TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The elements separable over the base form an intermediate field

Statement

For an algebraic extension $K / F$ , the set

$K_{s} : = {a \in K : a is separable over F}$

is an intermediate field between $F$ and $K$ .

Facts & Assumptions

Given: An algebraic extension $K / F$ and separable elements $a, b \in K$ .

[L1]

An algebraic extension generated by separable elements is separable (An algebraic extension generated by separable elements is separable).

[L2]

The subfield criterion requires $0, 1$ , closure under subtraction and multiplication, and inverses of nonzero elements (Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations).

Proof

technique · direct

1.1L1

The extension $F (a, b) / F$ is generated by separable elements, so [L1] makes every element of $F (a, b)$ separable over $F$ .

2.1step 1.1algebra

In particular, $a - b$ and $a b$ are separable, and if $a \neq 0$ then $a^{- 1}$ is separable. The elements $0$ and $1$ lie in $F$ and have linear minimal polynomials, so they are separable.

3.1step 2.1L2∎

The set $K_{s}$ therefore satisfies the subfield criterion [L2] and contains $F$ , so it is an intermediate field.

The elements separable over the base form an intermediate field

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources