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Minkowski bound for Gaussian integers

Example

Assume the Axiom of Choice. For K=Q(i) the Minkowski constant is MK=4/π<2, so every class in Cl⁡(Z[i]) has an integral representative of norm at most MK, hence of norm 1 and equal to Z[i] itself; the class group of the Gaussian integers is trivial and Z[i] is a principal ideal domain.

Facts & Assumptions

Given: The Axiom of Choice and the imaginary quadratic field K=Q(i) with OK=Z[i] and discriminant dK=−4.

[F1]

For the squarefree integer d=−1, which is 3(mod4), the quadratic-field formulas give OK=Z[i] and dK=4⋅(−1)=−4 (Integers in a quadratic field, Discriminant of a quadratic field).

[F2]

Signature: r1 is the number of field embeddings K→R fixing Q and r2 is the number of complex-conjugate pairs among the nonreal field embeddings K→C fixing Q, with r1+2r2=[K:Q] (Archimedean embeddings and signature).

[F3]

Minkowski bound: every class of Cl⁡(OK) contains an integral ideal b with Nb≤MK (Minkowski bound for ideal classes, The ideal class group).

[F4]

For a nonzero integral ideal a the norm Na=∣OK/a∣ is a finite positive integer, so Na≥1, and Na=1 forces a=OK (The absolute norm of an integral ideal, A nonzero number-field ideal has finite quotient).

[F5]

Gregory-Leibniz: π/4=1−1/3+R1 with R1=∫01x4/(1+x2) dx>0, hence π>8/3>2 (The Gregory-Leibniz series: pi over four equals 1-1/3+1/5-1/7+...).

[F6]

Under the Axiom of Choice, the ring of integers of a number field is a Dedekind domain (Rings of integers are Dedekind domains).

[F7]

Dedekind PID criterion: a Dedekind domain R is a principal ideal domain if and only if its ideal class group is trivial (A Dedekind domain is a PID exactly when its class group is trivial).

Proof

1.1F1F2algebra

By [F1], OK=Z[i] and dK=−4. Every field embedding K→C fixing Q sends i to a root of X2+1, that is, to ±i, and both of these are nonreal; so (r1,r2)=(0,1) with n=2 by [F2].

2.1F5step 1.1algebra

Minkowski constant: by step 1.1 and [F1], MK=(4/π)12!224=4π⋅12⋅2=4π; by [F5] π>8/3>2, so MK=4/π<2.

3.1F3F4step 2.1

Every class of Cl⁡(OK) contains an integral ideal b with Nb≤MK<2 by [F3] and step 2.1. By [F4] the norm Nb is a positive integer, so Nb=1 and therefore b=OK, which is principal; hence every class is the principal class and Cl⁡(Z[i]) is trivial.

4.1F6F7step 3.1

By [F6] the Gaussian integers Z[i]=OK form a Dedekind domain, so the criterion [F7] applies and the triviality of the class group from step 3.1 makes Z[i] a principal ideal domain.

5.1step 1.1step 2.1step 3.1step 4.1∎

In summary, K=Q(i) has MK=4/π<2, Cl⁡(Z[i]) is trivial, and Z[i] is a principal ideal domain.

Remarks

This is the smallest case of the Minkowski bound: the signature (0,1) contributes the factor 4/π rather than none, but the unique class bound 4/π<2 still falls below the smallest norm of a nonzero nonunit ideal, so the bound certifies that the class group is trivial without any further computation. The passage from a trivial class group to the principal ideal domain property uses that Z[i] is Dedekind, since a general domain with trivial class group need not be a PID.

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