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Main question : 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 $a\in R$, [EDIT1 : and $R\cap \mathfrak m$ is maximal in $R$] ?

[EDIT4:] Such an example, if it exists, can not be found if $R$ is a Dedekind ring, because the residue field of every place of $K={\rm quot}(R)$ is of the form $R/\mathfrak P$ for some maximal ideal $\mathfrak P$ of $R$ (this may be clearer after reading the next observations).

Motivation : I am trying to prove a difficult result (at least for me). A way to obtain it would be to show that if $\varphi$ is an epimorphism of an integral domain $R$ into a field $F$, and if $\tilde\varphi$ is a place of the fraction field of $R$, with residual field $\tilde F$ algebraic over $F$, then $\tilde F = F$. I have some doubts that such a miracle does occur; but this problem is not found 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.

[EDIT2:] Conversely, if the answer of the question is positive, then every place $\varphi$ that extends the canonical epimorphism $R[x]\to F'=R[x]/\mathfrak m $ to $K$ extends the canonical epimorphism $R\to F = R/(R\cap \mathfrak m)$ to $K$; furthermore, $\varphi(x)$ is algebraic over $F$ (since $F[\varphi(x)]= F[x+\mathfrak m]$ is a field), but $\varphi(x) \not\in F$ (else $R[x]/{\mathfrak m} = F$, hence $x \in a+\mathfrak m$ for some $a\in R$). So this is a counter example.

[EDIT3:] It is important to suppose that the residual field $\tilde F$ of $\varphi$ is algebraic over $F$, else it is not difficult to produce counter-examples. For example, take $R={\mathbb Q}[X,Y]$, and define the valuation $v$ of a polynomial $P(X,Y)$ to be the minimal "$\alpha$-degree" of the monomes in $P$, where the $\alpha$-degree of $X^iY^j$ is $i+\alpha j$, for some $\alpha > 0$. Then $v$ extends naturally to a valuation of $K={\mathbb Q}(X,Y)$, whose residual field can be shown to be ${\mathbb Q}(t)$ for some variable $t$. So, the place $\tilde \varphi$ corresponding to $v$ would be a counter example for the epimorphisme $\varphi: R\to \mathbb Q$ such that $\varphi(X)=\varphi(Y) = 0$.

N.B: This question is not a duplicate of a somewhat related (but very different) question posted in a previous thread

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    $\begingroup$ Hadn't you asked this a while ago? $\endgroup$ – Mariano Suárez-Álvarez Feb 18 '15 at 7:18
  • $\begingroup$ As I wrote at the end of the thread, this is not a duplicate question. $\endgroup$ – MikeTeX Feb 18 '15 at 14:31
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This answer is not mine: I asked Prof. Muhammad Zafrullah, a nice man, specialist of multiplicative ideal theory, that answers questions in his site. I have somewhat simplified and adapted his answer.

I first give a counter example to the "motivation problem", then derive the solution to the main question.

Let $M$ be a field, $L$ a finite non trivial algebraic extension of $M$. Put $R = M + XL[X]$, hence the fraction field of $R$ is $K=L(X)$. Then ${\frak p} = XL[X]$ is easily seen to be a maximal ideal of $R$, and we define the epimorphism $\varphi$ to be the natural surjection $R\to R/{\frak p} = M$. Now, the application defined by $\tilde \varphi(f(X)) = f(0)$ if $f\in L[X]$, and $\varphi(f(X)) = \infty$ otherwise is easily seen (and well known) to be a place of $K$, whose residual field is $L$, a finite non trivial extension of $M$. This is a counter example to the problem "motivation".

To derive a counter example to the main question: let $x \in K$ so that $\tilde\varphi(x)\in L$ and $\tilde\varphi(x)\not\in M$. then denoting by $\psi$ the restriction of $\tilde \varphi$ to $R[x]$, the ideal ${\frak m}=\psi^{-1}(0)$ is a maximal ideal of $R[x]$ (since $\tilde\varphi(R[x]) = M[\tilde\varphi(x)]$ is a field), and so is $R\cap {\frak m} = \varphi^{-1}(0) = \frak p$. Obviously, $x-a\not \in {\frak m}$ for every $a\in R$ otherwise $\tilde\varphi(x)\in M$.

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  • $\begingroup$ Such an answer may be more difficult if it were also demanded that $R$ be integrally closed. It would be nice to have an answer in this case. $\endgroup$ – MikeTeX Feb 26 '15 at 10:32
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Please, check if my answer is correct.

Let $R = \mathbb{Z}_{2\mathbb{Z}}$ be the localization of $\mathbb{Z}$ at the prime ideal $2 \mathbb{Z}$, $K = \mathbb{Q}$, and $x = \frac{1}{2}$.

Then $R[x] = \mathbb{Q}$ has only one maximal ideal, namely $(0)$.

And clearly for all $a \in R$ $$x-a \in (0) \Leftrightarrow x= a$$ which is not the case since $x \notin R$ while $a \in R$.

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  • $\begingroup$ Of course, your answer is correct. Unfortunately, I forgot to specify that the maximal ideal $\mathfrak m$ should be such that $R\cap \mathfrak m $ is maximal, otherwise, this would be of no help in the problem exposed in the section "Motivation". I have updated my question. Thank you for having fixed the flaw in the question. $\endgroup$ – MikeTeX Feb 18 '15 at 15:16

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