Relationship between index of a basis for a ring of integers and the integer span of the basis In my notes for number theory, after having proved the following:
Corollary: For a number field $K$, if $x_1,\dots,x_n\in\mathcal{O}_K$ is a basis for  $\mathcal{O}_K$ over $\mathbb{Q}$ and $M=\langle x_1,\dots,x_n\rangle_{\mathbb{Z}}$, then $\Delta(x_1,\dots,x_n)=(\mathcal{O}_K:M)^2D(K)$, where $D(K)$ is the discriminant of $K$ and $(\mathcal{O}_K:M)$ is the index of the subgroup $M$ in $\mathcal{O}_K$.
There is a line saying "in particular, if $\Delta(x_1,\dots,x_n)=c^2d$ with $d$ squarefree, $c,d \in \mathbb{Z}$ then $(\mathcal{O}_K:M) | c$ so $\mathcal{O}_K \subset \frac1cM$."
I understand why $(\mathcal{O}_K:M) | c$, but really can't work out why $\mathcal{O}_K \subset \frac1cM$. I would appreciate any explanation.

Here is a proof as to why $\mathcal{O}_K \subseteq \frac{1}{c} M$. Consider the quotient module $\mathcal{O}_K/M$; it has $( \mathcal{O}_K : M)$ number of elements. By Lagrange's Theorem, for  any element $[x]$ in the quotient we have $( \mathcal{O}_K : M)\cdot [x] = \overline{0}$. Since $(\mathcal{O}_K : M) \mid c$ we have $c \cdot [x] = \overline{0}$. But this is saying that $cx \in M$ and hence $\mathcal{O}_K \subseteq \frac{1}{c}M$.
