I encountered the following lemma in Qing Liu's Algebraic Geometry and Arithmetic Curves:
Let $f:X\to Y$ be a morphism of finite type between locally Noetherian schemes. Then $f$ is unramified if and only if for all $y\in Y$, the fibre $X_y$ is finite and $k(x)/k(y)$ is separable for all $x\in X_y$.
To test my understanding of this, I tried a quick example with number fields and found that I don't.
The extension of integer rings $\mathbb Z[i]/\mathbb Z$ is ramified at $2\mathbb Z$, so if $X := \operatorname{Spec}\mathbb Z[i]$, $Y := \operatorname{Spec}\mathbb Z$, and $y :=2 \mathbb Z$, then $$X_y = X\times_Y\operatorname{Spec}k(y) = \operatorname{Spec}(\mathbb Z[i]\otimes_{\mathbb Z}\mathbb F_2)$$ should be infinite, but $$\mathbb Z[i]\otimes_{\mathbb Z}\mathbb F_2 \cong \mathbb Z[X]/(X^2+1)\otimes_{\mathbb Z}\mathbb F_2 \cong \mathbb F_2[x]/(x^2+1) \cong \mathbb F_2[x]/(x+1)^2$$ which consists of one point. This makes perfect sense, since $2\mathbb Z[i] = (1+i)^2$; only one prime lies over $2$. More generally, the fibre over a prime $\mathfrak{p}$ in any number field $\mathcal{O}_K$ consists of the primes lying over $\mathfrak{p}$, and there are always finitely many; ramification actually reduces the number of such primes. The other condition, that $k(y)$ be separable over $k(x)$ is trivially satisfied in the case of number fields: $k(y)\cong\mathcal{O}_K/\mathfrak{p}$ is a finite field, hence perfect, so $k(x)/k(y)$ being finite is algebraic, hence separable.
Am I missing something? What distinguishes the fibres of ramified and unramified points? From the example, it seems instead of finiteness, perhaps reducedness of the fibres is the correct conclusion, is this correct?
Somewhat unrelated, but is there anything geometric that can be said about tame/wild ramification in the $p$-adic case?