LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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A base-field isomorphism extends across simple adjunctions of corresponding roots of an irreducible polynomial

Statement

Let $σ : F \to F^{'}$ be a field isomorphism, let $p \in F [x]$ be monic and irreducible, and put $p^{'} = σ_{*} p$ . If $α$ is a root of $p$ in an extension of $F$ and $β$ is a root of $p^{'}$ in an extension of $F^{'}$ , then there is a unique field isomorphism $\tilde{σ} : F (α) ⟶ F^{'} (β)$ extending $σ$ and satisfying $\tilde{σ} (α) = β$ .

Facts & Assumptions

Given: The fields, polynomial, roots, and isomorphism in the Statement.

[F1]

Coefficient transport along a field isomorphism is a polynomial-ring isomorphism and transports factorizations (A field isomorphism transports polynomials coefficientwise and carries roots, factorizations, and splitting to roots, factorizations, and splitting).

[F2]

The minimal polynomial of an algebraic element is the unique monic irreducible polynomial vanishing at it, and it divides every polynomial that vanishes there (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

[F3]

If an algebraic element has minimal polynomial of degree $n$ , its simple extension has the unique power basis $1, α, \dots, α^{n - 1}$ (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ ).

[F4]

Adjoining a root of a monic irreducible polynomial has the universal property that the root may be sent to any other root, uniquely over the base field (Universal property of adjoining a root of an irreducible polynomial).

Proof

technique · constructive

1.1

By [F1], $p^{'}$ is monic and irreducible. Since $p (α) = 0$ and $p^{'} (β) = 0$ , [F2] identifies $p$ and $p^{'}$ as the respective minimal polynomials. In particular they have the same degree $n$ .

F1F2

1.2

By [F3], each element of $F (α)$ has a unique form $\sum_{i < n} a_{i} α^{i}$ . Define $\tilde{σ} (\sum_{i < n} a_{i} α^{i}) = \sum_{i < n} σ (a_{i}) β^{i}$ . The quotient and root universal property in [F4], transported through [F1], shows this is a field homomorphism extending $σ$ and sending $α$ to $β$ .

F1F3F4construct

2.1

Repeating the construction for $σ^{- 1}$ , with $β$ and $α$ interchanged, gives an inverse. Hence $\tilde{σ}$ is an isomorphism. Its values on the unique power-basis expressions are forced, so it is unique.

F3step 1.2discharge-construct∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 45 results over 8 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

T. Judson, Abstract Algebra: Theory and Applications, Theorem 21.13 (standard reference, not scraped)