normal domains with algebraically closed quotient field - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T08:16:14Z http://mathoverflow.net/feeds/question/61155 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61155/normal-domains-with-algebraically-closed-quotient-field normal domains with algebraically closed quotient field Jose Capco 2011-04-09T17:38:05Z 2011-04-12T18:40:50Z <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#61163 Answer by Lubin for normal domains with algebraically closed quotient field Lubin 2011-04-09T21:21:03Z 2011-04-09T21:21:03Z <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>