Let $\left\{ 1 , \omega_1, \dots , \omega_{n1} \right\}$ be an integral basis for an order $\mathcal{O}$ of a number field of degree $n$ over $\mathbb{Q}$, let $\Delta $ be the discriminant of $\mathcal{O}$, and let $D \left( \omega_j \right)$ denote the discriminant of $\omega_j$. Is there a result that connects the discriminant of $\mathcal{O}$ to the discriminants of the generators of an integral basis for $\mathcal{O}$? e.i. something like \begin{equation} \Delta = \gcd \left\{ D\left( \omega_j \right) \right\}_{j=1}^{n1}, \end{equation} which would claim that the indices $a_j$ of $\mathbb{Z}[\omega_j]$ (for $j \geq 1$) in $\mathcal{O}$ are relatively prime? If the equation set out above is false in general, what is known about $\gcd \left\{ a_j \right\}_{j=1}^{n1}$?
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2$\begingroup$ There are cubic cyclic fields $F$ for which the index of $\mathbb{Z}[\alpha]$ is even for every $\alpha \in \mathcal{O}_F \backslash \mathbb{Z}$. In general the primes dividing every index are called ``common index divisors''. The following references should be helpful: K. Conrad, Rings of integers without a power basis math.uconn.edu/~kconrad/blurbs/gradnumthy/nopowerbasis.pdf and DummitKisilevsky, Indices in cyclic cubic fields archive.org/details/NumberTheoryAndAlgebra/page/n75 $\endgroup$ – François Brunault Oct 11 '18 at 17:11

$\begingroup$ Of course, I forgot those! I know them well. Thanks Francois. $\endgroup$ – Samuel Hambleton Oct 11 '18 at 21:32