Let $X$ be a scheme of finite type over $\mathrm{Spec}(A)$, where $A$ is a commutative ring. Let $U\subset X$ be any open subset, is $\Gamma(U,\mathscr{O}_{X})$ a finitely generated $\Gamma(X,\mathscr{O}_{X})$-algebra?
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$\begingroup$ (off topic) I stumbled upon the "with 1". The name ring include this. See arxiv.org/abs/1404.0135. $\endgroup$– Martin BrandenburgCommented Nov 15, 2022 at 9:35
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1$\begingroup$ Welcome new contributor. This is a perennial question on MathOverflow. In addition to the great answer below by user @Gro-Tsen, you can also refer to the following MathOverflow answer: mathoverflow.net/questions/240170/… There are many other examples. $\endgroup$– Jason StarrCommented Nov 15, 2022 at 12:15
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No. The following counterexamples are part of the folklore:
Even if $X$ is an affine variety over a field $k$ (so $\Gamma(X,\mathscr{O}_X)$ is certainly of finite type over $k$) and $U\subseteq X$ open, then $\Gamma(U, \mathscr{O}_X)$ can still fail to be of finite type over $k$ (and in particular, over $\Gamma(X, \mathscr{O}_X)$): see Ravi Vakil, “An example of a nice variety whose ring of global sections is not finitely generated”.
Even if $X$ is a connected projective variety over a field $k$ (so $\Gamma(X, \mathscr{O}_X) = k$ here) and $U\subseteq X$ open, then $\Gamma(U, \mathscr{O}_X)$ can fail to be noetherian (and in particular, of finite type over $k$): see Manuel Ojanguren, “Un ouvert bizarre” [in French, but it's only ½ page long].
(To be clear, here, “variety over $k$” := “reduced scheme of finite type over $k$”.)
This other question, which also links to the same two counterexamples, is also relevant. See also the context of Hilbert's 14th problem.
