Inverse limit of finite flat morphisms

Question

Suppose that an inverse limit of finite flat morphisms $X_k\to S$ of qcqs schemes, with affine transition maps, is $X\to S$, such that a closed fiber of $X\to S$ is finite.

Is $X\to S$ finite?

Answer 1

No (in general). Take $S = \mathrm{Spec}(A)$ and $X_k = \mathrm{Spec}(A[T]/(T^2))$, with affine transition maps given by $T \mapsto f T$ for some $f \in A$. The limit $X$ is the spectrum of $A \oplus A[f^{-1}] T$ (with $T^2 = 0$). Then $X \rightarrow S$ is an isomorphism (hence finite) over the closed subscheme $V(f)$ of $S$, while $X \rightarrow S$ is finite if and only if $A \rightarrow A[f^{-1}]$ is finite. To give a specific counterexample, just take $A = \mathbb{Z}$ and $f$ any integer $\geq 2$.

Answer 2

A different counterexample is given by taking $S = \operatorname{Spec} \mathbb F_p \coprod \operatorname{Spec} \mathbb C$, and letting $$X_k = \operatorname{Spec} \mathbb F_{p^k} \coprod \operatorname{Spec} \mathbb C.$$ Then $X = \operatorname{Spec} \bar {\mathbb F}_p \coprod \operatorname{Spec} \mathbb C$, which is not finite over $S$. But the fibre over $\operatorname{Spec} \mathbb C$ is finite.