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?
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]