Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-123/1/solution

Write . The field trace gives a nondegenerate -bilinear trace pairing
The inverse different, or codifferent, is the trace-dual lattice
It contains , because the trace of an element integral over belongs to the integrally closed ring . It is stable under multiplication by : if and , then . Nondegeneracy of the trace pairing and finite generation of show that this trace dual is a finitely generated -module spanning . Therefore it is a fractional ideal of .
Its inverse
is the different ideal. Since , every such lies in ; hence is an integral ideal of .
The discriminant ideal is locally generated by
where is a local -basis of . Equivalently, it is the image of the determinant of the trace pairing
This formulation makes the definition independent of a basis, since changing a basis multiplies its discriminant by the square of the determinant of the change-of-basis matrix.
The asserted identity of ideals can be checked after localization at every nonzero prime ideal of . We may therefore assume that is a discrete valuation ring and choose a basis of . Let be its trace-dual basis, so ; this is a basis of . If , then
Thus the determinant measuring the inclusion is . The determinant description of the norm of a fractional ideal consequently gives
Localization then proves the equality over the original Dedekind domain.
Finally, the determinant-of-pairing map identifies the invertible -module with the discriminant ideal. Hence in the ideal class group
up to the harmless inverse caused by the convention used to identify invertible modules with fractional ideals. In either convention the class is a square.

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