Is there a "geometric" interpretation of inert primes?

I am currently learning about étale morphisms and how they behave a lot like covering maps. Now if I'm in the example of the ring of integers of a number field, it is possible to think of it as a curve and if I take the ring of integers of a finite extension, this behaves similar as a ramified covering in Riemann surfaces theory. However, in the complex case, I think there is no analogue of the inertia phenomenon. I am trying to see wether this can still be given some kind of geometric meaning.

My current vague interpretation is that of it being some kind of "weighting" the points in the preimage of a given prime, as such it makes it possible for example to carry over the "degree formula" from topology (the degree of a map being the sum of local degrees), as over an unramified prime this is just the sum of the inertia degrees. I'm not really satisfied by this though, and want to know wether there are other or more precise Interpretations out there.

Also I would be interested wether there are other formulas like the one mentioned above, that have a meaning in this context, but maybe have to be adjusted by an inertia term.

• I think there is no real geometric (i.e. over $\mathbb{C}$) analogue, simply because every finite extension of an algebraically closed field is trivial. But perhaps the example of $\operatorname{Spec} \mathbb{C} [x] \to \operatorname{Spec} \mathbb{R} [x]$ is helpful? Commented Mar 5, 2014 at 19:07

If you've done any number theory, you are probably aware of the following classic formula:

Let $L/K$ be an extension of number fields, $\mathfrak{p}\in\text{Spec}(\mathcal{O}_K)$ and $\mathfrak{p}\mathcal{O}_L=\mathfrak{P}_1^{e_1}\cdots\mathfrak{P}_n^{e_n}$. Define $f_i$ to be $\,[\mathcal{O}_L/\mathfrak{P}_i:\mathcal{O}_K/\mathfrak{p}]$. Then,

$$\sum_i e_if_i =[L:K]$$

This is actually just a more general fact about curves, manifesting itself in the world of algebraic number theory. Suppose $\pi:C\to C'$ is a non-constant map of regular pure-dimensional curves. Let $p\in C'$ and let $\{q_1,\ldots,q_n\}=\pi^{-1}(p)$. Let $t$ be a uniformizer of $\mathcal{O}_{C',p}$. Then,

$$\deg\pi=\sum_i \text{val}_{q_i}(\pi^\ast(t))[k(q_i):k(p)]\qquad (\ast)$$

which is precisely the type of generalization of both the local degree formula in complex analysis, and the first formula from number theory.

As for how to think about inert points, I can only give you my (possibly flawed) intuition. When we do modern algebraic geometry, we are constantly torn between the classical intuitive world of Hartshorne chapter 1, and the sometimes unituitive Grothendieck style world.

In particular, one of the biggest differences is that in the classical world, we can actually see all of our points. I don't just mean that in the modern language we deal with primes instead of maximals. No, I literally mean that all of our closed points are visible. Let's take, for example, the scheme $\mathbb{A}^1_\mathbb{R}$. Intuitively, this is the result of taking $\mathbb{A}^1_\mathbb{C}$, the plane, and folding it over the real axis--identifying conjugate points.

You see that passing from $\mathbb{A}^1_\mathbb{C}$ to $\mathbb{A}^1_\mathbb{R}$ we "lost" some of our points. Well, we didn't lose them, but we glued some of them together so that we can't literally see them individually any more, we can only see pairs of them. Thus, we have something similar to "hidden points". These are points which, if we moved to the correct geometric setting (algebraically closed) we would see, but when we are dealing with more arithmetic cases we cannot--they are glued together.

Any truly geometric theorem should not be able to distinguish between a scheme $X/k$ and the scheme $\overline{X}/\overline{k}$. So, when we take formulae like $(\ast)$, and apply them to non-geometric schemes, we see weird terms come up, which come precisely from the formula thinking purely in the geometric case. So, for example, an inert prime of $L/K$ is a prime that, by all rights, should have $[L:K]$ preimages, but all of which are "invisible" because we're in an arithmetic setting.

Thus, formulae like $(\ast)$ are counting something close to what we expect (something like the number of preimage points) and saying it should be globally constant. But, because of possible pathologies in both the maps of schemes we are considering, and the schemes themselves precisely what is meant by preimage is changed. We must look for hidden points (the extension of residue fields), and we must look for doubled points (the ramification index).

This type of intuition, about seeing hidden points, persists a lot through modern algebraic geometry. Another place it commonly rises is in the inclusion of the word "universal" to a lot of definitions. For example, one might intuit that an etale morphism of varieties is an open embedding if and only if it is injective. This is incorrect, and it's incorrect precisely because of hidden points.

Namely, an open embedding is a geometric concept (it's preserved under base change to an algebraic closure), but the notion of injectivity is not (e.g. $\text{Spec}(\mathbb{C})\to\text{Spec}(\mathbb{R})$). Thus, to have a fighting chance at making etale morphisms be open embeddings, we shouldn't require injectivity but injectivity after any base change (known as radical or radiciel). And, indeed, it is true that an etale morphism is an open embedding if and only if it's radiciel.

• Very nice answer! Commented Mar 5, 2014 at 23:29
• wow, thanks for the effort, that is a really cool answer! I will still keep the question open a little bit because I want to see if some other formulas/examples generalizing geometric facts and involving the inertia degree come up. Commented Mar 6, 2014 at 2:19
• @Brenin Thanks you :) Commented Mar 6, 2014 at 6:39
• @jorst No problem. Commented Mar 6, 2014 at 6:41