# Definition of Artin map

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.

• If $\mathfrak{p}$ is ramified, then its Artin symbol is not well-defined; it is a coset of the inertia group of $\mathfrak{p}$. Commented Apr 21, 2015 at 12:20
• @rogerl: Thank you for your reply! Could please give smoe more details? If $\mathfrak p$ is ramified then the Artin symbol why is not defined ? Thank you and sorry if this a silly question... Commented Apr 21, 2015 at 12:27

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 $$\mathfrak{P}$$.
• One last question related to the definition of the Artin symbol: In the case of $I$ being trivial, i.e $\mathfrak p$ is unramified in $B$. How can we conclude that the Artin symbol satisfies the condition $((L/K) / \mathfrak \beta ) (\alpha) \equiv \alpha ^(q) mod \mathfrak \beta$ for all $\alpha \in \mathcal O_B$ ? And why is this unique ? Thank's! Commented Apr 21, 2015 at 12:57
• Because that is exactly how the Artin symbol is defined: it is an element that lifts $x\mapsto x^q$. It is unique because if $I$ is trivial, then $D\cong Gal(B/\mathfrak{P}\big/A/\mathfrak{p})$, so the lift is unique. Commented Apr 21, 2015 at 13:05