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Assuming Choice, a base-field embedding extends across every algebraic extension
Statement
Assume the Axiom of Choice. Let be algebraic, let be algebraically closed, and let be a field embedding. Then extends to a field embedding . The proof uses Zorn's lemma.
Facts & Assumptions
Given: The Axiom of Choice, an algebraic extension , an algebraically closed field , and an embedding .
Assuming Choice, a nonempty poset in which every chain has an upper bound has a maximal element (Zorn's lemma).
A chosen root of an irreducible polynomial induces the unique embedding of the corresponding simple root extension (Universal property of adjoining a root of an irreducible polynomial).
A field isomorphism transports coefficients, evaluation, and roots of polynomials (A field isomorphism transports polynomials coefficientwise and carries roots, factorizations, and splitting to roots, factorizations, and splitting).
Relative embeddings are field embeddings agreeing with the specified map on the base (-homomorphisms and -embeddings of field extensions).
Every nonconstant polynomial over an algebraically closed field has a root (An algebraically closed field: every nonconstant polynomial has a root in the field).
Proof
Let consist of pairs where is an intermediate field and is an embedding extending , ordered by extension of the domain and map. The pair shows that is nonempty.
The union of a chain of such pairs has the union of the domains as an intermediate field and the union of the compatible maps as an embedding into , so every chain has an upper bound.
By Zorn's lemma [L1], choose a maximal pair in .
If , choose . It is algebraic over . Transport its minimal polynomial through the isomorphism using [L3], choose a root in by [L5], and use [L2] to extend to an embedding of .
Step 4.1 contradicts maximality unless . Thus on the maximal domain is the required extension of ; the use of Choice is precisely [L1].
Depends on
- Zorn's lemma
- Universal property of adjoining a root of an irreducible polynomial
- A field isomorphism transports polynomials coefficientwise and carries roots, factorizations, and splitting to roots, factorizations, and splitting
- $F$-homomorphisms and $F$-embeddings of field extensions
- An algebraically closed field: every nonconstant polynomial has a root in the field
Used by
- Assuming Choice, any two algebraic closures are base-isomorphic Corollary
- Assuming Choice, conjugates in an algebraic closure are related by a base automorphism Corollary
- Assuming Choice, real algebraic numbers embed properly in an algebraic closure of ℚ Example
- FALSE: an algebraic closure is unique up to a unique base-field isomorphism False statement
- Assuming Choice, separable closures exist and are base-isomorphic Theorem
- Pure inseparability and its conjugate, embedding, and separable-degree criteria Theorem
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 39 results over 6 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.
Sources
- J. S. Milne, Fields and Galois Theory, Chapter 6 (standard reference, not scraped)
- P. L. Clark, Field Theory, Chapter 4 (standard reference, not scraped)