Let $p_i$ be distinct primes. $Q(\sqrt{p_1},\sqrt{p_2},\dots)$'s integral closure is $A$.

$\textbf{Q:}$Is $A$ noetherian or artinian?(For number field, the artinian part is easy as you have DVR and 1-1 correspondence of primes with localization.) I think it is not artinian by considering $(\sqrt{\prod_{i\leq n}p_i})$ forming descending chain. I could not construct an obvious ascending sequence. I could try take $A_i$ be integral closure of $Q(\sqrt{p_1},\dots,\sqrt{p_i})$ and use ramification theory of $p_i$ here. However the ring of coefficient changes and I need to extend the ramified ideals to $A$ which does not guarantee the chain is proper ascending.

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    $\begingroup$ Do you mean $\mathbb Z[\sqrt{p_i}]$ instead of $\mathbb Q[\sqrt{p_i}]$? $\endgroup$ – punctured dusk May 18 '18 at 16:26
  • $\begingroup$ @Lubin. The troublesome part is $(p_1,p_2)\subset (\sqrt{p_1},\sqrt{p_2},\sqrt{p_3})=I_3$ but $p_1,p_2$ are distinct. Hence $1=(p_1,p_2)$. This stabilizes at the step 3. This is in analogy with $Q(2^{1/2},2^{1/4},\dots)$. I tried this method and it does not produce me a proper ideal. $\endgroup$ – user45765 May 18 '18 at 17:07
  • $\begingroup$ @barto It is $Q(\{\sqrt{p_i}\}_i)\cap\mathcal{A}$. The set of all algebraic integers. $\endgroup$ – user45765 May 18 '18 at 17:08
  • $\begingroup$ Yes, of course, @user45765 . I was being worse than stupid. $\endgroup$ – Lubin May 19 '18 at 3:23

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