Let $L/\mathbb{Q}$ be a finite Galois extension of degree n, let $\mathcal{O}_{L}$ be the ring of integers of $L$, By Dedekind lemma we have that
$\mathfrak{p}=\mathfrak{b}_{1}^{e}...\mathbb{b}_{g}^{e}$ where $ \mathfrak{b}_{i} \in \mathcal{O}_{L}$ and $\mathfrak{p}$ $\in $$\mathbb{Z}$ Since this extension is Galois we have $gfe=[L:\mathbb{Q}]=n$ where $e$ is the ramification index and $f$ is the inertia degree which is given by $f_{i}=[\mathcal{O}_{L}/\mathfrak{b} :\mathbb{Z}/\mathfrak{p}]$
we know that $\mathfrak{p}$ is ramified if $e >1 $,
I am asking is there a geometric way to know whether a prime ideal is ramified or not? if it exists can you apply it on this picture (give any example)
Also, I want to know what is the geometric interpretation of the inertia degree and the ramification index ?
Note that if $k'/k$ is any finite separable extension of fields, then $k' \otimes_k \bar{k} \cong \bar{k}^{[k' : k]}$. So, given an extension of number fields $L/K$, a nonzero prime ideal $\mathfrak{p} \subseteq \mathcal{O}_K$ with residue field $k = \mathcal{O}_K/\mathfrak{p}$ is unramified in the extension $\mathcal{O}_L/\mathcal{O}_K$ if and only if the geometric fiber $\operatorname{Spec}(\mathcal{O}_L/\mathfrak{p} \otimes_{k} \bar{k})$ (that is, the base change of the fiber, which is a $k$-scheme, to the algebraic closure $\bar{k}$) has exactly $[L : K]$ closed points.
This exactly mirrors the situation for branched coverings of Riemann surfaces, where a branch point is exactly one where the fiber has fewer points than the degree of the covering map. The difference is that since we're dealing with non-algebraically-closed fields, we might have to extend the residue field in order to "see" the full size of the fiber.
From this perspective, here are geometric interpretations of inertia degree and ramification index of a prime $\mathfrak{b} \in \operatorname{Spec} \mathcal{O}_L$: