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CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13
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Stem fields of a monic irreducible polynomial are uniquely F-isomorphic when their distinguished roots are matched

Statement

Let pF[x] be monic and irreducible. If L=F(α) and M=F(β) with p(α)=p(β)=0, then there is a unique F-isomorphism LM sending α to β.

Facts & Assumptions

Given: The polynomial, extensions, and distinguished roots in the statement.

[F1]

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

[F2]

A root of a monic irreducible polynomial determines a unique homomorphism from its adjoining quotient, fixing F and matching the root (Universal property of adjoining a root of an irreducible polynomial).

[F3]

A simple algebraic extension generated by a root is isomorphic to its minimal-polynomial quotient (A simple algebraic extension is its minimal-polynomial quotient and has power basis 1,a,,an1 and degree n).

Proof

technique · direct
1.1

By [F1], the minimal polynomials of α and β divide p; monic irreducibility of all three makes both minimal polynomials equal to p.

F1algebra
2.1

Using [F2] and the quotient identifications in [F3], obtain F-homomorphisms φ:LM and ψ:ML with φ(α)=β and ψ(β)=α.

F2F3step 1.1
3.1

The composite ψφ fixes F and α, so uniqueness in [F2] makes it the identity on L; similarly φψ is the identity on M.

F2step 2.1
4.1

Hence φ is an F-isomorphism, and the same uniqueness clause shows that no other such isomorphism can send α to β.

F2step 3.1

Depends on

Used by

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Sources