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Let $U$ and $V$ be irreducible varieties and $f\colon V\rightarrow U$ be a proper surjective morphism.
Assume, $f^{-1}(\eta)$ is irreducible ($\eta$ is the generic point of $U$). $\require{AMScd}$ \begin{CD} f^{-1}(\eta) @>\displaystyle >> V\\ @V \displaystyle \ V V\ @VV \displaystyle{f} V\\ \operatorname{Spec}(k(\eta)) @>> \displaystyle > U \end{CD} I want to prove

Let $\phi\in H^{0}(V,\mathcal{O}_{V})$, then $\phi \in K(U)$ ($K(U)$ is the function field of $U$).

Intuitively, $\phi\mid_{f^{-1}(\eta)}$ is constant, so $\phi\in k(\eta)$. But, I don't understand how to prove rigorously.
I think I have to consider flat base change, $\require{AMScd}$ \begin{CD} f^{-1}(\eta)\otimes \overline{k(\eta)} @>>>f^{-1}(\eta) @>\displaystyle >> V\\ @V \displaystyle \ V V\ @VV \displaystyle{f\mid_{f^{-1}(\eta)}} V@VV\displaystyle{f} V\\ \operatorname{Spec}(\overline{k(\eta)}) @>>> \operatorname{Spec}(k(\eta)) @>> \displaystyle > U \end{CD}

I'm stuck. Thanks in advance.

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    $\begingroup$ What's the problem? As you write, $\phi_{|f^{-1}(\eta )}\in H^0(f^{-1}(\eta ),\mathscr{O})=f^*k(\eta )=f^*K(U)$. $\endgroup$
    – abx
    Commented Dec 24, 2020 at 9:54
  • $\begingroup$ Thanks. You mean since $f^{-1}(\eta)$ is proper irreducible over $k(\eta)$, $f^{*}\colon k(\eta)\rightarrow H^{0}(f^{-1}(\eta),\mathcal{O})$ is an isomorphism ? $\endgroup$
    – Aoki
    Commented Dec 24, 2020 at 15:23
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    $\begingroup$ Yes, that's what I mean. $\endgroup$
    – abx
    Commented Dec 24, 2020 at 15:49

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