CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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An element is algebraic over $F$ if and only if its simple extension $F (a) / F$ is finite

Statement

For an element $a$ of an extension $K / F$ ,

$a is algebraic over F ⟺ F (a) / F is finite .$

If $a$ is algebraic with minimal polynomial of degree $n$ , then $[F (a) : F] = n$ .

Facts & Assumptions

Given: A field extension $K / F$ and an element $a \in K$ .

[L1]

If $a$ is algebraic with minimal polynomial of degree $n$ , then $1, a, \dots, a^{n - 1}$ is a basis of $F (a) / F$ and its degree is $n$ (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ ).

[L2]

Every element of a finite extension is algebraic over the base (Every finite field extension is algebraic).

Proof

technique · direct

1.1

If $a$ is algebraic, [L1] gives a finite power basis of $F (a)$ and the stated degree.

givenL1

1.2

Conversely, if $F (a) / F$ is finite, [L2] says every element of $F (a)$ , in particular $a$ , is algebraic over $F$ .

givenL2

2.1

These are the two implications of the equivalence.

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 42 results over 7 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)

An element is algebraic over FF if and only if its simple extension F(a)/FF(a)/F is finite