Ramified primes, as defined by Neukirch

Question

In ch. I, §8, p. 49 of the English translation of $\textit{Algebraic Number Theory}$, Neukirch states the following (I paraphrase, but retain his notation):

Let $L/K$ be an extension of number fields, and let $\mathcal{O}$ and $\mathcal{o}$ be their respective rings of integers.

Then every prime ideal $\mathfrak{p} \subset \mathcal{o}$ possesses a unique factorisation

$$\mathfrak{p} = \mathfrak{P}_1^{e_1}...\mathfrak{P}_1^{e_1}$$

where the $\mathfrak{P}_i$ are prime ideals in $\mathcal{O}$.

He then says that the prime ideal $\mathfrak{P}_i$ in the above factorisation is said to be $\textbf{unramified}$ over $K$ if $e_i = 1$ ($\textit{i.e.}$ if the prime only appears once in the factorisation) $\underline{\textit{and}}$ the residue field extension $(\mathcal{O}/\mathfrak{P}_i)/(\mathcal{o}/\mathfrak{p})$ is separable.

But $(\mathcal{O}/\mathfrak{P}_i)/(\mathcal{o}/\mathfrak{p})$ is a finite extension of finite fields, so will it not always be separable? Why is this included in the definition?

Below I have included the excerpt from the book I am referring to.

Thank you for your attention.

$\textbf{Addendum:}$ In response to the comments below, include two more excerpts from ch. I, §9, p. 58 and p. 59 respectively.

From p. 58:

From p. 59:

I should explain that his notation is $\kappa(\mathfrak{P}) := \mathcal{O}/\mathfrak{P}$ and $\kappa(\mathfrak{p}) := \mathcal{o}/\mathfrak{p}$.

Now if the assumption was only included for formal reasons (because he did not want to assume that finite fields are perfect), then why would he repeatedly refer to the separability of the residue field extension as a "special case"?

Answer 1

I took a look at the corresponding section in the book. Although the end goal is of course to apply all this to number fields, this is actually not the assumption in that section. Neukirch works at the level of generality of Dedekind domain, not rings of integers of number fields. And in that more general case, the residue fields don't have to be finite, and the inertia extension does not have to be separable.

Answer 2

Given your interest in L-functions you should look at the finite extensions of the $p$-adic integers $\Bbb{Z}_p$. The main point is Hensel lemma: if $f\in \Bbb{Z}_p[x]$ and $f(a)=0\bmod p,f'(a)\ne 0\bmod p$ then there is $b\in \Bbb{Z}_p$ such that $b=a\bmod p, f(b)=0$. This is what makes $p$-adic fields much simpler than $\Bbb{Q}$, because we can use this to classify all the extensions and connect some apparently very different fields. In the general setting we replace $\Bbb{Z}_p$ by $\varprojlim O/\mathfrak{p}^n$ where $\mathfrak{p}$ is a non-zero invertible prime ideal of $O$ (corresponding to a discrete valuation). The unramified condition is what we need to ensure the extension is generated by the lift of some roots of separable polynomials $\bmod \mathfrak{p}$. And we can understand $O$ by looking at $\varprojlim O/\mathfrak{p}^n$ for each prime ideal.

Answer 3

In number fields there is a basic theorem that the ramified primes are the prime factors of the discriminant (prime ideal factors if the base field is bigger than $\mathbf Q$). If you extend this result to a finite extension of fraction fields of Dedekind domains, where residue fields are no longer automatically finite (really, no longer automatically perfect) then you discover that the prime factors of the discriminant ideal are the primes downstairs whose prime factorization upstairs has all exponents equal to $1$ and where the residue field extensions at all primes upstairs are separable. That is the simplest reason to define unramified primes in a general setting (not just number fields) to include the peculiar extra property of separable residue field extensions.