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An algebraic extension containing a root of every nonconstant base polynomial is algebraically closed
Statement
Let be algebraic. If every nonconstant polynomial in has a root in , then is algebraically closed. One root-adjoining extension suffices; no iterated tower of root extensions is required.
Facts & Assumptions
Given: An algebraic extension containing a root of every nonconstant polynomial over .
The one-step root condition over a perfect base makes an algebraic extension algebraically closed (The one-step root condition makes an algebraic extension of a perfect field algebraically closed).
In positive characteristic, the elements with a suitable -power in the base form a perfect intermediate field whose polynomials retain the root condition in (The elements with a th power in the base form a perfect subfield carrying the one-step root condition).
Every characteristic-zero field is perfect (Fields of characteristic zero, finite fields, and algebraically closed fields are perfect).
Proof
If has characteristic zero, [L3] makes it perfect and [L1] makes algebraically closed.
If has characteristic , let be the perfect intermediate field from [L2]. The extension is algebraic because is algebraic, and [L2] gives the one-step root condition over , so [L1] again makes algebraically closed.
The characteristic-zero and positive-characteristic cases exhaust all fields and establish the conclusion without repeating the root-extension construction.
Depends on
Used by
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 48 results over 10 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, Proposition 6.5 (standard reference, not scraped)
- P. L. Clark, Field Theory, Theorem 4.9 (standard reference, not scraped)