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Let $L/K$ be an extension of number fields (i.e. finite extensions of $\mathbb{Q}$). The ramification of finite places, i.e. prime ideals, is controlled by the module of Kähler differentials $\Omega_{L/K}$. But what about infinite places? We say that an infinite place $\sigma: K\to \mathbb{R}$ ramifies if it extends to two distinct places $\sigma, \bar \sigma: L\to \mathbb{C}.$ So my question is, can we somehow enhance Kähler differentials so that they tell us something about ramificatioin of those infinite places? Or do we have some other algebraic gadget for that?

Tom Perutka
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  • What do you mean with two distinct places ? The embedding $\Bbb{Q\to R}$ extends in $6$ ways ($3$ absolute values) to an embedding $\Bbb{Q}(e^{2i\pi /7})\to \Bbb{C}$ – reuns Nov 11 '20 at 16:15
  • Yeah, this was said in a bad way. I meant to say that a real infinite prime (i.e. embedding $K\to \mathbb{R}$) ramifies if it extends to a complex prime (i.e. two conjugate embeddings $\sigma, \bar \sigma: L\to \mathbb{C}$). – Tom Perutka Nov 11 '20 at 21:04
  • And you meant $\Omega_{O_L/O_K}$ as $\Omega_{L/K}=0$. What happens when we try to take the completion (in some way or another) ? – reuns Nov 11 '20 at 21:30
  • That's a good question. If we have an infinite place $\tau$ of $L$ lying above an infinite place $\sigma$ of $K$ and we look at the completions, we get that $\sigma$ ramifies iff the degree of $L$ completed at $\tau$ over $K$ completed at $\sigma$ is two. So I guess we don't need to consider anything more complicated. – Tom Perutka Nov 12 '20 at 09:18
  • Since we are considering $O_L$-modules, $\Bbb{Z}[\sqrt{-d}]/\Bbb{Z}$ is already complete, in the other cases the completion of $O_L$ will be $\Bbb{R}$ or $\Bbb{C}$ – reuns Nov 12 '20 at 14:03

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At the following link you will find some info on Kahler differentials and rings of integers in number fields:

Why isn't every finite locally free morphism etale?

Note: Any finite extension $K \subseteq L$ is separable, hence $\Omega^1_{L/K}=0$ is zero.

Citation: "For any finite extension $K \subseteq L$ of number fields, it follows the morphism $\pi: S:=Spec(\mathcal{O}_L) \rightarrow T:=Spec(\mathcal{O}_K)$ is always integral, locally free (and ramified when $K=\mathbb{Q}$). Hence it is seldom etale. There is an open subscheme $U \subseteq T$ with the property that $\pi_U:\pi^{-1}(U) \rightarrow U$ is etale. The fiber $\pi^{-1}((p))$ in formula F1 above is unramified iff $l_i=1$ for all $i$. The module of Kahler differentials $\Omega:=\Omega^1_{\mathcal{O}_L/\mathcal{O}_K}$ is zero on the open subscheme $\pi^{-1}(U)$. Note moreover that the field extension

$\mathcal{O}_K/\mathfrak{p} \subseteq \mathcal{O}_L/\mathfrak{q}$

is always separable since it is a finite extension of finite fields (here the ideals $\mathfrak{p}, \mathfrak{q}$ are maximal ideals). The ramified prime ideals in $\mathcal{O}_K$ are related to the discriminant of the extension $L/K$ (see in "Algebraic number theory", page 49, by Neukirch). In Neukirch he defines the discriminant using a basis for $L$ over $K$, but it can also be defined using the module of Kahler differentials $\Omega$."

hm2020
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