point determines a morphism of $K$-algebras $H^0({{\mathscr{O}}}_Y)\to K$, and this forces $H^0({{\mathscr{O}}}_Y)$ to be $K$.

\[03.07.2018–2\]Let $Y$ be a flat and proper $A$-scheme. If $H^0({{\mathscr{O}}}_{Y{\otimes}_AK})=K$, then $H^0({{\mathscr{O}}}_Y)=A$. In particular, if $Y{\otimes}_AK$ is reduced and connected, and has a $K$-point, then $H^0({{\mathscr{O}}}_Y)=A$.

We know that $H^0({{\mathscr{O}}}_Y)$ is a finite $A$-algebra and that $H^0({{\mathscr{O}}}_Y){\otimes}K=K$ by flat base-change [@hartshorne77 III.9.3, p. 255] together with the assumption. Then, $A\subset H^0({{\mathscr{O}}}_Y)\subset K$. Since $A$ is normal, we must have $A=H^0({{\mathscr{O}}}_Y)$. The verification of the last claim follows immediately from Lemma \[03.07.2018–1\].

\[03.05.2018–1\]Let $Y$ be a proper and $H^0$-flat $A$-scheme. Let ${{\mathscr{F}}}$ be a vector bundle on $Y$ whose base-changes ${{\mathscr{F}}}_K$ and ${{\mathscr{F}}}_k$ are trivial. Then ${{\mathscr{F}}}$ is trivial.

It is not difficult to see that the function $y\mapsto{\mathrm{rank}}\,{{\mathscr{F}}}_y$ has to be constant and equal to $n$, say. From the equalities $$\begin{split}
\dim_kH^0(Y_k,{{\mathscr{F}}}_k)&=\dim_kH^0(Y_k,{{\mathscr{O}}}_{Y_k})^{{\oplus}n}\\
&=\dim_KH^0(Y_K,{{\mathscr{O}}}_{Y_K})^{{\oplus}n}\\
&=\dim_KH^0(Y_K,{{\mathscr{F}}}_K),
\end{split}$$ and Corollary 2 of [@mumford70 §5], we conclude that the canonical morphism $$H^0(Y,{{\mathscr{F}}}){\longrightarrow}H^0(Y_k,{{\mathscr{F}}}_k)$$ is surjective. Pick sections $s_1,\ldots,s_n\in H^0(Y,{{\mathscr{F}}})$ such that $\{s_{i}|_{Y_k}\}_{i=1}^n$ is a basis of ${{\mathscr{F}}}_k$. Then, for each point $y$ of the closed fibre, Nakayama’s Lemma shows that $\{s_{i,y}\}_{i=1}^n$ is a basis of the free ${{\mathscr{O}}}_{y}$-module ${{\mathscr{F}}}_y$ so that ${\mathrm{Supp}}({{\mathscr{F}}}/\sum{{\mathscr{O}}}_Ys_i)$ is a closed subset of $Y$ disjoint from 