What is the proper terminology for this? Given a purely real, rational integer $p$ that is prime in $\mathbb{Z}$, we know very well that it ramifies in $\mathbb{Q}(\sqrt{pm})$ (where $m$ is a nonzero integer coprime to $p$), it is inert in some of the other quadratic rings and it splits in the others.
In a ring of degree $4$, things are of course a bit more complicated than that. For example, in $\mathcal{O}_{\mathbb{Q}(\zeta_8)}$, we see that $2$ is ramified, since it's ramified in each of the three intermediate fields ($\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt{-2})$ and $\mathbb{Q}(\sqrt{2})$).
Furthermore, we see that $(1 - \zeta_8)(1 + \zeta_8) = 1 - i$ and $(1 - {\zeta_8}^3)(1 + {\zeta_8}^3) = 1 + i$. Is this what they call "ramifies completely"?
Turning our attention to $3$, we see that it is not prime in $\mathcal{O}_{\mathbb{Q}(\zeta_8)}$, because, although it does not split in two of the intermediate fields, it does split in $\mathbb{Z}[\sqrt{-2}]$. I may have overlooked something, but as far as I can tell, the equation $x^4 + b x^3 + c x^2 + d x \pm 3 = 0$ has no solutions  in $\mathcal{O}_{\mathbb{Q}(\zeta_8)}$.
If I'm right, this would mean that $3$ does not split as "completely" as $2$ ramifies. Assuming I'm correct in these assertions, am I using the correct terminology? And if not, what is the correct terminology?
 A: Here's the general situation for number fields: Let $L/K$ be a degree $n$ extension of number fields, let $\newcommand{\OO}{\mathcal{O}}\OO_K$ and $\OO_L$ be their rings of integers, and let $P$ be a nonzero prime ideal of $\OO_K$. Ramification theory concerns the factorization of the ideal $P \OO_L$ (and its implications for the structure of the Galois group); a good reference for this is chapter 1 of Neukirch's Algebraic Number Theory.
Since $\OO_L$ is a Dedekind domain, we can uniquely factor $P \OO_L$ as a product of prime ideals
$$P \OO_L = Q_1^{e_1} \cdot \ldots \cdot Q_r^{e_r}$$
for some distinct nonzero prime ideals $Q_1, \dots, Q_r$ of $\OO_L$ and some positive integers $e_1, \dots, e_r$. (This factorization of ideals is unique up to ordering, even if $\OO_L$ doesn't have unique prime factorization of elements.)
The exponent $e_i$ is called the ramification index of $Q_i$. We also define the inertia degree of $Q_i$ to be $f_i = [\OO_L/Q_i : \OO_K/P]$, the degree of the extension of residue fields. We have
$$ n = e_1 f_1 + \dots + e_r f_r.$$
(If $L$ is a Galois extension of $K$, then $f_i = f_j$ and $e_i = e_j$ for all $i, j$. In this case, we usually refer to them as the "inertia degree $f$ of $P$ in $L/K$" and the "ramification index $e$ of $P$ in $L/K$", and we have $n = efr$.)
There are three extreme cases:


*

*If $e = n$ (that is, $P \OO_L = Q^e$), then we say $P$ is totally ramified.

*If $f = n$ (that is, $P \OO_L$ is already a prime ideal), then we say $P$ is inert.

*If $r = n$ (that is, $P \OO_L = Q_1 \cdot \ldots \cdot Q_n$ for distinct $Q_1, \dots, Q_n$), then we say $P$ splits completely.


There are also various intermediate cases:


*If $e_i > 1$ for some $i$, then we say $Q_i$ and $P$ are ramified in $L/K$.

*If $e_i = 1$ for all $i$, then we say $Q_i$ and $P$ are unramified.

*If $1 < r < n$, then I don't think there's a universally accepted term for it, but I'd personally say something like "splits partially". ($P$ could be ramified or unramified in this case.)

A: The terminology of "ramify" and "split" carries over, but now needs the adjectives "completely" and [placeholder for others to add appropriate adjectives].
The ring of algebraic integers of $\mathbb{Q}(\sqrt{2} + \sqrt{3})$ provides richer examples, in my opinion. We see that $2$ ramifies, since it ramifies in each of the intermediate fields ($\mathbb{Q}(\sqrt{2}), \mathbb{Q}(\sqrt{3}), \mathbb{Q}(\sqrt{6})$; if the second one of these gives you any doubt, remember that, in $\mathbb{Z}[\sqrt{3}]$, $\langle 1 - \sqrt{3} \rangle \subseteq \langle 1 + \sqrt{3} \rangle$ and vice-versa, so in fact $\langle 1 - \sqrt{3} \rangle = \langle 1 + \sqrt{3} \rangle$ and $\langle 2 \rangle$ is a square ideal).
Thus we say that $2$ "ramifies completely." Furthermore, the [quartic integer of norm $2$ I haven't found it yet]
The situation with $3$ is a bit more interesting than that. Obviously $3$ ramifies in $\mathbb{Z}[\sqrt{3}]$ and $\mathbb{Z}[\sqrt{6}]$, but it's actually inert. Thus we say that $3$ ramifies in a [someone else please fill in right term]
