LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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 $f(ma) = m f(a)$ for $m \in \mathbb{Z}$ , 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 $f(a - b) = f(a) - f(b)$ ;
$f(ma) = m\,f(a)$ for every $a \in R$ and every $m \in \mathbb{Z}$ , the multiples being those of Integer multiples in a ring: $(m + n)a = ma + na$ , $m(a + b) = ma + mb$ , $(ma)b = m(ab) = a(mb)$ and $(ma)(nb) = (mn)(ab)$ for all $m, n \in \mathbb{Z}$ 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 $f(x+y) = f(x)+f(y)$ , $f(xy) = f(x)f(y)$ 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]

$f(xy) = f(x)f(y)$ 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]

$g(yz) = g(y)g(z)$ 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 $n \in \mathbb{Z}$ ; read additively, $h(0) = 0$ , $h(-a) = -h(a)$ and $h(na) = n\,h(a)$ (A group homomorphism automatically satisfies $f(e) = e'$ and $f(g^{-1}) = f(g)^{-1}$ , and $f(g^{n}) = f(g)^{n}$ for every $n \in \mathbb{Z}$ ; 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 $ma$ in a ring is the integer power of $a$ in its additive group, read additively (Integer multiples in a ring: $(m + n)a = ma + na$ , $m(a + b) = ma + mb$ , $(ma)b = m(ab) = a(mb)$ and $(ma)(nb) = (mn)(ab)$ for all $m, n \in \mathbb{Z}$ 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 $vu = 1$ or $uv = 1$ 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 $ab \in T$ 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 $ab \in S$ 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

1.1

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.

A1L1

1.2

Claim 5: for $x, y \in R$ , $(g \circ f)(x+y) = g(f(x)+f(y)) = g(f(x)) + g(f(y))$ and $(g\circ f)(xy) = g(f(x)f(y)) = g(f(x))g(f(y))$ , while $(g \circ f)(1_R) = g(1_S) = 1_T$ . The identity map satisfies the three conditions trivially.

A1A2A3A4A5A6

1.3

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.

A2A3L4

2.1

Claim 1: [L2] read additively gives $f(0_R) = 0_S$ and $f(-a) = -f(a)$ ; hence $f(a-b) = f(a + (-b)) = f(a) + f(-b) = f(a) - f(b)$ .

step 1.1A1L2

2.2

Claim 2: by [L3] the multiple $ma$ 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.

step 1.1L2L3

3.1

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 $f(a)f(b) = f(ab) \in f(R)$ by [A2]. So $f(R)$ satisfies the subring criterion.

step 2.1A2A3L5

4.1

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

step 1.2step 2.1step 2.2step 1.3step 3.1∎

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 $n \in \mathbb{Z}$ ; 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 $n \in \mathbb{Z}$ ; for monoid homomorphisms preservation of the identity must be assumed with additive notation; nothing is proved twice.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 57 results over 18 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

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

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