# If $\mathrm{disc}(\alpha_1,\alpha_2,\cdots,\alpha_n)$ is square-free, then $\{ \alpha_1, \alpha_2, \cdots, \alpha_n\}$ is an integral basis

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.

• If $d = disc(\alpha_1, \alpha_2, \ldots, \alpha_n)$ is square free, then, due to the relation between $d$ and the discriminant of the integral basis that you found, you must have $(det(m_{ij}))^2 = 1$. Can you continue from this point? – Tob Ernack Jul 12 '18 at 18:11
• yeah if I prove this straight I can use that fact. I will try this way as well. – Andrew Jul 12 '18 at 18:15
• If you want to prove this the other way, then starting from the assumption that $\{\alpha_1, \ldots, \alpha_n\}$ is not an integral basis, you must have that the $\mathbb{Z}$-submodule it generates be a proper submodule of the ring of integers. Something like the Smith normal form should allow you to show the determinant of the change-of-basis matrix is a nonunit and hence $d$ is not square-free. – Tob Ernack Jul 12 '18 at 18:22
• I don’t understand the problem, @TobErnack. If the lattice $L$ spanned by the $\alpha$s is proper, the transition matrix must be noninvertible, so its determinant is an integer $\ne\pm1$. No Smith. – Lubin Jul 12 '18 at 19:09
• @Lubin you are of course right, I realize now my proposition was akin to killing a fly with a bazooka. – Tob Ernack Jul 12 '18 at 19:50