Squarefree Discriminant Gives an Integral Basis

Theorem

Let α1,,αnOK be a basis for a number field K. If disc(α1,,αn) is squarefree, then α1,,αn is also 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 α1,,αnOK be a basis for a number field K, and let β1,,βn be an integral basis of OK. Then, using this result, we have that

disc(α1,,αn)=det(M)2disc(β1,,βn)

where M=(mij)1i,jnMn(Q) is given by

αi=j=1nmijβj.

However since disc(α1,,αn)=det(M)2 is squarefree, this means that both det(M)2 and disc(β1,,βn) must be squarefree. Since the only squarefree square is 1, this means that det(M)2=1 and hence

disc(α1,,αn)=disc(β1,,βn)=disc(K)

which implies that α1,,αn is an integral basis.