Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
I don't understand what "of $A$" means in condition 3. –  Gerry Myerson Apr 10 '11 at 0:07
    
an integral element of $A$ that is a member of $K$ –  Jose Capco Apr 10 '11 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 '11 at 22:25
    
yes integral over $A$. Thanks for the correction :) ... english is just too difficult :) –  Jose Capco Apr 11 '11 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 '11 at 7:28

1 Answer 1

up vote 10 down vote accepted

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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.