Squarefree Discriminant Gives 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

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

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

which implies that ${\alpha }_1,\dots ,{\alpha }_n$ is an integral basis.