I am trying to understand geometrically the ramification of primes in a finite separable field extension.
Let $A$ be a Dedekind domain with fraction field $K$ and $L/K$ a finite separable field extension of degree $n$, and let $B$ be the integral closure of $A$ in $L$, which is then also a Dedekind domain. Let $X=\text{Spec}(B)$ and $Y=\text{Spec}(A)$, and consider the projection $X\to Y$ which sends a prime $\mathfrak{q}\in X$ over $\mathfrak{p}\in Y$ to $\mathfrak{p}$ (i.e., the map of affine schemes corresponding to the inclusion $A\to B$).
Let $\mathfrak{p}\in Y$ be a non-zero prime ideal in $A$. The fiber over $\mathfrak{p}$ is then the set of prime ideals $\mathfrak{q}_{i}\in X$ such that $$ \mathfrak{p}B=\mathfrak{q}_{1}^{e_{1}}\cdots \mathfrak{q}_{g}^{e_{g}}$$ is the corresponding prime factorization in $B$ (uniquely determined, because $B$ is a Dedekind domain). Let $f_{i}=[\kappa(\mathfrak{q}_{i}):\kappa(\mathfrak{p})]$ be the corresponding residue degrees. Then:
$$\sum_{i=1}^{g}f_{i}e_{i}=n$$
We say that $\mathfrak{q}_{i}$ is ramified above $\mathfrak{p}$ if the ramification index $e_{i}$ is strictly greater than 1, and unramified if $e_{i}=1$. We say that $\mathfrak{p}$ is unramified in $B$ if all $e_{i}=1$; that $\mathfrak{p}$ splits in $B$ if all $e_{i}=f_{i}=1$; and we say that $\mathfrak{p}$ is inert in $B$ if $g=e_{1}=1$.
We can picture both $X$ and $Y$ as algebraic curves and $X$ as a ramified cover of $Y$. I am trying to understand the precise geometric meaning of all the previous definitions in terms of this geometric picture.
My professor said that ramified primes above $\mathfrak{p}$ are those points $\mathfrak{q}_{i}\in X$ where the fiber over $\mathfrak{p}$ meets $X$ in a tangent point (I understand that by fiber he means the vertical line above $\mathfrak{p}$ when we picture $Y$ as a straight line). I tried to match this interpretation with the example of $A=\mathbb{Z}$ and $B=\mathbb{Z}[i]$, based on the picture of $\text{Spec}(\mathbb{Z}[X])$ of Mumford:
Prime ideals in $\mathbb{Z}[i]=\mathbb{Z}[X]/(X^{2}+1)$ should correspond to prime ideals containing $(X^{2}+1)$. That is, they should correspond to points in the curve $V(X^{2}+1)$ drawn in the picture.
The picture above $(2)$ matches this interpretation perfectly, since $(2)$ ramifies in $\mathbb{Z}[i]$. And so does the picture above $(5)$, since $(5)$ splits into two primes in $\mathbb{Z}[i]$. But now I have a problem with the picture above $(3)$. There should be only one point, namely $(3,X^{2}+1)\in X$, but I guess Mumford just doesn't draw it. But even if he did, how would he draw this point without creating a singularity on the curve? Because, being the spectrum of a Dedekind domain, $X$ should be a regular curve, right? So how would we draw this point?
But if you draw this point in the naive way, introducing a singularity, suddenly the residue degree seems to make some geometric sense to me: it seems to be the number of parameters that we need to describe the curve around the corresponding point. How much of this naive impression is true? Is this just a coincidence? Because algebraically this doesn't make much sense to me. One should be able to describe the curve locally with a single parameter, because all the local rings are DVR. I will try to summarize all these doubts in the following question:
What is the precise geometric meaning of the ramification index and the residue degree?
Finally, if possible, I would like to hear some clarification comments on the source that started all these doubts (initially I was naively happy with my professor's interpretation): Neukirch's Algebraic Number Theory. In chpater I, Section 13 (One-dimensional Schemes) he draws this ramification situation:
I have several problems with this picture.
- Shouldn't $X$ be a regular curve?
- Why aren't the ramification points tangent, as my professor claimed them to be?
- If this is really the picture, how do we distinguish geometrically between inert points and ramification points? For example, the point on the right. Is it inert or is it ramified?
Neukirch says that this picture is only a fair rendering of the algebraic situation when the residue fields in $A$ are algebraically closed, but this doesn't help much to answer the previous questions.
Thanks for the spending the time to read this long question. I am quite confused with this issue and I wanted to express my doubts as clear as possible.