I am looking for an integral domain $A$ with the following properties:

  1. $A$ is not integrally closed
  2. $A$ has a quotient field $K$ that is algebraically closed and that has characteristic 0
  3. There is an integral element $x\in K$ (over $A$) such that $A[x]$ is integrally closed.

Can someone help to tell me if the above is even possible?

Edit: Lubin easily gave me an example. Now I want to consider the case when I replace the condition 2. by:

2'. $A$ has a quotient field $K$ that is real closed.

  • $\begingroup$ Remark: assume $A$ fullfills all the requirements and let $m$ be a maximal ideal of $A$. There exists a maximal ideal $n$ of $B:=A[x]$ lying over $m$. Then $B/n$ is algebraically closed. On the other hand $B/n = A/m [x+n]$, hence by Schreier's theorem either $A/m = B/n$ or $A/m$ is real closed. $\endgroup$ – Hagen Apr 12 '11 at 7:28
  • $\begingroup$ Yes this also follows if I change K in 2. to be real closed. Then we get the same conditions: For any maximal prime ideal $\mathfrak m$ of $A$ we get $A/\mathfrak{m}$ is either real or algebraically closed. $\endgroup$ – Jose Capco Apr 12 '11 at 19:09

Try this: Let $B_0$ be the ring of real algebraic integers, and let $B=B_0[1/2]$, so the ring of real algebraic numbers integral except possibly at $2$. But $B[i]$ is equal to the ring of algebraic numbers integral except possibly at $2$, and this is integrally closed. And so we take $A=B[3i]$, not integrally closed, and of course the fraction field is algebraically closed.

  • $\begingroup$ This looks really nice! $\endgroup$ – Todd Trimble Apr 9 '11 at 22:47
  • $\begingroup$ Wow, yes indeed. Can we do the same thing if the field were real closed? $\endgroup$ – Jose Capco Apr 10 '11 at 11:09
  • 1
    $\begingroup$ I'll bet a nickel that that's not possible. But no more than that! $\endgroup$ – Lubin Apr 10 '11 at 16:36

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