Finite fiber of scheme morphism is zero-dimensional?

Question

Let $X$ and $Y$ be locally Noetherian schemes and $f:X\rightarrow Y$ be an étale morphism of finite type. Let $x\in X$ and $y=f(x)$. I would like to know why the fiber $X_y$ is a zero-dimensional scheme. This is statement without justification in Liu's Algebraic Geometry. By a prior result, we know that $X_y$ is finite and reduced when the hypothesis above hold. This seems to intuitively be enough to insure it is zero-dimensional (compare to manifolds consisting of a finite number of points, for example), but I don't know how to prove it using material in Liu.

I've pieced together a proof of this using theorems in the Stacks project that I haven't seen in Liu, but Liu passes over it in a sentence without explanation. Is there an easy way to see it is true?

Answer 1

Let $f : X \to Y$ be a morphism locally of finite presentation.

1. $f : X \to Y$ is unramified if and only if all its fibres are unramified, so we may assume $Y = \operatorname{Spec} k$.
2. $f : X \to \operatorname{Spec} k$ is unramified if and only if $X$ is a disjoint union of affine schemes of the form $\operatorname{Spec} K$ where $K$ is a finite separable field extension of $k$.
3. So if $f : X \to \operatorname{Spec} k$ is unramified, then $X$ is discrete as a topological space.

To see (2), one may also assume $X$ is affine. Then the problem is reduced to commutative algebra – more precisely, to the structure theorem for artinian rings.