Unramification of a prime ideal in an order of a finite Galois extension of an algebraic number field 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 $K$.
 A: 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!
A: 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.
