I am trying to connect my geometric intuition for ramification with the geometric picture/ intuition for immersions in differential topology. It seems that a formally unramified map of schemes is analogous to an immersion of manifolds. But there are some questions I have left about how these definitions compare- I want to connect the geometric pictures of these definitions.
Let $L/K$ be an extension of algebraic number fields. Take a prime $\mathfrak{p}$ in $K$ which factors as $\prod_{i = 1}^n \mathfrak{q}_i^{e_i}$ in $L$. If $e_i = 1$ then we say that $L/K$ is unramified. If we think in terms of the map of schemes $\text{Spec}(\mathcal{O}_L) \rightarrow \text{Spec}(\mathcal{O}_K)$, we have a map of curves whose fibers $$\text{Spec}(\mathcal{O}_L / \mathfrak{p} \mathcal{O}_L) \cong \amalg_{i = 1}^n \text{Spec}(\mathcal{O}_L / \mathfrak{q}_i)$$ are disjoint unions of points. It is most desirable for the number of these points to be equal to the degree of the extension $[L : K]$, but this number could be less, a case which we call "ramified":
Unramified then corresponds to the situation where we don't have points like the red ones above. So this suggests a picture for ramified maps of schemes in terms of their fibers: a ramified map could have branches coming together. Could I have an analogous example of a ramified map of smooth manifolds - one where the fibers branch out?
So my question is this: how is ramification like not being an immersion, visually and in terms of the fibers? It seems like an immersion of smooth manifolds is close to requiring that the fibers have dimension $0$, while "unramified" for schemes is saying that the fibers have dimension $0$ and do not consolidate in branches. Is this at all correct?