Definition of Artin map

Question

Let $L/K$ be an unramified Abelian extension. Then the Artin symbol $ \left ( (L/K) / \mathfrak p \right) $ is defined for all prime ideals $\mathfrak p$ of $\mathcal O_K$(because in an Abelian extension, the Artin symbol depends only on the underlying prime).

Now, pick a fractional ideal $\mathfrak a = \prod_{i=1}^r \mathfrak p_i^{r_i} $. Then we can define the Artin symbol $ \left ( (L/K) / \mathfrak a \right) $ to be the product: $$ \left ( (L/K) / \mathfrak a \right) = \prod_{i=1}^r \left ( (L/K) / \mathfrak p_i \right)^{r_i} $$

Then the Artin map is defined as follows:

$$ \left( \frac{L/K}{ \cdot} \right) : I_K \to \text{Gal}(L/K) $$

Question: Why do we need the extension to be unramified ? Or in other words, if the extension $L/K$ is ramified, then the Artin map is not well-defined on all fractional ideals $I_K$ ?

It seems like I don't understand something either in the definition of the Artin map, or some basic fact about fractional ideals.

Any help would be really appreciated.

Thank you in advance.

Answer 1

If $\mathfrak{P}$ is a prime lying over $\mathfrak{p}$, we get an exact sequence \begin{equation*} 0\to I\to D\to Gal(B/\mathfrak{P}\big/A/\mathfrak{p})\to 0, \end{equation*} where $D$ is the decomposition group (the stabilizer of $\mathfrak{P}$ in the Galois group), and $I$ is the kernel of the map, the inertia group.

If the residue field is finite, $A/\mathfrak{p}$ is isomorphic to $\mathbb{F}_q$ for some rational prime $q$, so the Galois group on the right is isomorphic to $\mathbb{Z}/f\mathbb{Z}$. The Frobenius element of that group is the automorphism $x\mapsto x^q$, which fixes $A/\mathfrak{p}$.

The Artin symbol is essentially a lifting of the Frobenius element. So if $I$ is nontrivial (which it turns out happens exactly if $\mathfrak{P}$ is ramified over $\mathfrak{p}$), this lift is defined only up to an $I$-coset.

If $I$ is trivial, then you potentially get a different lift for each prime $\mathfrak{P}$ lying over $\mathfrak{p}$; it turns out that these lifts are in the same conjugacy class of $Gal(L/K)$. Thus if $Gal(L/K)$ is Abelian, as you point out, the lift depends only on $\mathfrak{p}$, not on $\mathfrak{P}$.