Showing if an étale morphism is finite by looking at the fibers Let $f:X\to S$ be an étale morphism of schemes, with $S=\text{Spec}(A)$, for $A$ a local domain, with residue field $k$ and field of fractions $K$. Denote by $$\eta:=\text{Spec}(K)\hookrightarrow S \hookleftarrow s:=\text{Spec}(k)$$
the open and closed points.

Question: Can we decide that $f$ is finite by just looking at the fibers $X_\eta\to \eta$ and $X_s\to s$ ?

I know that $f$ could be not finite, for example it if is an open immersion (e.g. $X=\eta$). But in this case the special fiber will be empty.
I suspect (but I don't have a reference) that a necessary condition for $X\to S$ to be finite is that $X_{\overline \eta}$ and $X_{\overline s}$, the base change to the separable closures, have the same number of points (i.e. the degree of $X_\eta\to \eta$ and $X_s\to s$ are equal).

Is this true? Is this condition sufficient?

 A: Unfortunately, this is false. Let $Y$ be the affine line with two origins over a field $k$. I claim that the projection map $Y\to \Bbb A^1_k=\operatorname{Spec} k[t]$ is etale because it is locally an isomorphism - on the affine open set which is the complement of one of the origins, this map is the identity. So $Y\to \Bbb A^1_k$ is etale, and thus the base change $Y_{\operatorname{Spec} k[t]_{(t)}}\to \operatorname{Spec} k[t]_{(t)}$ is etale. Letting $X=Y_{\operatorname{Spec} k[t]_{(t)}}\sqcup\operatorname{Spec} k(t)$, we see that the generic and special fiber over any field extension both have two points, but this map is not finite since it is not affine.
On the other hand, we can't tell it apart from the obvious finite double cover $\operatorname{Spec} k[t]_{(t)}\sqcup \operatorname{Spec} k[t]_{(t)}\to \operatorname{Spec} k[t]_{(t)}$, which has exactly the same fibers. Somehow the moral is that etale and statements about fibers are both local conditions, while finite is a global condition, and we need more to get the desired conclusion.
