ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06

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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 $disc (Z [\sqrt{5}]) = 2^{2} \cdot 5 = 20$ .

Depends on

Used by

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

Dependency tree · two levels

10 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Algebraic Number Theory, Example 2.11 (standard reference, not scraped)