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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!

bellcircle
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    Show that the discriminant of your ring is odd. This proves that the only possible denominators must be odd, and this may be excluded by looking at the traces of the elements in question. –  Oct 31 '17 at 10:06
  • I think that this can be proven by induction, using the claim in this solution. I believe this is the same as what GreginGre was suggesting in one of his comments below. – Tristan Phillips Oct 31 '17 at 23:44

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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.

GreginGre
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  • https://doi.org/10.1006/jnth.1995.1036, http://people.math.carleton.ca/~williams/papers/pdf/194.pdf – lhf Oct 30 '17 at 10:16
  • I'm afraid the paper is not I wanted, since $p_i$'s in my questions are all ordinary integers, so my question is not related to quartic field. In fact, the case $n=2$ is given as an exercise in the book of Marcus, Daniel, Number Fields, Springer, 51-52, 1977. – bellcircle Oct 30 '17 at 10:45
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    Ok, after a bit more thinking about your question, i think the result is true. The proof is based on the following well-known fact: if $K,L$ are two number fields which are linearly disjoint over $\mathbb{Q}$, and such that $gcd(d_K,d_L)=1,$ then $O_{KL}\simeq O_K\otimes_{\mathbb{Z} O_K$. Then you should obtain your result by induction, if I'm not mistaken. You need to prove in the induction step that the prime divisors of the discriminant of $\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_r})$ are$p_1,\ldots,p_r$, I didn't try, so you should check I'm not mistaken. – GreginGre Oct 30 '17 at 13:11