Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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.

Module endomorphisms form a ring under pointwise addition and composition

Statement

For every left R-module M, pointwise addition and composition make EndR(M) a unital ring with identity idM. See The endomorphism ring EndR(M) under addition and composition.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a left R-module M, define EndR(M):=HomR(M,M). Addition is pointwise and multiplication is composition, (fg)(m):=f(g(m)). The ring laws and the identity endomorphism are established in prop-endomorphisms-form-a-ring. (The endomorphism ring EndR(M) under addition and composition).

[L2]

For left R-modules M,N, the set HomR(M,N) of module homomorphisms is an abelian group under pointwise addition, with zero the zero homomorphism and inverse (f)(m)=f(m) (def-module-homomorphism-kernel-image-and-cokernel, def-group). (The abelian group HomR(M,N) and maps induced by pre- and postcomposition).

[L3]

For left R-modules M,N, a function f:MN is an R-module homomorphism if f(m+m)=f(m)+f(m) and f(rm)=rf(m) for all m,mM and rR (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · direct
1.1

If f,gEndR(M) then fg is again a module homomorphism, since (fg)(m+n)=f(g(m)+g(n))=(fg)(m)+(fg)(n) and (fg)(rm)=f(rg(m))=r(fg)(m). So composition is a binary operation on EndR(M).

L1L3givenalgebra
1.2

Taking N=M in [L2] makes (EndR(M),+) an abelian group, with the zero homomorphism as neutral element and (f)(m)=f(m) as the inverse of f.

L1L2
1.3

Composition is associative: for all m, (f(gh))(m)=f(g(h(m)))=((fg)h)(m).

L1algebra
1.4

Both distributive laws hold. For all m, (f(g+h))(m)=f(g(m)+h(m))=f(g(m))+f(h(m))=(fg+fh)(m), where the middle equality is additivity of f; and ((g+h)f)(m)=g(f(m))+h(f(m))=(gf+hf)(m) directly from pointwise addition.

L1L2L3algebra
1.5

The identity map idM satisfies idM(m+n)=m+n and idM(rm)=rm, so it lies in EndR(M), and fidM=f=idMf for every f.

L1L3algebra
2.1

Steps 1.1 through 1.5 are exactly the axioms of a unital ring for (EndR(M),+,,idM). For the zero module the only map 00 is id0, so EndR(0) has one element and is the one-element ring, in which the identity coincides with the zero element. This proves the stated claim.

step 1.1step 1.2step 1.3step 1.4step 1.5algebra

Depends on

Used by

Cited to discharge well-definedness by The endomorphism ring End_R(M) under addition and composition.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 18 results over 15 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