## 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 $k$th 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.