Let $K,L$ be two number fields such that $L$ is a quadratic extension of $K$, then can we always find an element $\alpha\in \mathcal{O}_L$ such that $\mathcal{O}_L=\mathcal{O}_K[\alpha]$, i.e., $\{1,\alpha\}$ is a relative integral basis for $L/K$?
When $K=\mathbb{Q}$, this assertion is true due to a classical result. But is it always true for general $K$?
Thanks very much!
Let $L/K$ be your quadratic extension of number fields, with rings of integers $O_L , O_K$, and denote by $\Delta=\Delta(L/K)$ the ideal discriminant. Then :
1)@Lord Shark the Unknown. $O_L$ is $O_K$-free iff $\Delta=(D)$ and $L=K(\sqrt D)$, see [M].
2) Let $L=K(\sqrt a)$ and consider the ideal decompositions $(a)=\mathcal A^2.\mathcal B$ and $\Delta=\mathcal C^2.\mathcal D$, where $\mathcal B$ and $\mathcal D$ are integral square free ideals. Then $L/K$ admits an integral basis iff $\mathcal B=\mathcal D$ and $\mathcal A$ and $\mathcal C$ belong to the same ideal class, see [M].
3) In [F], Fröhlich has defined, for any finite extension of number fields $L/K$, a new kind of "idelic" discriminant $\mathfrak D(L/K) \in J_K/{U_K}^2$, where $J_K$ is the group of idèles and $U_K$ the group of unit idèles of $K$. The classical ideal discriminant $\Delta(L/K)$ is then the image of $\mathfrak D(L/K)$ under the natural map $J_K/{U_K}^2 \to J_K/{U_K}$. The advantage of the new discriminant shows up immediately with the following theorem : $L/K$ admits an integral basis iff $\mathfrak D(L/K)$ is "principal", i.e. is the image of a principal idèle $d$ of $J_K$ under the projection $J_K\to J_K/{U_K}^2$. Back to the quadratic case, suppose that $L/K$ admits an integral basis. Then there exists $b\in K^*$ s.t. $d\equiv b^2$ mod $4$ and $(1, \frac {b+\sqrt d}{2})$ is an integral basis, see [F], p.24.
[F] A. Frohlich, Discriminants of algebraic number fields, Math. Zeit. 74 (1960), 18-28
[M] H. B. Mann, On integral bases, Proc. AMS 9 (1958), 167-172