I have an algebraic number ring $\mathbb{A}$ of degree $n$ and discriminant $d$. Suppose I have a set $S = \{\alpha_1,\dots \alpha_n\}$ with the property that:
- Every $\alpha_i$ is an algebraic integer
- As $\mathbb{Z}$-module, it is rank $n$
- The discriminant of the set is equal to $d$ (that is, the square of the determinant of the matrix generated by the $n$ embeddings of the set into the complex plane.)
Then is it always true that $S$ is also an integral basis for $\mathbb{A}$?
I can see that it is true for the quadratic case $\mathbb{Q}[\sqrt{r}]$. And I think you can make some argument that you can invert the basis over $\mathbb{Z}$ without changing the discriminant. But I'm not confident enough as it stands. Help appreciated.