normal domains with algebraically closed quotient field

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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.

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I don't understand what "of $A$" means in condition 3.  – Gerry Myerson Apr 10 2011 at 0:07
an integral element of $A$ that is a member of $K$ – Jose Capco Apr 10 2011 at 11:21
@Jose, but if $x$ is an element ${\it of\ }A$ then $A[x]$ is just $A$. Maybe you mean $x$ is integral ${\it over\ }A$? – Gerry Myerson Apr 10 2011 at 22:25
yes integral over $A$. Thanks for the correction :) ... english is just too difficult :) – Jose Capco Apr 11 2011 at 4:25
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. – Hagen Apr 12 2011 at 7:28
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1 Answer

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.

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This looks really nice! – Todd Trimble Apr 9 2011 at 22:47
Wow, yes indeed. Can we do the same thing if the field were real closed? – Jose Capco Apr 10 2011 at 11:09

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I'll bet a nickel that that's not possible. But no more than that! – Lubin Apr 10 2011 at 16:36
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