 Solution

For a basis of a module $m_1,\dots ,m_d$ of the full-rank module $M$, define its discriminant of elements of a number field byThe field discriminant is this determinant for an integral basis of ${\mathcal{O}}_K$. A change of integral basis with determinant $\pm 1$ does not change it. Write $m_j={\sum }_i{b}_{ij}{e}_i$ in an integral basis and let $B=(b_{ij})$. The two trace matrices satisfy $T_M=B^{\mathsf{T}}{T}_{{\mathcal{O}}_K}B$. The index of the lattice is $r=|\det B|$, as follows for example from Smith normal form. ThereforeMultiplicative closure makes $M$ an order when it contains one, but the determinant argument itself requires only full finite index, so it also covers the stated module hypothesis without that extra condition.