normal domains with algebraically closed quotient field - MathOverflow most recent 30 from http://mathoverflow.net2013-05-23T08:16:14Zhttp://mathoverflow.net/feeds/question/61155http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/61155/normal-domains-with-algebraically-closed-quotient-fieldnormal domains with algebraically closed quotient fieldJose Capco2011-04-09T17:38:05Z2011-04-12T18:40:50Z
<p>I am looking for an integral domain $A$ with the following properties:</p>
<ol>
<li>$A$ is not integrally closed</li>
<li>$A$ has a quotient field $K$ that is algebraically closed and that has characteristic 0</li>
<li>There is an integral element $x\in K$ (<em>over</em> $A$) such that $A[x]$ is integrally closed.</li>
</ol>
<p>Can someone help to tell me if the above is even possible? </p>
<p><strong>Edit</strong>: Lubin easily gave me an example. Now I want to consider the case when I replace the condition 2. by:</p>
<p>2'. $A$ has a quotient field $K$ that is real closed.</p>
http://mathoverflow.net/questions/61155/normal-domains-with-algebraically-closed-quotient-field/61163#61163Answer by Lubin for normal domains with algebraically closed quotient fieldLubin2011-04-09T21:21:03Z2011-04-09T21:21:03Z<p>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.</p>