Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16
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.

(R/I)R(R/J)R/(I+J) for ideals of a commutative ring

Example

Let I,J be ideals of a commutative ring R. There is a canonical isomorphism

(R/I)R(R/J)R/(I+J).

This includes I=0, J=0, I=R, and J=R.

Facts & Assumptions

Given: A commutative ring R and ideals I,JR.

[L1]

For a right R-module M, MRR/JM/JM (MRR/IM/IM naturally).

[L2]

I+J={i+j:iI, jJ} is an ideal (The sum I+J and product IJ of two-sided ideals).

[L3]

A module homomorphism induces an isomorphism from its quotient by its kernel to its image (First isomorphism theorem for modules: M/kerfimf).

Verification

technique · direct
1.1

Apply [L1] with M=R/I to obtain (R/I)R(R/J)(R/I)/J(R/I).

givenL1
1.2

The map q:R/IR/(I+J) given by q(r+I)=r+(I+J) is well-defined and surjective. Its kernel consists of the classes i+j+I=j+I with iI and jJ, which is exactly J(R/I).

givenL2algebra
2.1

By [L3], step 1.2 induces (R/I)/J(R/I)R/(I+J); composing with step 1.1 proves the displayed isomorphism.

step 1.1step 1.2L3
3.1

If I=0 or J=0, the formula reduces to the appropriate tensor-unit isomorphism. If either ideal is R, then both sides are zero. Thus all stated boundary cases are included.

step 2.1L2

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 31 results over 9 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