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I have been under the impression for several years that if $L/K$ is an extension of number fields, then an infinite place of $K$ is said to ramify in $L$ if it comes from a real embedding of $K$ which extends to complex embeddings of $L$.

I am currently reading Neukirch's magisterial Algebraic Number Theory and he seems to be saying something inconsistent with this on p. 184. He writes:

... This convention suggests that we consider $e$ as being an infinite prime number, and the extension $\mathbb{C}\mid \mathbb{R}$ as being unramified with inertia degree 2...

... If $L\mid K$ is a finite extension of $K$, then we denote the primes of $L$ by $\mathfrak{P}$ and write $\mathfrak{P}\mid\mathfrak{p}$ to signify that the valuations in the class $\mathfrak{P}$, when restricted to $K$, give those of $\mathfrak{p}$. In the case of an infinite prime $\mathfrak{P}$, we define the inertia degree, resp. the ramification index, by $$ f_{\mathfrak{P}\mid\mathfrak{p}} = [L_\mathfrak{P}:K_\mathfrak{p}],\text{ resp. } e_{\mathfrak{P}\mid\mathfrak{p} = 1}$$

Isn't he saying that an infinite prime is always unramified then? (And either inert or split?)

If this is what he's saying, then was my previous understanding wrong? Or is Neukirch using a different convention than everybody else? If so, what's the thinking behind the different conventions?

If on the other hand there is no conflict between Neukirch and my previous understanding, then what am I misunderstanding about what he's saying?

Basically, what's going on around here?

Thanks in advance for any help you can offer.

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    $\begingroup$ Seems like a dupe of math.stackexchange.com/questions/769178/… but that has no actual answer either. Though a comment there suggests it is just convention. But a more definite answer would be good give this is a recurring issue. $\endgroup$
    – quid
    May 28, 2015 at 17:45

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