Let $p_i$'s be distinct primes such that $p_i \equiv 1\;(\mathrm{mod}\; 4)$ for every $i=1,...,k$. It is well-known that $\dfrac{1+\sqrt{p_i}}{2}$'s are algebraic integers and the number ring $\mathbb{A} \cap \mathbb{Q}(\sqrt{p_i})$ is generated by $\{1,\dfrac{1+\sqrt{p_i}}{2} \}$, where $\mathbb{A}$ is the ring of all algebraic integers. Here 'generated' means 'generated as a free abelian group(or $\mathbb{Z}$-module).'
I am struggling with the generalization such that the number ring $R=\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_k}) \cap \mathbb{A}$ has integral basis $\mathcal{B}=\{\prod_i \alpha_i^{\epsilon(i)}: \epsilon \in \{0,1\}^{\{1,...,k\}} \}$ where $\alpha_i=\dfrac{1+\sqrt{p_i}}{2}$.
I already figured out that the above set is $\mathbb{Z}$-linearly independent from the fact that $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_k})$ has degree $2^k$ over $\mathbb{Q}$. Now it remains to prove that $\mathcal{B}$ actually generates $R$.
Since $R$ is not a vector space but a free abelian group, the linear independence does not guarantee that $\mathcal{B}$ is a basis. This is where I stuck. Does anyone have ideas? Any advices or hints will help a lot!
If the answer is not available to be posted here, then any links or bibliographies are welcome!
Thank you for your help!
If I remember correctly, the generalisation you want does not hold ! I remember a paper where is the case $n=2$ is solved, and it's already extremely difficult.
It's the paper of Huard, Spearman,Williams, Integral bases for quartic fields with quadratic subfields, J. Number theory 51 (1), 87-102, 1995
I can't remember the results precisely, but i'm 85% sure that the answer is more complicated than the simple guess we could make.