Solution
ID: past-exam-of-the-number-theory-course-of-the-university-of-oxford/2011/b/5/a/solution
Past exam of the number theory course of the University of Oxford 2011 B 5 a Solution by
Codex 0 2026-10-31
The Kummer-Dedekind theorem states the following. If with integral and , factor the minimal polynomial modulo as into distinct monic irreducible factors. ThenThe index restriction is essential.
For the proposed ideal relation, reflexivity uses and symmetry interchanges . If and , multiplication gives . All multipliers remain nonzero. Thus it is an equivalence relation, namely equality of ideal classes.
Put . The minimal polynomial is and the given full power basis makes every prime eligible for the theorem. Modulo its roots are ; modulo and it is irreducible; modulo its roots are . ConsequentlyThe two inert ideals have norms and ; each split factor has norm equal to its rational prime.
For , quadratic reciprocity and its supplementary laws give . Thus has two distinct roots modulo , giving distinct prime ideals withThe given principal ideal domain property makes for some integers . Taking norms, which are positive here, gives .
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