Suppose I have a morphism of schemes $f: X \to Y$, and the fibre $f^{-1}(y)$ as a scheme over $k(y)$, the residue field of $y$, is a union of finite set of points $\operatorname{Spec} k_i$, $k_i$ finite separable over $k(y)$. Let $K$ be the algebraically closed field.
I want to deduce that this is equivalent to the geometric fibre of $y$ is a finite set of reduced points.
What I have done: First $$ (\bigcup_i \operatorname{Spec} k_i) \times_{\operatorname{Spec} k(y)} \operatorname{Spec} K = \bigcup_i (\operatorname{Spec} k_i \times_{\operatorname{Spec} k(y)} \operatorname{Spec} K). $$ Then $$ \operatorname{Spec} k_i \times_{\operatorname{Spec} k(y)} \operatorname{Spec} K = \operatorname{Spec} k_i \otimes_{k(y)} K = \operatorname{Spec} K $$ From this it seems that the geometric fibre is then a disjoint union $\operatorname{Spec} K$'s, and has nothing to do with $k_i$'s being separable, just that it is contained in $K$... I think I am missing something here. Any clarification would be appreciated. Thank you.