Let $K\DeclareMathOperator{\disc}{disc}$ be a number field of degree $n$ and $\left\{ \alpha_1, \alpha_2, \cdots, \alpha_n \right\}$ be a subset of ring of integer of $K$. Prove if $ \disc(\alpha_1, \alpha_2, \cdots, \alpha_n)$ is a square-free integer, then $\left\{ \alpha_1, \alpha_2, \cdots, \alpha_n \right\}$ is an integral basis.
My Thought:
Let $d= \disc(\alpha_1, \alpha_2, \cdots, \alpha_n)$. Since $d$ is a square-free integer, $d \not= 0 \Rightarrow \left\{\alpha_1, \alpha_2, \cdots, \alpha_n \right\}$ is linearly independent over $\mathbb{Q}$.
As $\deg(K)=n$, this implies $\left\{\alpha_1, \alpha_2, \cdots, \alpha_n \right\}$ is a basis of $K$ over $\mathbb{Q}$.
Let $\left\{\beta_1, \beta_2, \cdots, \beta_n\right\}$ be integral basis of $K$ (every number field has an integral basis).
Since $\alpha_i \in \mathbb{Q}$, there exists unique representation of the form
$\alpha_i=m_{i1}\beta_1+ \cdots +m_{in}\beta_n$ where $m_{ij} \in \mathbb{Z}$.
Then $d=(\det(m_{ij}))^2 \disc(\beta_1, \beta_2, \cdots, \beta_n)$
but $\det(m_{ij}) \in \mathbb{Z}$ as $m_{ij} \in \mathbb{Z}$ and
$ \disc(\beta_1, \beta_2, \cdots, \beta_n) \in \mathbb{Z}$ as $\beta_i \in O_K$.
I want to assume $\left\{ \alpha_1, \alpha_2, \cdots, \alpha_n \right\}$ is not an integral basis and then show $d$ is not square-free.
Where did I go wrong? I should not get a contradiction unless I used a result of assumption that $\left\{ \alpha_1, \alpha_2, \cdots, \alpha_n \right\}$ is not an integral basis.