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A base-field embedding carries an algebraic element to a conjugate
Statement
Let be an -embedding and let be algebraic over . Then is conjugate to over . In particular, an -endomorphism of a splitting field permutes the distinct roots of every base polynomial that splits there.
Facts & Assumptions
Given: An -embedding and an element algebraic over .
A field isomorphism transports polynomial evaluation and carries roots to roots after applying the induced coefficient map (A field isomorphism transports polynomials coefficientwise and carries roots, factorizations, and splitting to roots, factorizations, and splitting).
An endomorphism of a splitting field fixing the base permutes the finite set of distinct roots of the defining polynomial (Every -endomorphism of a splitting field permutes the distinct roots and is an automorphism).
Conjugate elements are the roots of the same minimal polynomial over the base (Conjugate algebraic elements over a field).
Proof
Regard as an isomorphism . Let be the minimal polynomial of . Since fixes , [L1] gives .
Thus is a root of and is conjugate to by [L3].
When is a splitting field, [L2] strengthens this root preservation to a permutation of the distinct roots.
Depends on
Used by
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 38 results over 9 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
- 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)