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Does there exist a commutative ring extension $A\subsetneq B$ satisfying the following conditions:

  1. $A$ is a normal local domain and $B$ is a regular domain with the same dimension;
  2. $A$ and $B$ have the same fraction field;
  3. $B$ is finitely generated as an $A$-algebra.
G.-S. Zhou
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    Hi. You can take $B = k[x,y]$ polynomial ring over a field, and $A = k[x,xy]$. Since $y = xy/x$, they have the same field of fractrions, these two rings are regular (so normal), and $A[y] \to B$ gives item 3. – Youngsu Dec 15 '18 at 20:08
  • Hi. We need $A$ to be local. – G.-S. Zhou Dec 16 '18 at 02:40
  • You can localize $A, B$ at the complement of the maxial ideal $(x,xy)A$. – Youngsu Dec 16 '18 at 07:58

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