In an algebraic number field, a basis of algebraic integers which has squarefree determinant is integral

Let A = \{\alpha_i\}_{i=1}^n be a basis for an algebraic number field \mathbb{Q}(\theta) \neq \mathbb{Q} consisting of algebraic integers such that \Delta[A] is squarefree. (Recall that since A consists of algebraic integers, \Delta[A] is a rational integer.) Prove that A is an integral basis.

Let B = \{\beta_i\} be an integral basis for \mathbb{Q}(\theta). (B exists by Theorem 6.9 in TAN.) Since each \alpha_i is an algebraic integer, there exist rational integers c_{i,j} such that \alpha_i = \sum_j c_{i,j} \beta_j for each i.

Let \theta_k be the conjugates of \theta and let \alpha_i = p_i(\theta) and \beta_i = q_i(\theta). Now p_i(\theta) = (\sum c_{i,j} q_j)(\theta), so that certainly \alpha_i^{(k)} = \sum_j c_{i,j} \beta_j^{(k)}, where ^{(k)} denotes the kth conjugate of \alpha_i and \beta_j. In particular, we have V(A) = M V(B), where M = [c_{i,j}]. So \Delta[A] = \mathsf{det}(M)^2 \Delta[B]. Certainly \mathsf{det}(M)^2 is a rational integer.

Since \Delta[A] is squarefree, in fact \mathsf{det}(M)^2 = 1, and so \Delta[A] = \Delta[B]. Since B is an integral basis, it has minimal discriminant among the bases of \mathbb{Q}(\theta) (by Theorems 6.9 and 6.10 in TAN), as does A. So A is an integral basis.

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