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Let $A\to B$ be integral morphism of commutative rings, and $A$ be noetherian ring. Can you provide an example that $B$ is not noetherian? How about integral extension $A\subset B$ (i.e. the morphism is monomorphism)?

I tried to make an counterexample using non-noetherian rings $k[x_1, \cdots ]$ or $\prod^\infty \mathbb{Z}/2\mathbb{Z}$, but I failed.

nessy
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    https://math.stackexchange.com/a/2390771 – Yai0Phah May 11 '20 at 10:00
  • What about $\mathbb Z\subset\mathbb Z[2^{1/2},\dots,2^{1/2^n},\dots]$? – user26857 May 11 '20 at 20:58
  • @user26857 It is clear this is an integral extension of noetherian ring $\mathbb{Z}$. I'm having trouble understanding the latter ring is non-noetherian. Is it easy? – nessy May 12 '20 at 01:46
  • In brief, both of the top answers at the link resolve this question too. The mentioned ring of integers of $K$ is exactly the integral closure of $\Bbb Z$ in $K$ - in particular, it's an integral ring extension, and $\Bbb Z$ is obviously Noetherian. But the examples in the link show explicit contradictions: the first gives a chain of ideals which does not stabilize, and the second shows the ring violates a known property of noetherian domains. – KReiser May 12 '20 at 08:24
  • @KReiser Thanks, I understand the second example at the link. However I don't know how to tackle with the first one. Is it easy to show that (1) ring of integers of $K$ is exactly the mentioned ring, and (2) $I_j\subset I_{j+1}$ is a strict inclusion. (I have no prerequisite knowledge about algebraic number theory. Is it obvious by general theory or something?) – nessy May 12 '20 at 16:14
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    I think it's relatively straightforwards. Certainly $\Bbb Z[2^{1/2^m}]$ where $m$ ranges over the positive integers is integral over $\Bbb Z$, and one can write its integral closure as the union of the integral closures of the truncated versions of this ring (ie stop $m$ at $m_0$). Computing each finite step isn't so bad, since it's a quadratic extension of the previous step. (If you've never seen the computation of the integral closure of $\Bbb Z\subset \Bbb Q(\sqrt{d})$ for $d$ an integer, you might want to take a look at that.) To show the strict inclusion, one may use the 2-adic valuation. – KReiser May 12 '20 at 18:32
  • @KReiser Thanks for your comment. It seems to make sense. – nessy May 13 '20 at 02:56

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