By Robin Chapman

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If all ideals of OK are principal, for instance if K = Q, then so are all fractional ideals, for the fractional ideal βI = βγ if I = γ . We define the sum and product of fractional ideals in the same way as for ideals. In particular if β ∈ K, β = 0 then β 1/β = 1 = OK , so that principal fractional ideals are invertible. We shall show that all fractional ideals are invertible. We start with an alternative characterization of fractional ideals. 7 Let K be a number field. Then I is a fractional ideal of K if and only if • I is a nonzero subgroup of K under addition, 31 • if β ∈ I and γ ∈ OK then γβ ∈ I, and • there is a nonzero η ∈ K such that β/η ∈ OK for each β ∈ I.

It suffices to show that |J : P J| = N (P ). ). Let β ∈ J with β ∈ / P J. ). Let γ1 , . . , γm be a system of coset representatives for P in OK , so that m = N (P ). We shall show that βγ1 , . . , βγm form a system of coset representatives for P J in J. If δ ∈ J then δ = ξβ + η with ξ ∈ OK and η ∈ P J as J = β + P J. Now ξ − γk ∈ P for some k, and so β(ξ − γk ) ∈ P J. Hence δ ≡ βγk (mod P J) so that each coset of P J in J is represented by some βγk . We need to show that the βγk represent distinct cosets of P J.

Define a sequence β0 , β1 , β2 , . . of elements of OK as follows. Let β0 = 1. Suppose that βn has been defined. Then choose βn+1 to be some number in OK 53 with βn+1 > βn and |N (βn+1 )| < A. Consider the ideals β0 , β1 , β2 , . .. 10 only a finite number of different ideals can occur. Hence there exist j < k with βj = βk . Thus βk = ξβj where ξ ∈ U (OK ), and ξ > 1 as βk > βj . In fact the structure of the unit group of OK is easy to determine. 3 Let K be a real quadratic field. There exists η ∈ OK such that η > 1 and such that every unit in OK has the form ±η j where j ∈ Z.