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

  • 1
    $\begingroup$ 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

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 $\tilde\varphi(x)\in M$.

  • $\begingroup$ 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

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 $a \in R$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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