LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A ring homomorphism between fields is a field homomorphism in the published sense, and every such map is injective

Statement

Let $F$ and $G$ be fields (Field), regarded as rings by Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring, and let $f : F \to G$ be a function. Then:

$f$ is a ring homomorphism (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ) if and only if $f$ is a field homomorphism (Field homomorphism and embedding); the two definitions impose the same three conditions;
every such $f$ is injective (Injection, surjection, bijection).

So "ring homomorphism between fields" and "field homomorphism" name the same maps, and no second notion of homomorphism of fields is introduced.

Facts & Assumptions

Given: Fields $F$ and $G$ , with zeros $0_{F}, 0_{G}$ and identities $1_{F}, 1_{G}$ , and a function $f : F \to G$ (Field).

[L1]

$F$ and $G$ are commutative rings with $1 \neq 0$ , their ring operations, zeros and identities being the field ones (Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring, Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L2]

A ring homomorphism is a map satisfying (RH1) $f (x + y) = f (x) + f (y)$ , (RH2) $f (x y) = f (x) f (y)$ and (RH3) $f (1_{F}) = 1_{G}$ (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ).

[L3]

A field homomorphism is a map satisfying $φ (x + y) = φ (x) + φ (y)$ , $φ (x y) = φ (x) φ (y)$ and $φ (1_{F}) = 1_{G}$ (Field homomorphism and embedding).

[L4]

A ring homomorphism satisfies $f (a - b) = f (a) - f (b)$ (A ring homomorphism satisfies $f (0) = 0$ , $f (- a) = - f (a)$ and $f (m a) = m f (a)$ for $m \in Z$ , carries units to units, and has a subring as its image; composites of ring homomorphisms are ring homomorphisms).

[L5]

In a ring, $0 \cdot x = x \cdot 0 = 0$ (In any ring $0 \cdot a = a \cdot 0 = 0$ , $(- a) b = a (- b) = - (a b)$ , $(- a) (- b) = a b$ , $(- 1) a = - a$ and $a (b - c) = a b - a c$ ).

[L6]

In a field, every $x \neq 0$ has an inverse $x^{- 1}$ with $x^{- 1} x = 1$ , and $1_{G} \neq 0_{G}$ (Field).

[L7]

A map is injective when $f (a) = f (b)$ forces $a = b$ (Injection, surjection, bijection).

Proof

technique · direct

1.1

By [L1] the ring structures on $F$ and $G$ are the field structures, so the expressions $f (x) + f (y)$ , $f (x) f (y)$ , $1_{F}$ and $1_{G}$ mean the same thing in [L2] and in [L3].

2.1

Claim 1: the three conditions of [L2] and the three conditions of [L3] are the same three equations, so $f$ satisfies one triple exactly when it satisfies the other.

step 1.1L2L3

3.1

Let $f$ be such a map and suppose $f (a) = f (b)$ with $a \neq b$ . Then $a - b \neq 0_{F}$ , and $f (a - b) = f (a) - f (b) = 0_{G}$ by [L4].

step 2.1L4

4.1

Put $c : = (a - b)^{- 1}$ , which exists by [L6]. Then $1_{G} = f (1_{F}) = f (c (a - b)) = f (c) f (a - b) = f (c) \cdot 0_{G} = 0_{G}$ , using (RH3), (RH2) and [L5]. This contradicts $1_{G} \neq 0_{G}$ .

step 3.1L2L5L6

5.1

Hence $f (a) = f (b)$ forces $a = b$ , so $f$ is injective; with step 2.1 this proves both claims.

step 2.1step 4.1L7∎

Remarks

This is one of the page's bridges to the published vocabulary. Together with Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring and Every commutative division ring is a field, so "field" and "commutative division ring" name the same structures and the published definition and the ring-theoretic one agree, it means that everything proved on this page about rings, subrings and ring homomorphisms applies to the library's fields with no translation, and that a reader meeting Field homomorphism and embedding and Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ is meeting one notion twice, not two notions.
The injectivity proof here uses no ideals. The Remarks of Field homomorphism and embedding sketch the standard kernel-is-an-ideal argument; ideals are not defined on this page, and the argument above needs only an inverse and $0 \cdot x = 0$ .
Injectivity really does need $1_{G} \neq 0_{G}$ , that is, it needs the target to be a field rather than an arbitrary ring. The zero map from a field to the one-element ring satisfies (RH1), (RH2) and (RH3) there, since $1 = 0$ in that ring, and it is not injective.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

31 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

Ring homomorphism (Wikipedia) (standard reference, not scraped)
Field (mathematics) (Wikipedia) (standard reference, not scraped)
Thomas W. Judson, Abstract Algebra: Theory and Applications, §16.5: Ring Homomorphisms and Ideals (standard reference, not scraped)