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Maximal residue-injective subfields exist
Statement
Assume the Axiom of Choice.
Let be an equicharacteristic local ring. Then there exists a subfield that is maximal, under inclusion, among subfields whose residue map to is injective.
Facts & Assumptions
Given: An equicharacteristic local ring and the Axiom of Choice.
The residue field's prime field embeds in , so the family of residue-injective subfields is nonempty (The prime field lifts in the equicharacteristic case).
Assuming the Axiom of Choice, every nonempty poset in which every chain has an upper bound has a maximal element (Zorn's lemma).
Proof
Let be the set of subfields for which the residue map is injective. By [L1], is nonempty. Order by inclusion.
If is a chain, then is again a subfield of : closure under the field operations is inherited from some chain member containing the finitely many elements involved. Its residue map is still injective, because a nonzero element of the union already lies in one chain member where injectivity holds. Thus every chain in has an upper bound in .
By [L2], the poset has a maximal element. That is exactly a maximal residue-injective subfield of .
Depends on
Used by
Dependency tree · two levels
12 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)