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Lemma 6 — the relative norm-class group
Statement
Let be CM, let be its conjugation, put , and write . Let consist of pairs in which is a fractional ideal of and generates . Identify
Multiplication of the two entries makes the equivalence classes a group, and
Proof
There is an exact sequence
where the first and last maps are norms, a unit maps to , and maps to the ideal class of . Thus
The norm of a unit of is its square. Hence the first cokernel in (4) is a quotient of
which has at most elements by Dirichlet's unit theorem.
For the second cokernel, let denote the idele groups. The natural surjections from ideles to ordinary ideal classes commute with norms: at a finite prime this follows from the valuation formula for the local norm. They therefore induce a surjection
The second inequality of global class field theory bounds the order of the left group by . It follows that the class-group norm cokernel has order at most two. Substitution in (4) proves (2).
This uses only the bound needed for the proof. The source's additional claim that the cokernel has order two whenever is unramified at all finite places is not valid for ordinary ideal class groups without accounting for infinite places, and is not used here.
Source and dependency scope
This is Lemma 6 on physical p. 6 of the arXiv v1 manuscript. The exact sequence and its two cardinality estimates are reconstructed. Dirichlet's unit theorem and the class-field theory norm-index bound remain external inputs. The latter is Theorem 5.1(a) of Chapter VII in J. S. Milne, Class Field Theory, version 4.03 (August 6, 2020), printed p. 212 (physical p. 221). It applies because is a quadratic Galois extension. The quotient argument and source qualification above were supplied by this compilation; they are not an author-issued correction.
Used by. [[discrete_geometry/sawin_2026_explicit_lower_bound_unit_distance_problem/lemma_7|Lemma 7]].
Bears on. Problem 90.