Let $m$ be a square free integer, $\mathbb{Q}(\sqrt{m})$ a quadratic field extension of $\mathbb{Q}$, $\Delta$ is its discriminant and $O_{\mathbb{Q}(\sqrt{m})/\mathbb{Q}}$ its ring of integers. We have the following results (see Lenstra, On the calculation of regulators and class numbers of quadratic fields):

1. Every ideal of $O_{\mathbb{Q}(\sqrt{m})/\mathbb{Q}}$ is of the form $d\times (\mathbb{Z}a+\mathbb{Z}\frac{b+\sqrt{\Delta}}{2})$ with $d, a, b\in\mathbb{Z}$ and $c=\frac{b^{2}-\Delta}{4a}\in\mathbb{Z}$.

2. Let $I_{1}=\mathbb{Z}a_{1}+\mathbb{Z}\frac{b_{1}+\sqrt{\Delta}}{2}$ and $I_{2}=\mathbb{Z}a_{2}+\mathbb{Z}\frac{b_{2}+\sqrt{\Delta}}{2}$ be two ideals of $O_{\mathbb{Q}(\sqrt{m})/\mathbb{Q}}$, $c_{1}=\frac{b_{1}^{2}-\Delta}{4a_{1}}$ and $c_{2}=\frac{b_{2}^{2}-\Delta}{4a_{2}}$ . Then the ideal $I_{3}=I_{1}\times I_{2}$ is of the following form $I_{3}=d_{3}\times (\mathbb{Z}a_{3}+\mathbb{Z}\frac{b_{3}+\sqrt{\Delta}}{2})$ with

$$ \left{ \begin{array}{l}

d_{3}=gcd(a_{1},a_{2},\frac{b_{1}+b_{2}}{2}),\vspace{0.2cm}\

a_{3}=\frac{a_{1}a_{2}}{d_{3}^{2}},\vspace{0.2cm}\

b_{3}=\frac{\alpha a_{1}b_{2}+\beta a_{2}b_{1}+\delta \frac{b_{1}b_{2}+\Delta}{2}}{d_{3}},

\end{array} \right. $$

and $d_{3}=\alpha a_{1}+\beta a_{2}+\delta \frac{b_{1}+b_{2}}{2}; \alpha, \beta, \delta\in\mathbb{Z}$.

These results are related to classical Gauss composition law. My question is if these results remain true when we replace $\mathbb{Q}(\sqrt{m})/\mathbb{Q}$ by any quadratic extension of number fields $\mathbb{L}/\mathbb{K}$ such that $\mathbb{K}$ is of the class number one? and where can I find the proofs if they exist.

Thank you.