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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.
Algebraicity is transitive in towers of field extensions
Statement
If , the extension is algebraic, and is algebraic, then is algebraic.
Facts & Assumptions
Given: A tower with and algebraic, and an element .
Finitely many algebraic generators produce a finite extension (An extension generated by finitely many algebraic elements is finite).
An algebraic element generates a finite simple extension (An element is algebraic over if and only if its simple extension is finite).
Finite degrees multiply in a tower (Tower law for finite extensions: ).
Every finite extension is algebraic (Every finite field extension is algebraic).
Algebraicity means satisfying a nonzero polynomial over the base (Algebraic and transcendental elements and algebraic extensions).
Proof
Since is algebraic over , choose a nonzero polynomial with value zero at .
Every coefficient is algebraic over . Hence is finite over by [L1].
The same polynomial lies in , so is algebraic over and [L2] makes finite. The tower law [L3] makes finite.
By [L4], is algebraic over . Since was arbitrary, is algebraic.
Depends on
- An extension generated by finitely many algebraic elements is finite
- An element is algebraic over $F$ if and only if its simple extension $F(a)/F$ is finite
- Tower law for finite extensions: $[L:F]=[L:K][K:F]$
- Every finite field extension is algebraic
- Algebraic and transcendental elements and algebraic extensions
Used by
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 36 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
- A. W. Knapp, Basic Algebra, 2nd ed., Chapter IX, Section 1 (standard reference, not scraped)