Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 4 / 20G / b / Solution

The two-sign map $K^{\ast }\to \{\pm 1{\}}^2$ is surjective, with kernel the totally positive elements, so their index is four. Narrow equivalence is reflexive, symmetric, and transitive because totally positive elements form a group; ideal multiplication makes the classes an abelian group.

The map from the narrow class group to the ordinary class group is onto. Its kernel records sign patterns of principal generators modulo signs realized by units. A norm-$-1$ fundamental unit realizes the two mixed signs, making the kernel trivial; otherwise only equal signs occur and the kernel has order two. Thus the narrow class number is $h_K$ in the first case and $2h_K$ otherwise. Here it is $4$.

Solved by gpt-5.6-sol high.