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Is the following proposition true? If yes, how would you prove this?

Proposition Let $K$ be an algebraic number field. Let $L/K$ be a finite Galois extension. Let $A$ and $B$ be the rings of algebraic integers in $K$ and $L$ respectively. Let $G$ be the Galois group of $L/K$. Let $C$ be a subring of $B$ such that $A \subset C \subset B$. Suppose $\sigma(C) = C$ for every $\sigma \in G$. Let $\mathfrak{p}$ be a prime ideal of $A$. By the lying-over theorem, there exists a prime ideal $\mathfrak{P}$ of $C$ lying over $\mathfrak{p}$. Let $H$ = {$\sigma \in G$; $\sigma(x) \equiv x$ (mod $\mathfrak{P}$) for all $x \in C$}. Suppose $H = 1$. Then $\mathfrak{p}$ is unramified in $L$.

Motivation Let $K, A, L, B$ be as above. Let $n = [L : K]$. Let $f(X)$ be a monic polynomial of degree $n$ in $A[X]$. Suppose that $f(X)$ has no multiple roots. Suppose $L/K$ is the splitting field of $f(X)$. Let $\alpha_1,\dots,\alpha_n$ be the roots of $f(X)$. Let $C = A[\alpha_1,\dots,\alpha_n]$. Then $C$ satisfies the conditions of the above proposition. Sometimes it's easy to verify $H = 1$. Hence we can conclude that $\mathfrak{p}$ is unramified in $L$.

Application Let $f(X)$ be a monic polynomial of degree $n$ in $\mathbb{Z}[X]$. Let $d$ be the discriminant of $f(X)$. Suppose $d \neq 0$. Let $L$ be the splitting field of $f(X)$ over $\mathbb{Q}$. Let $G$ be the Galois group of $L/\mathbb{Q}$. Let $\alpha_1,\dots,\alpha_n$ be the roots of $f(X)$ in $L$. Let $C = \mathbb{Z}[\alpha_1,\dots,\alpha_n]$. Let $p$ be a prime number not dividing $d$. Let $P$ be a prime ideal of $C$ lying over $p$. Let $H$ = {$\sigma \in G$; $\sigma(x) \equiv x$ (mod $P$) for all $x \in C$}. Let $\sigma \in G -$ {1}. There exists $i$ such that $\sigma(\alpha_i) \neq \alpha_i$. Since the discriminant of $f(X)$ is not divisible by $P$, $\alpha_1,\dots,\alpha_n$ are distinct mod $P$. Hence $\sigma$ does not belong to $H$. Hence $H = 1$. Hence $p$ is unramified in $L$ by the above proposition.

Example Let $f(X) = X^3 + 2X + 1 \in \mathbb{Z}[X]$. Let $L$ be the splitting field of $f(X)$ over $\mathbb{Q}$. The discriminant of $f(X)$ is -59, where 59 is a prime number. Hence if $p \neq 59$ is a prime number, $p$ is unramified in $L$ by the above result.

Let $K = \mathbb{Q}(\sqrt{-59})$. The class number of $K$ is 3. $L/K$ is an abelian extension of degree 3. By the above result, the only prime number which can be ramified in $L$ is 59. It can be proved by another method that a prime of $K$ lying over 59 is unramified in $L$. Hence $L$ is the Hilbert class field of $K$.

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@Hurkyl and others, I opened a meta on this. Let us discuss there. Here is the link: meta.math.stackexchange.com/questions/4936/… –  Makoto Kato Aug 17 '12 at 22:29
@PualGarret Dear Paul Garrett, There is always a prime of $C$ lying over a prime of $A$ by the lying-over theorem. Regards, –  Makoto Kato Aug 18 '12 at 0:01
As a graduate student who works in algebraic number theory, I really don't see how this is any more "localized" than any number of other questions on this site, including many questions I've asked. I personally do not know the answer to the question, and have not thought about it, but I think it's a perfectly reasonable question for this site. –  Keenan Kidwell Aug 18 '12 at 3:00
@Hurkyl Please keep in mind that the charter of MSE is to welcome mathematical questions at all levels. This includes questions that may be of interest to a minority of our members. Please don't vote to close these questions for reasons of topicality. –  Bill Dubuque Aug 18 '12 at 3:21
@All Please refrain from discussing meta-level matters on the main site. There is now a meta thread on this topic. Please continue any non-mathematical discussion there. –  Bill Dubuque Aug 18 '12 at 3:23
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2 Answers

up vote 12 down vote accepted

Not an answer, but too long to fit into a comment. Clearly the assumption, $H=1$, implies that the field of fractions $F$ of the intermediate ring $C$ is all of $L$, for otherwise $Gal(L/F)$ would be a non-trivial subgroup of $H$.

Assume that $\mathfrak{P}=\mathfrak{P}_B\cap C$ for some prime ideal $\mathfrak{P}_B$ (necessarily lying over $\mathfrak{p}$). Let $G_T=G_T(\mathfrak{P}_B\vert\mathfrak{p})$ be the corresponding inertia group: $$ G_T=\{\sigma\in G\mid \sigma(\mathfrak{P}_B)=\mathfrak{P}_B,\ \forall\,x\in B: \sigma(x)\equiv x\pmod{\mathfrak{P}_B}\}. $$

Assume that $\sigma\in G_T$. Let $x\in C$ be arbitrary. Then $\sigma(x)-x\in C$ and $\sigma(x)-x\in\mathfrak{P}_B$, so $\sigma(x)-x\in C\cap\mathfrak{P}_B=\mathfrak{P}$. Therefore $\sigma\in H$.

As we work under the assumption that $H$ is trivial, we can conclude that $G_T=1.$ Therefore $$ e(\mathfrak{P}_B\vert \mathfrak{p})=|G_T|=1. $$ So $\mathfrak{P}_B\vert \mathfrak{p}$ is unramified, and the claim follows (by Galois theory of number fields the ramification indices over all the primes extending $\mathfrak{p}$ are equal).

So the question is reduced to asking whether all the primes of $C$ lying over $\mathfrak{p}$ are gotten by restricting a prime of $B$. The mapping from the primes of $B$ to those of $C$ is not necessarily injective, but the question is about its surjectivity. Edit: Makoto Kato points out that as $B$ is the integral closure of $C$, the surjectivity follows from general theory.

As an example of the case, where this mapping is not injective I proffer $K=\mathbb{Q}$, $L=K[\sqrt3]$, $\mathfrak{p}=13$, $C=\mathbb{Z}[13\sqrt3]$. The prime $13$ splits in $L$ as $$ 13=(4+\sqrt3)(4-\sqrt3), $$ but both prime ideals, $\langle4+\sqrt3\rangle$ and $\langle4-\sqrt3\rangle$ of $B$ intersect $C$ in $$ \mathfrak{P}=\{a+b\cdot13\sqrt3\mid a,b\in\mathbb{Z},\ a\equiv0\pmod{13}\}. $$

Anyway, my (possibly confused) thinking here is that $G_T$ would always inject into $H$. For an automorphism to be in $G_T$ we are to some extent asking more (in comparison to $H$), so if we can ascertain a trivial inertia group at the level of $C$, we should expect a trivial inertia group at the level of $B$ as well.

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Since $B$ is integral over $C$, there exists a prime ideal $\mathfrak{P}_B$ of $B$ lying over $\mathfrak{P}$. So I think your answer solved the question. –  Makoto Kato Aug 18 '12 at 8:17
Oops :-) ${}{}{}$ –  Jyrki Lahtonen Aug 18 '12 at 8:52
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If, as is apparently the case, the goal is a criterion for non-ramification in the actual ring of integers of a field extension $K/k$, perhaps Galois, the intermediate object $\mathfrak o_k[\alpha_1,\ldots,\alpha_n]$ need not be a ring, and we need not address questions of ideal theory there. Namely, to my mind, the convenient point is that ramified primes divide the different, whose norm is the discriminant. A discriminant is defined for any $\mathfrak o_k$-submodule, not only subring, of $\mathfrak o_K$, and is divisible by the true discriminant.

Thus, in many examples of interest, one computes the discriminant of an obvious $\mathfrak o_k$-sublattice of $\mathfrak o_K$, finds that a prime $\mathfrak p$ of $\mathfrak o_k$ already does not divide that discriminant, so can't divide the true discriminant, so no prime lying over it can divide the different.

The discriminants themselves, being determinants, have useful short-cuts for their computation, depending on the specifics of the situation.

Edit: an easy example... in $\mathbb Q({\root 3 \of 10})$, letting $\alpha$ be a cube root of $10$, the obvious subring $\mathbb Z[\alpha]$ happens not to be the integral closure. Nevertheless, the discriminant of this lattice, $\det\pmatrix{1 & \alpha & \alpha^2\cr 1 & \omega\alpha & \omega^2\alpha^2 \cr 1 & \omega^2\alpha & \omega\alpha^2}=\alpha^3\cdot (\omega-1)(\omega^2-1)$. Thus, the only possible ramification is at $3$ and $10=2\cdot 5$.

Even better known is the discussion of quadratic extensions over $\mathbb Q$, where $2$ is certainly not always ramified, and, indeed, is unramified when the ring of integers is index-two larger than one might have naively though, e.g., includes ${1+\sqrt{d}\over 2}$.

I'm certainly not claiming that every discriminant is this easily evaluated!

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I don't understand how your method proves the fact described in the application above. Regards, –  Makoto Kato Aug 18 '12 at 21:14
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