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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?

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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 $\geq 2$.

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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.

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    $\begingroup$ Yes, the connectedness of $S$ is clearly a necessary condition, and I just assumed OP forgot to require that, so I gave a counterexample with $S$ connected. $\endgroup$
    – js21
    May 3, 2018 at 8:33
  • $\begingroup$ @js21: I agree that your example is much more interesting. I wanted to note that the question as stated actually has trivial counterexamples as well. – Remy "minimal counterexamples" van Dobben de Bruyn. $\endgroup$ May 3, 2018 at 9:06

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