Alphabeta MathHow statement and proof provenance work
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Assuming Choice, every field has an algebraic closure
Statement
Assuming the Axiom of Choice, every field has an algebraic closure.
Facts & Assumptions
Given: A field and the Axiom of Choice.
Assuming Choice, there is an algebraic extension containing a root of every nonconstant polynomial over (Assuming Choice, every field has an algebraic extension containing roots of all nonconstant base polynomials).
Every algebraic extension with that one-step root property is algebraically closed (An algebraic extension containing a root of every nonconstant base polynomial is algebraically closed).
An algebraic closure is an algebraic extension that is algebraically closed (An algebraic closure of a field).
Proof
Use [L1] to construct an algebraic extension containing a root of every nonconstant base polynomial.
By [L2], this same field is already algebraically closed.
Thus is an algebraic closure by [L3].
Depends on
Used by
- The separable degree [K:F]ₛ as a count of embeddings into an algebraic closure Definition
- 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
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 46 results over 11 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)