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Number Fields Rings of Integers and Discriminants — Examples

1 · Prerequisites

2 · Summary

The examples calculate integral bases and discriminants while keeping the index of a power order explicit. The pure-cubic entry records that a squarefree-discriminant maximality certificate is unavailable there; no nonmonogenic field is asserted without a complete source.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Integers of Q

Example

$O_{Q} = Z$ and $d_{Q} = 1$ .

Facts & Assumptions

Given: A rational number $a / b$ in lowest terms.

Verification

technique · direct

1.1given

Its monic minimal polynomial is $X - a / b$ , integral only when $b = 1$ .

2.1step 1.1algebra∎

The basis $(1)$ has trace Gram determinant $1$ .

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Gaussian and Eisenstein integral bases

Example

$O_{Q (i)} = Z [i]$ , $d = - 4$ ; and $O_{Q (\sqrt{- 3})} = Z [(1 + \sqrt{- 3}) / 2]$ , $d = - 3$ .

Facts & Assumptions

Given: The quadratic integer and discriminant formulas (Integers in a quadratic field, Discriminant of a quadratic field).

Verification

technique · direct

1.1given

Insert $d = - 1$ and $d = - 3$ in the given integral-basis formula.

2.1step 1.1given∎

Insert the same residues modulo $4$ in the given discriminant formula.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

The integral basis and discriminant of Q(sqrt 5)

Example

Put $K = Q (\sqrt{5})$ and $ω = (1 + \sqrt{5}) / 2$ . Then $O_{K} = Z [ω]$ , $(1, ω)$ is an integral basis, and $disc (K) = 5$ . The suborder $Z [\sqrt{5}]$ has basis $(1, \sqrt{5})$ , index $2$ , and discriminant $20$ .

Facts & Assumptions

Given: $K = Q (\sqrt{5})$ and $ω = (1 + \sqrt{5}) / 2$ .

[F1]

For squarefree $d \neq 1$ , the quadratic integral-basis formula applies (Integers in a quadratic field).

[F2]

The corresponding quadratic-field discriminant is $d$ when $d \equiv 1 (m o d 4)$ (Discriminant of a quadratic field).

[F3]

An order $A \subseteq O_{K}$ satisfies $disc (A) = [O_{K} : A]^{2} disc (K)$ (Order-index discriminant formula).

Verification

technique · direct

1.1F1F2given

Since $5 \equiv 1 (m o d 4)$ , [F1] gives $O_{K} = Z [ω]$ ; [F2] gives $disc (K) = 5$ .

2.1F3step 1.1algebra∎

In the basis $(1, ω)$ , one has $\sqrt{5} = 2 ω - 1$ , so the change matrix to $(1, \sqrt{5})$ has determinant $2$ . Thus $[O_{K} : Z [\sqrt{5}]] = 2$ , and [F3] gives $disc (Z [\sqrt{5}]) = 2^{2} \cdot 5 = 20$ .

ExampleConstruction: AI-generatedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

A pure cubic power basis: why the squarefree-discriminant certificate does not apply

Example

Let $α = \sqrt[3]{2}$ and $K = Q (α)$ . The power basis $(1, α, α^{2})$ has discriminant $- 108$ . Consequently the squarefree power-basis criterion does not certify that $Z [α]$ is the full ring of integers. This example records the boundary of that criterion; it makes no claim here about the actual index of $Z [α]$ .

Facts & Assumptions

Given: $α = \sqrt[3]{2}$ and its polynomial $f (X) = X^{3} - 2$ .

[F1]

The discriminant of a power basis is the polynomial discriminant of the minimal polynomial (Power-basis and polynomial discriminants).

[F2]

The squarefree criterion concludes maximality only when the power-basis discriminant is squarefree (Squarefree power discriminant criterion).

Verification

technique · direct

1.1given

Eisenstein's criterion at $2$ makes $f$ irreducible over $Q$ ; thus it is the monic minimal polynomial of the integral element $α$ and $(1, α, α^{2})$ is a $Q$ -basis of $K$ .

2.1F1step 1.1algebra

The cubic discriminant formula gives $disc (f) = - 27 (- 2)^{2} = - 108$ . By [F1] this is the discriminant of the displayed power basis.

3.1F2step 2.1∎

Since $- 108$ is divisible by $2^{2}$ and $3^{3}$ , it is not squarefree. Therefore the hypothesis of [F2] fails, so that criterion supplies no maximality conclusion in this example.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

The nonmaximal quadratic order Z[sqrt 5] inside O_Q(sqrt 5)

Example

For $K = Q (\sqrt{5})$ , the subring $A = Z [\sqrt{5}]$ is an order in $O_{K} = Z [(1 + \sqrt{5}) / 2]$ . It has index $2$ and $disc (A) = 20 = 2^{2} \cdot 5$ .

Facts & Assumptions

Given: $K = Q (\sqrt{5})$ and $A = Z [\sqrt{5}]$ .

[F1]

An order is a unital full-rank subring of $O_{K}$ (Order in a number field).

[F2]

The integral basis $(1, ω)$ with $ω = (1 + \sqrt{5}) / 2$ has field discriminant $5$ (The integral basis and discriminant of Q(sqrt 5)).

[F3]

The order-index discriminant formula applies to $A \subseteq O_{K}$ (Order-index discriminant formula).

Verification

technique · direct

1.1F1F2given

Since $\sqrt{5} = 2 ω - 1$ , $A$ is the subring generated by an integral element and has full-rank basis $(1, \sqrt{5})$ ; hence [F1] makes it an order.

2.1F2F3step 1.1algebra∎

Relative to $(1, ω)$ , the basis $(1, \sqrt{5})$ has determinant $2$ . Thus $[O_{K} : A] = 2$ , and [F3] with [F2] gives $disc (A) = 2^{2} \cdot 5 = 20$ .

ExampleConstruction: AI-generatedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Index 2 obstructs reading factorisation from Z[sqrt 5] modulo 2

Example

For $A = Z [\sqrt{5}] ⊊ O_{K}$ with $K = Q (\sqrt{5})$ , reduction of the power polynomial $X^{2} - 5$ modulo $2$ gives $(X + 1)^{2}$ . But $O_{K} / 2 O_{K} ≅ F_{2} [X] / (X^{2} + X + 1)$ , which is a field. Thus the repeated factor in the nonmaximal power order is not a factorisation assertion about $O_{K}$ .

Facts & Assumptions

Given: $K = Q (\sqrt{5})$ , $A = Z [\sqrt{5}]$ , and $ω = (1 + \sqrt{5}) / 2$ .

[F1]

The order $A$ has index $2$ in $O_{K}$ (The nonmaximal quadratic order Z[sqrt 5] inside O_Q(sqrt 5)).

[F2]

The index-discriminant formula detects this nonmaximality (Order-index discriminant formula).

Verification

technique · direct

1.1F1givenalgebra

In $A / 2 A$ , the class of $\sqrt{5}$ satisfies $X^{2} - 5 \equiv X^{2} + 1 = (X + 1)^{2}$ , so $A / 2 A ≅ F_{2} [X] / ((X + 1)^{2})$ has a nonzero nilpotent.

2.1F1step 1.1algebra

The element $ω$ satisfies $ω^{2} - ω - 1 = 0$ . Hence $O_{K} / 2 O_{K} ≅ F_{2} [X] / (X^{2} + X + 1)$ ; the polynomial has no root in $F_{2}$ , so this quotient is a field.

3.1F2step 1.1step 2.1∎

The two quotient rings cannot agree, and [F2] identifies the reason as the index divisible by $2$ . Therefore reduction of the power polynomial in $A$ cannot by itself describe factorisation in the maximal order.

RemarkRemark: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Nonmonogenic example source obligation

The planned concrete nonmonogenic-field slot is deliberately not asserted. The opened sources contain only a general warning here; a named field needs a separate complete source and proof before it can be authored.

Number Fields Rings of Integers and Discriminants — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Integers of Q

Example

Facts & Assumptions

Verification

Gaussian and Eisenstein integral bases

Example

Facts & Assumptions

Verification

The integral basis and discriminant of Q(sqrt 5)

Example

Facts & Assumptions

Verification

A pure cubic power basis: why the squarefree-discriminant certificate does not apply

Example

Facts & Assumptions

Verification

The nonmaximal quadratic order Z[sqrt 5] inside O_Q(sqrt 5)

Example

Facts & Assumptions

Verification

Index 2 obstructs reading factorisation from Z[sqrt 5] modulo 2

Example

Facts & Assumptions

Verification

Nonmonogenic example source obligation

Sources