LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

How statement and proof provenance work

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 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

Statement

Let $R$ and $S$ be rings (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides) and $f : R \to S$ a ring homomorphism (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ). Then:

$f (0_{R}) = 0_{S}$ and $f (- a) = - f (a)$ for every $a \in R$ ; consequently $f (a - b) = f (a) - f (b)$ ;
$f (m a) = m f (a)$ for every $a \in R$ and every $m \in Z$ , the multiples being those of Integer multiples in a ring: $(m + n) a = m a + n a$ , $m (a + b) = m a + m b$ , $(m a) b = m (a b) = a (m b)$ and $(m a) (n b) = (m n) (a b)$ for all $m, n \in Z$ and $a, b \in R$ ;
if $u \in R^{\times}$ then $f (u) \in S^{\times}$ and $f (u)^{- 1} = f (u^{- 1})$ ;
the image $f (R) = {f (a) : a \in R}$ is a subring of $S$ (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication);
if $g : S \to T$ is a ring homomorphism then so is $g \circ f : R \to T$ , and the identity map of $R$ is a ring homomorphism.

Facts & Assumptions

Given: Rings $R$ , $S$ with zeros $0_{R}$ , $0_{S}$ and identities $1_{R}$ , $1_{S}$ , and a function $f : R \to S$ with $f (x + y) = f (x) + f (y)$ , $f (x y) = f (x) f (y)$ and $f (1_{R}) = 1_{S}$ . For claim 5, let $T$ be a ring and let $g : S \to T$ be a ring homomorphism (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ).

[A1]

$f (x + y) = f (x) + f (y)$ for all $x, y \in R$ .

[A2]

$f (x y) = f (x) f (y)$ for all $x, y \in R$ .

[A3]

$f (1_{R}) = 1_{S}$ .

[A4]

$g (y + z) = g (y) + g (z)$ for all $y, z \in S$ (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ).

[A5]

$g (y z) = g (y) g (z)$ for all $y, z \in S$ (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ).

[A6]

$g (1_{S}) = 1_{T}$ (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ).

[L1]

$(R, +, 0_{R})$ and $(S, +, 0_{S})$ are abelian groups, and by [A1] the map $f$ is a homomorphism of these groups in the sense of Monoid homomorphism and group homomorphism (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L2]

A group homomorphism $h$ satisfies $h (e) = e^{'}$ , $h (g^{- 1}) = h (g)^{- 1}$ and $h (g^{n}) = h (g)^{n}$ for every $n \in Z$ ; read additively, $h (0) = 0$ , $h (- a) = - h (a)$ and $h (n a) = n h (a)$ (A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed, Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L3]

The integer multiple $m a$ in a ring is the integer power of $a$ in its additive group, read additively (Integer multiples in a ring: $(m + n) a = m a + n a$ , $m (a + b) = m a + m b$ , $(m a) b = m (a b) = a (m b)$ and $(m a) (n b) = (m n) (a b)$ for all $m, n \in Z$ and $a, b \in R$ , Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L4]

A unit of a ring is an invertible element of its multiplicative monoid, its inverse is unique, and a single equation $v u = 1$ or $u v = 1$ determines it (Left inverse, right inverse, and invertible element of a monoid, In a monoid, a left inverse and a right inverse of the same element are equal; hence an invertible element has exactly one inverse, and it is two-sided, The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring).

[L5]

Subring criterion: $T \subseteq S$ is a subring exactly when $1_{S} \in T$ and $a - b \in T$ and $a b \in T$ for all $a, b \in T$ (Subring criterion: $S \subseteq R$ is a subring if and only if $1_{R} \in S$ and $a - b \in S$ and $a b \in S$ for all $a, b \in S$ ; and an intersection of subrings is a subring, Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication).

Proof

technique · direct

1.1A1L1

By [L1] the map $f$ is a homomorphism from the additive group of $R$ to the additive group of $S$ , so [L2] applies to it.

1.2A1A2A3A4A5A6

Claim 5: for $x, y \in R$ , $(g \circ f) (x + y) = g (f (x) + f (y)) = g (f (x)) + g (f (y))$ and $(g \circ f) (x y) = g (f (x) f (y)) = g (f (x)) g (f (y))$ , while $(g \circ f) (1_{R}) = g (1_{S}) = 1_{T}$ . The identity map satisfies the three conditions trivially.

1.3A2A3L4

Claim 3: let $u \in R^{\times}$ with inverse $u^{- 1}$ , so $u u^{- 1} = 1_{R} = u^{- 1} u$ . Applying $f$ and using [A2] and [A3], $f (u) f (u^{- 1}) = f (1_{R}) = 1_{S} = f (u^{- 1}) f (u)$ . So $f (u)$ has the two-sided inverse $f (u^{- 1})$ in $S$ , hence $f (u) \in S^{\times}$ , and $f (u)^{- 1} = f (u^{- 1})$ because inverses in a monoid are unique.

2.1step 1.1A1L2

Claim 1: [L2] read additively gives $f (0_{R}) = 0_{S}$ and $f (- a) = - f (a)$ ; hence $f (a - b) = f (a + (- b)) = f (a) + f (- b) = f (a) - f (b)$ .

2.2step 1.1L2L3

Claim 2: by [L3] the multiple $m a$ is the integer power of $a$ in $(R, +, 0_{R})$ , and $m f (a)$ is the integer power of $f (a)$ in $(S, +, 0_{S})$ ; so the claim is the third part of [L2] read additively.

3.1step 2.1A2A3L5

Claim 4: $1_{S} = f (1_{R}) \in f (R)$ by [A3]; for $f (a), f (b) \in f (R)$ we have $f (a) - f (b) = f (a - b) \in f (R)$ by step 2.1 and $f (a) f (b) = f (a b) \in f (R)$ by [A2]. So $f (R)$ satisfies the subring criterion.

4.1step 1.2step 2.1step 2.2step 1.3step 3.1∎

Claims 1 to 5 are established in steps 2.1, 2.2, 1.3, 3.1 and 1.2.

Remarks

Claim 1 is free, claim 3 is not. $f (0_{R}) = 0_{S}$ comes from additivity alone, because the additive structure is a group and cancellation is available there (A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed). The multiplicative analogue would be $f (1_{R}) = 1_{S}$ , and it is not free: it is axiom (RH3) of Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ , and the step proving claim 3 above uses it twice.
The image is a subring, not merely a non-unital one, and that too rests on (RH3): without it the image would still be closed under subtraction and multiplication but might miss $1_{S}$ , and the companion page's map $n \mapsto (n, 0)$ has exactly that image.
Claim 2 is a dictionary entry, not new arithmetic. The multiples on both sides are the additive powers of Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , so the statement is the power law of A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed with additive notation; nothing is proved twice.

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A ring homomorphism satisfies f(0)=0f(0)=0, f(−a)=−f(a)f(−a)=−f(a) and f(ma)=mf(a)f(ma)=mf(a) for m∈Zm∈Z, carries units to units, and has a subring as its image; composites of ring homomorphisms are ring homomorphisms

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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