Squarefree Discriminant Gives an Integral Basis

Theorem

Let \(\alpha_1, \dots, \alpha_n \in \mathcal{O}_\mathbb{K}\) be a basis for a number field \(\mathbb{K}\). If \(\mathrm{disc}(\alpha_1, \dots, \alpha_n)\) is squarefree, then \(\alpha_1, \dots, \alpha_n\) is also an integral basis.

The converse of this is not true, and thus cannot be used to exclude possible integral bases. See integral basis algorithm.

Proof

Let \(\alpha_1, \dots, \alpha_n \in \mathcal{O}_\mathbb{K}\) be a basis for a number field \(\mathbb{K}\), and let \(\beta_1, \dots, \beta_n\) be an integral basis of \(\mathcal{O}_\mathbb{K}\). Then, using this result, we have that

\[ \mathrm{disc}(\alpha_1, \dots, \alpha_n) = \det(M)^2\mathrm{disc}(\beta_1, \dots, \beta_n)\]

where \(M = (m_{ij})_{1 \leq i, j \leq n} \in \mathcal{M}_n(\mathbb{Q})\) is given by

\[ \alpha_i = \sum_{j = 1}^n m_{ij}\beta_j.\]

However since \(\mathrm{disc}(\alpha_1, \dots, \alpha_n) = \det(M)^2\) is squarefree, this means that both \(\det(M)^2\) and \(\mathrm{disc}(\beta_1, \dots, \beta_n)\) must be squarefree. Since the only squarefree square is \(1\), this means that \(\det(M)^2 = 1\) and hence

\[ \mathrm{disc}(\alpha_1, \dots, \alpha_n) = \mathrm{disc}(\beta_1, \dots, \beta_n) = \mathrm{disc}(\mathbb{K})\]

which implies that \(\alpha_1, \dots, \alpha_n\) is an integral basis.