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:


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:

enter image description here

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:

enter image description here

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.


2 Answers 2


From my perspective, inertia does not really have a geometric flavor. In other words, looking at $\mathbb{Z}[i]$ and its fiber over $(3)$ is hard to picture because it is very far from our usual intuition of topological spaces. If I really need to imagine what is going on, I would say that the fiber has two points (indeed $\mathbb{Z}[i]\otimes_{\mathbb{Z}}\mathbb{F}_3$ is of dimension 2 over $\mathbb{F}_3$) but these two points lies in an algebraic extension.

Perhaps a better way to have some geometric intuition is to start with more geometric objects than $\mathbb{Z}[i]$.

So let's have a look at $Y=\operatorname{Spec}A$ with $A=\mathbb{R}[y]$ and $X=\operatorname{Spec}B$ with $B=A[x]/(x^2-y)$. Here $Y$ is the vertical line on our usual plane, and $X$ is the parabola $y=x^2$. The map $X\rightarrow Y$ is given by $(x,y)\mapsto y$. Thus the fibers over $y_0$ are all the points of the parabola lying in the horizontal line $y=y_0$.

So here are the different fibers :

  • if $y_0>0$ then there are two such points $(\sqrt{y_0},y_0)$ and $(-\sqrt{y_0},y_0)$. Algebraically : the ideal $(y-y_0)$ in $B$ decomposes as $(x-\sqrt{y_0})(x+\sqrt{y_0})$. So $(y-y_0)B$ is split (in particular it is unramified).
  • if $y_0=0$, then there is only one point $(0,0)$. Algebraically the ideal $(y-y_0)$ in $B$ decomposes as $(x-x_0)^2$. This is an example of ramification, and as your professor said, you can see that there is a ramification geometrically since then the line $y=0$ is tangent to the parabola.
  • if $y_0<0$, then there is no real points, but there is two complex ones : $(i\sqrt{-y_0},y_0)$ and $(i\sqrt{-y_0},y_0)$. These complex points does not comes from two different ideals in $B$ and in fact the ideal $(y-y_0)B$ is prime and the residue field is an extension of degree 2 of $A/(y-y_0)$. So $(y-y_0)$ is inert in $B$.

See, this is not easy to picture it : if we draw only real points, these inert primes should have empty fibers. If we draw complex points, well, $\mathbb{C}$ being algebraically closed, there is no inertia...

Here the map $X\rightarrow Y$ is of degree 2, so a prime in $Y$ is either split, ramified or inert. Of course if the degree is greater than 2, the fiber over each prime may be a bit of everything. For example, look at $X=\operatorname{Spec}A[x]/(y-P(x))$ where $P$ is your favorite polynomial of degree 3 or 4, and draw horizontal lines.

Now let's have a look at the second picture. I remember that when I first met them, I asked myself questions similar to yours.

  • Yes $X$ should be regular, what you see is not $X$.
  • Inertia is not represented at all, this is a picture of complex points.
  • What looks like a singularity is a representation of a point where locally the map $X\rightarrow Y$ is given by $z\mapsto z^e$ where $e$ is the number of branches going through the point.

The idea of this diagram is to represent the behavior of a solution of an equation $P_{y_0}(x)=0$ where $y_0$ varies.

So take again the parabola $y=x^2$. A diagram of the situation has an X shape. Let $y_0\in Y$ be any point (except 0). Then there is above two solutions to the equation $x^2=y_0$ (remember that we look at complex point and we forgot about inertia). Choose one, say $x_0=\sqrt{y_0}$. Then as $y_0$ varies, $x_0$ varies continuously, until we reach $y_0=0$. Then if $y_0$ keeps moving $x_0$ can choose two different paths. We have an embranchment point. And that is why we may want to draw it with a singularity.

If we want to add inertia, we could imagine points like those of Mumford's picture : inert primes do not make embranchment points, so there should be several line which does not meet through a single point (single because we have only a single prime to represent several points in an algebraic extension). This is what you see above $(3)$ in Mumford's picture for example.

  • $\begingroup$ Thank you, you really made my day with that geometric example 👏🏻 $\endgroup$
    – seldon
    Feb 17, 2019 at 17:00

I believe the following holds: Given any finite extension $K:=\mathbb{Q} \subseteq L$ with rings of integers $\mathcal{O}_L$, we get a canonical map

R1. $\pi: S:=Spec(\mathcal{O}_L)\rightarrow T:=Spec(\mathbb{Z})$.

The extension $\mathbb{Q} \subseteq L$ is always ramified, hence there are primes $p \in \mathbb{Z}$ where the fiber $\pi^{-1}((p))$ is non-reduced. The map $A:=\mathbb{Z} \subseteq B:=\mathcal{O}_L$ is an integral extension of rings and $\mathcal{O}_L$ is a finite rank free $\mathbb{Z}$-module. Hence $A\subseteq B$ is a flat map of rings. The relative module of Kahler differentials $\Omega^1_{B/A}$ is non-zero, hence the map $\pi$ is not etale. The support $Z\subseteq S$ of $\Omega^1_{B/A}$ is a closed subscheme, and since $\pi(Z):=W\subseteq T$ is a closed subscheme (the map $\pi$ is finite), we get an open set $U:=T-W$. I believe the induced map $\pi_U: \pi^{-1}(U) \rightarrow U$ is etale, since $\pi_U$ is flat and $\Omega^1_{\pi^{-1}_U(U)/U}=0.$ Here you need a relative version of Hartshorne Exercise III.10.3. Maybe the stacks project does this, but I do not have a precise reference.

The "geometric" picture seems to be: Over $U$ it follows the number of closed points in the fiber is constant and the fiber rings are reduced. Outside of $U$ this is no longer true.

You find a related question here:

Why isn't every finite locally free morphism etale?


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