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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
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A field's prime subfield is isomorphic to Q in characteristic zero and to Fp in characteristic p

Statement

Let F0 be the prime subfield of a field F.

  1. If char⁡F=p>0, then F0 is isomorphic to Fp=Z/p.
  2. If char⁡F=0, then F0 is isomorphic to Q.

Each isomorphism sends 1 to 1F.

Facts & Assumptions

Given: A field F and its prime subfield F0.

[L1]

The prime subfield is the intersection of all subfields of F and hence the smallest one (The prime subfield as the intersection of all subfields).

[L2]

The characteristic of F is zero or prime (The characteristic of a field is zero or a prime number).

[L3]

For prime p, the quotient Z/p is a field (For every prime p, the two operations on Z/p make it a field).

[L4]

The rational numbers form a field (The rationals form a field).

[L5]

A field homomorphism preserves addition, multiplication and 1 (Field homomorphism and embedding); it is therefore injective, its kernel being an ideal of a field that does not contain 1.

Proof

technique · direct
1.1givenL2L3L5algebra

Suppose char⁡F=p>0. The map Z/p→F given by [n]↦n⋅1F is well defined, is a field homomorphism, and is injective; its image is a subfield contained in every subfield of F.

1.2givenL2L4L5algebra

Suppose char⁡F=0. The map Z→F, n↦n⋅1F, is injective. Sending a rational class a/b with b≠0 to (a⋅1F)(b⋅1F)−1 is well defined and gives an injective field homomorphism Q→F.

2.1step 1.1L1

By [L1], that image equals F0, proving the first classification.

2.2step 1.2L1

Its image is a subfield and every subfield of F contains all integer multiples of 1F and their nonzero quotients. Hence the image is contained in every subfield and equals F0 by [L1].

3.1step 2.1step 2.2L2∎

Steps 2.1 and 2.2 exhaust the alternatives in [L2].

Depends on

Used by

Dependency tree · two levels

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Sources