{u}$ . For every nonzero polynomial $q (T) \in K [T]$ , the residue of $q (u)$ is $q (\overline{u}) \in ρ (K) (\overline{u})$ . Since $\overline{u}$ is transcendental over $ρ (K)$ , this residue is nonzero, so $q (u) \notin m$ and therefore is a unit of $A$ .

2.1step 1.1givenconstruct

Hence evaluation at $u$ defines an injective homomorphism $K (T) ⟶ A, r (T) ⟼ r (u),$ because every denominator evaluates to a unit by step 1.1. Let $K^{'}$ be its image. Then $K^{'}$ is a subfield of $A$ , and its residue image contains $ρ (K)$ together with $\overline{u}$ .

3.1L1step 2.1algebra

If an element of $K^{'}$ has zero residue, its representing rational function has zero value at the transcendental element $\overline{u}$ , so the rational function is zero. Thus the residue map is injective on $K^{'}$ . By [L1], this is exactly the desired enlargement step.

4.1step 3.1∎

Therefore every transcendental residue element adjoins across a maximal residue-injective subfield.

Depends on

Used by

Dependency tree · two levels

6 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)