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FALSE: the isomorphism between two splitting fields that fixes the base field is unique

Statement

False statement. If $E$ and $E^{'}$ are splitting fields of the same polynomial over $F$ , there is exactly one isomorphism $E \to E^{'}$ that fixes $F$ pointwise.

Facts & Assumptions

Given: The splitting field $E = Q (\sqrt{2})$ of $x^{2} - 2$ over $Q$ .

[F1]

Any two splitting fields of the same nonzero polynomial are isomorphic over the base field (Any two splitting fields of a polynomial are isomorphic over the base field).

[F2]

The splitting field of $x^{2} - 2$ over $Q$ is $E = Q (\sqrt{2})$ , with roots $\sqrt{2}$ and $- \sqrt{2}$ (The splitting field of $x^{2} - 2$ over $Q$ is $Q (\sqrt{2})$ , with roots $\pm \sqrt{2}$ ).

[F3]

An isomorphism of base fields extends across simple adjunctions when a chosen root is sent to a corresponding root of the transported irreducible polynomial (A base-field isomorphism extends across simple adjunctions of corresponding roots of an irreducible polynomial).

[F4]

Eisenstein's criterion proves a primitive integer polynomial irreducible under its prime-divisibility hypotheses (Eisenstein criterion over the integers).

Refutation

technique · counterexample

1.1

Take both splitting fields to be $E = E^{'} = Q (\sqrt{2})$ for $x^{2} - 2$ . The identity map is one $Q$ -isomorphism $E \to E$ .

1.2

The polynomial $x^{2} - 2$ satisfies [F4] with the prime $2$ , so it is irreducible. Apply [F3] to the identity on $Q$ and the corresponding roots $\sqrt{2}$ and $- \sqrt{2}$ . It gives a $Q$ -automorphism $τ : E \to E$ satisfying $τ (\sqrt{2}) = - \sqrt{2}$ .

F2F3F4

2.1

The roots are distinct: if $\sqrt{2} = - \sqrt{2}$ , then $2 \sqrt{2} = 0$ , and multiplication by $1 / 2$ would give $\sqrt{2} = 0$ , contradicting $(\sqrt{2})^{2} = 2$ . Hence $τ$ is not the identity. Thus there are at least two base-fixing isomorphisms, refuting uniqueness while leaving the existence result [F1] intact.

F1F2step 1.1step 1.2algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

18 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

T. Judson, Abstract Algebra: Theory and Applications, Corollary 21.14 (standard reference, not scraped)