Alphabeta Math
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6 results · all verified · 0 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 6 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Ideals and Quotient Rings: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Null rational sequences form a maximal ideal in the ring of rational Cauchy sequences

Example

Null rational sequences form a maximal ideal in the ring of rational Cauchy sequences.

Facts & Assumptions

Given: The ring CC of rational Cauchy sequences and its subset NN of null sequences.

[L1]

Maximal ideals are maximal among proper ideals (Prime ideals and maximal ideals in a commutative ring).

[L2]

Rational Cauchy sequences form a commutative ring CC (Cauchy sequences form a commutative ring).

[L3]

NN is an ideal of CC (Null sequences form an ideal).

[L4]

The null ideal NN is maximal in CC (The null ideal is maximal).

Verification

technique · direct
1.1

The ambient object is the commutative ring CC of [L2], and NN is an ideal by [L3].

L1L2L3L4given
2.1

The maximality statement in [L4] is precisely maximality in the ideal order of [L1].

step 1.1L1L2L3L4given
3.1

Thus null rational sequences give a maximal ideal.

step 2.1
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The reals are the quotient of rational Cauchy sequences by the maximal ideal of null sequences

Example

The reals are the quotient of rational Cauchy sequences by the maximal ideal of null sequences.

Facts & Assumptions

Given: The Cauchy-sequence ring CC and the null ideal NN.

[L1]

The real numbers are defined as the quotient of rational Cauchy sequences by null sequences (The real numbers).

[L2]

Rational Cauchy sequences form the ring CC (Cauchy sequences form a commutative ring).

[L3]

NN is an ideal of CC (Null sequences form an ideal).

[L4]

NN is maximal in CC (The null ideal is maximal).

[L5]

Quotienting a commutative ring by a maximal ideal gives a field (R/MR/M is a field if and only if MM is a maximal ideal).

[L6]

The constructed real numbers form a field (The reals form a field).

Verification

technique · direct
1.1

By [L1], the underlying set and operations of R\mathbb R are the quotient C/NC/N.

L1L2L3L4L5L6given
2.1

Since [L3] and [L4] make NN a maximal ideal, [L5] also identifies C/NC/N as a field.

step 1.1L1L2L3L4L5L6given
3.1

This is the stated quotient realisation.

step 2.1
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

For every integer n>1n>1, nZn\mathbb Z is a maximal ideal of Z\mathbb Z if and only if nn is prime

Example

For every integer n>1n>1, nZn\mathbb Z is a maximal ideal of Z\mathbb Z if and only if nn is prime.

Facts & Assumptions

Given: An integer n>1n>1.

[L2]

A quotient is a field exactly for a maximal ideal (R/MR/M is a field if and only if MM is a maximal ideal).

[L3]

A maximal ideal in a commutative ring is prime (Every maximal ideal of a commutative ring is prime).

[L6]

Z\mathbb Z is a commutative ring (The integers form a commutative ring).

Verification

technique · direct
1.1

If nn is prime, [L5] and [L1] make Z/nZ\mathbb Z/n\mathbb Z a field, so [L2] makes nZn\mathbb Z maximal.

L1L2L3L4L5L6given
2.1

If nZn\mathbb Z is maximal, [L3] makes it prime; applying this to a factorisation n=abn=ab gives the Euclid property and [L4] makes nn prime.

step 1.1L1L2L3L4L5L6givenalgebra
3.1

The two implications prove the example.

step 2.1
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

The zero ideal of Z\mathbb Z is prime but not maximal

Statement refuted

Every prime ideal of a commutative ring is maximal.

Facts & Assumptions

Given: The zero ideal (0)(0) in Z\mathbb Z.

[L1]

A prime ideal is proper and satisfies the zero-product implication; a maximal ideal has no proper intermediate ideal (Prime ideals and maximal ideals in a commutative ring).

[L2]

The ideal criterion verifies subtraction closure and absorption (Ideal criteria and intersections of ideals).

[L3]

Z\mathbb Z is a nonzero commutative ring (The integers form a commutative ring).

[L4]

Integer cancellation implies ab=0ab=0 only if a=0a=0 or b=0b=0 (The integers have no zero divisors; multiplicative cancellation).

Counterexample

technique · direct
1.1

The set (0)(0) is a proper ideal, and [L4] shows that ab(0)ab\in(0) implies a(0)a\in(0) or b(0)b\in(0).

L1L2L3L4givenalgebra
2.1

The ideal 2Z2\mathbb Z satisfies (0)2ZZ(0)\subsetneq2\mathbb Z\subsetneq\mathbb Z.

step 1.1L1L2L3L4givenalgebra
3.1

Hence (0)(0) is prime but not maximal, refuting the statement.

step 2.1
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

2Z2\mathbb Z is an ideal of Z\mathbb Z but is not a subring under the library's unital convention

Example

2Z2\mathbb Z is an ideal of Z\mathbb Z but is not a subring under the library's unital convention.

Facts & Assumptions

Given: The subset 2ZZ2\mathbb Z\subseteq\mathbb Z.

[L1]

A two-sided ideal is an additive subgroup with multiplication absorption (Left, right and two-sided ideals).

[L2]

The ideal criterion is subtraction closure and absorption (Ideal criteria and intersections of ideals).

[L4]

Z\mathbb Z is a ring with identity 11 (The integers form a commutative ring).

Verification

technique · direct
1.1

Differences of even integers are even, and multiplying an even integer by any integer remains even, so 2Z2\mathbb Z is an ideal.

L1L2L3L4givenalgebra
2.1

The identity 11 does not belong to 2Z2\mathbb Z.

step 1.1L1L2L3L4given
3.1

Thus the missing identity rules out 2Z2\mathbb Z as a unital subring.

step 2.1
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

R×{0}R\times\{0\} is the kernel of R×SSR\times S\to S, so (R×S)/(R×{0})S(R\times S)/(R\times\{0\})\cong S

Example

R×{0}R\times\{0\} is the kernel of R×SSR\times S\to S, so (R×S)/(R×{0})S(R\times S)/(R\times\{0\})\cong S.

Facts & Assumptions

Given: Rings R,SR,S and the coordinate projection p:R×SSp:R\times S\to S, p(r,s)=sp(r,s)=s.

[L2]

A ring homomorphism preserves operations and identity (Ring homomorphism: additive, multiplicative, and required to send 11 to 11).

[L3]

A ring-homomorphism kernel is a two-sided ideal (The kernel of a ring homomorphism is a two-sided ideal).

[L4]

The first ring isomorphism theorem gives quotient-by-kernel isomorphisms (First isomorphism theorem for rings: R/kerfimfR/\ker f\cong\operatorname{im}f).

[L5]

Ideals are additive subgroups with absorption (Left, right and two-sided ideals).

Verification

technique · direct
1.1

Coordinatewise operations make pp a surjective ring homomorphism.

L1L2L3L4L5givenalgebra
2.1

Its kernel is exactly {(r,0):rR}=R×{0}\{(r,0):r\in R\}=R\times\{0\}, which is therefore an ideal.

step 1.1L1L2L3L4L5givenalgebra
3.1

The kernel calculation yields (R×S)/(R×{0})S(R\times S)/(R\times\{0\})\cong S.

step 2.1

Sources