Math 254a: Rings of Integers and Dedekind Domains

Proof. Let (α1, . . . , αn) be any Q-basis for K; we first claim there for each i, there exists a non-zero di ∈ Z such that diαi ∈ OK . Indeed, it is easy to check that it is enough to let di be the leading term of any integer polynomial satisfied by αi. Thus, for any non-zero β ∈ I, we find that (βd1α1, . . . , βdnαn) is a Q-basis for K contained in I. It remains to show that such a basis with minimal discriminant over Q is in fact a Z-basis for I. Note that because the discriminant of elements in OK is an integer, there is a minimal one. Take any α ∈ I, and write α = ∑ n i=1 aiαi, for ai ∈ Q. We want to show that ai ∈ Z for all i. Suppose not; without loss of generality, we may assume a1 6∈ Z. Write a1 = m+ǫ, with 0 < ǫ < 1 and m = ⌊a1⌋, and replace α1 by α ′ 1 = α − mα1, which is still in I. We then obtain a new basis inside I, and the determinant of the transition matrix is ǫ, so by the changeof-basis formula of the previous section, this new basis has smaller discriminant, contradicting minimality.