# Maximal ideal not containing specific expression

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1. Feb 12, 2015

### coquelicot

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.

2. Feb 12, 2015

### mathwonk

what are you trying to prove?

3. Feb 13, 2015

### coquelicot

Hello mathwonk,

I am extending the ramification theory of Dedekind domains and valuation rings to a general setting: let $A$ be an integrally closed domain, $K$ its field of fractions, $L$ a Galois extension of $K$, finite or infinite, and $B$ the integral closure of $A$ in $L$. Assume that $p$ is a maximal ideal of $A$, and $P$ is a maximal ideal of $B$ above $A$.
Denote $F_A = A/p$ and $F_B = B/P$, so $F_B/F_A$ is a normal field extension.

A great deal of definitions and results of ramification theory "passes" under this setting: Decomposition group, inertia group, theorem of surjectivity of the decomposition group to the group of automorphisms in the residual field, unramified extensions, tamely ramified extensions, group of ramification, and most of the corresponding theorems. The result is fairly nice, and implies immediately the corresponding results for Dedekind extension and valuation extensions.

I have more or less finished the task, but there remain two rather inelegant issues, and the whole work would not be serious if I leave them as is: Assuming $L/K$ finite,
1) must $F_B/F_A$ be finite ?;
2) assuming $F_B/F_A$ finite, must $[F_B:F_A]$ (the inertia degree) divide $[L:K]$ ?

I can show that 1) is necessary if $F_B/F_A$ is separable, or if $A$ is noetherian;
I can shown that 2) is necessary if $F_B/F_A$ is separable, or if ${\rm char}(F_B)$ does not divide $[L:K]$.
Of course, a proof of 2) would imply 1).

I am trying to prove 2), which implies 1) in the general case. I have reduced the problem to the following minimal form:
Assume that ${\rm char}(F_A) = \pi>0$, and that $[L:K]$ is of prime degree $\pi$ ($L$ is not assumed to be Galois here, but this changes nothing in the setting for this problem). Assume furthermore that $F_B/F_A$ is purely inseparable.
Does 1) hold in this case ?

It would be a waste of time to try proving the general claim: here is the essential difficulty, and if it can be solved, I can solve the two general problems above. If there is a counter example, then at least it will be licit to say "suppose that $[F_B:F_A]$ is finite" and a similar assertion about the divisibility by the residual degree.

To sum-up, what I am trying to prove is the following: Assume that $A$ is an integrally closed domain, $K$ its fraction field, $L$ an extension of $K$ of prime degree $\pi$, $p$ a maximal ideal of $A$ containing $\pi$, $P$ a maximal ideal of $B$ above $A$, $F_A$, $F_B$ the corresponding residual fields. Assume $F_B/F_A$ purely inseparable.
Must the relation $[F_B:F_A]\leq \pi$ hold ? (this will allow answering 2) above, and the question " must $[F_B:F_A]$ be finite?" would allow answering 1) above)

As I said previously, a way to prove this would be to prove the following claim, which is probably false but is interesting for its own and I would like at least to know a counter example :
Let $A$ be an integral domain of fraction field $K$, and $\varphi$ an epimorphism of $A$ into a field $F_A$. It is known that $\varphi$ can be extended to a place of $K$, with (finite) values into an algebraic extension $F'$ of $F_A$. Is it possible that $F' \not=F_A$? (a negative answer would imply my theorem). In fact, to prove my theorem, I need only the following: "There exists an extension of $\varphi$ to a place of $K$ whose residual field is equal to $F_A$". But I think that if the first question has a counter example, this last claim is very unlikely to hold.

P.S : I have spent 3 weeks to try to solve this problem: I feel I am exploding.

Last edited: Feb 13, 2015