Alphabeta MathHow statement and proof provenance work
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Every algebraic extension of a perfect field is separable
Statement
If is algebraic and is perfect, then is separable.
Facts & Assumptions
Given: An algebraic extension with perfect.
Every algebraic element has a monic irreducible minimal polynomial over the base (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).
Every nonconstant irreducible polynomial over a perfect field is separable (Perfect fields: every irreducible polynomial is separable).
An extension is separable when every one of its elements has separable minimal polynomial over the base (Separable algebraic elements and separable extensions).
Proof
For each , [L1] supplies its irreducible minimal polynomial over , and [L2] makes that polynomial separable.
Thus every element of is separable over , so is separable by [L3].
Depends on
Used by
- Every finite extension of a perfect field is simple Corollary
- The one-step root condition makes an algebraic extension of a perfect field algebraically closed Lemma
- Differentials of a separably generated field extension Theorem
- Finitely generated extensions of a perfect field are separably generated Theorem
- Jacobian rank detects regularity at closed points Theorem
- Over a perfect field, every endomorphism has a unique commuting semisimple-plus-nilpotent decomposition, polynomial in the endomorphism Theorem
- The complex numbers are algebraically closed Theorem
Dependency tree · two levels
8 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
- P. L. Clark, Field Theory, Chapters 3 to 5 (standard reference, not scraped)
- J. S. Milne, Fields and Galois Theory, Chapters 2, 3, and 5 (standard reference, not scraped)