coquelicot
- 295
- 67
May there exist an integral domain R, with fraction field K, that fulfills the following condition:
there exists x\in K, x\not \in R and a maximal ideal \frak m of R{[}x{]}, such that \frak m does not contain x-a for any a\in R ?
Motivation : I am trying to prove a difficult result. A way to obtain it would be to show that if \varphi is an epimorphism of an integral domain R into a field F, then the residual field of every place \tilde \varphi extending \varphi to the fraction field of R, with finite values into an algebraic closure of F, is equal to F. I have some doubts that such a miracle does occur; but this problem is not available in the literature.
Now, if the answer of the asked question is negative, then we are done, taking the restriction of \tilde\varphi to R{[}x{]} in the (allegedly) absurd supposition that such an extension of \varphi exist.
there exists x\in K, x\not \in R and a maximal ideal \frak m of R{[}x{]}, such that \frak m does not contain x-a for any a\in R ?
Motivation : I am trying to prove a difficult result. A way to obtain it would be to show that if \varphi is an epimorphism of an integral domain R into a field F, then the residual field of every place \tilde \varphi extending \varphi to the fraction field of R, with finite values into an algebraic closure of F, is equal to F. I have some doubts that such a miracle does occur; but this problem is not available in the literature.
Now, if the answer of the asked question is negative, then we are done, taking the restriction of \tilde\varphi to R{[}x{]} in the (allegedly) absurd supposition that such an extension of \varphi exist.