Equality of discriminants of integral bases (statement in Ireland and Rosen, A Classical Introduction to Modern Number Theory) I'm doing independent study and need assistance.
This is taken from Ireland and Rosen's A Classical Introduction to Modern Number Theory,
Chapter 12. 
Let F/Q be an algebraic number field, D the ring of integers in F, and A an ideal of D. In a discussion after Proposition 12.2.2 (if the discriminant of an ideal is minimal then that ideal is spanned by an integral basis), the authors state that it follows from Proposition 12.1.2 (if you have two bases, the discriminant of one is equal to a determinant times the other) that the discriminant of any integral basis for of an ideal of D is constant. I'm trying to prove how this follows.
thanks,

Just to put this in an answer.
Let $\{\omega_1,\ldots,\omega_n\}$ and $\{\omega_1',\ldots,\omega_n'\}$ be two integral bases for $D$. By definition, this means that $D$ is a free $\mathbb{Z}$-module, and that these are two bases for $D$. Thus, by definition there must exists some $M\in\text{GL}_n(\mathbb{Z})$ such that $M(\omega_i)=\omega'_i$. Note then that 
$$M[\sigma_i(\omega_j)]=[\sigma_i(\omega_j')]$$
and so 
$$\begin{aligned}\text{disc}(\omega_1',\ldots,\omega_n') &=\det([\sigma_i(\omega_j')])^2\\ &=\det(M[\sigma_i(\omega_j)])^2\\ &=\det(M)^2\det([\sigma_i(\omega_j)])^2\\ &= \det(M)^2\text{disc}(\omega_1,\ldots,\omega_n)\end{aligned}$$
But, note that since $M\in\text{GL}_n(\mathbb{Z})$ that $\det(M)\in\mathbb{Z}^\times=\{\pm 1\}$. Thus, $\det(M)^2=1$, and so $\text{disc}(\omega_1,\ldots,\omega_n)=\text{disc}(\omega_1',\ldots,\omega_n')$ as desired.
