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)
Pick now 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
$$ \left( \frac{L/K}{ \cdot} \right) : I_K \to \text{Gal}(L/K) $$
Furthermore, there is a product formula $$ \prod_{v\in V^k}\left( \frac{L/K}{v} \right)=1 $$
See https://math.stackexchange.com/posts/1244811/ for original post. An answer in that post said:
The Artin symbol is essentially a lifting of the Frobenius element. So if the inertia group $I(\frak p)$ is nontrivial (which it turns out happens exactly if $\frak P$ is ramified over $\frak p$), this lift is defined only up to an $I(\frak p)$-coset.
My question is: what happens to the ramified primes in that formula? E.g. if there is only one ramified prime $\frak q$, then is $\displaystyle\left( \frac{L/K}{\frak q} \right)$ defined to be the identity element of the Galois group, or the unique element (in the inertia group coset) to make the product formula hold? Or are ramified primes simply excluded from the product formula?