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Motivation: When I was young(er), I was once shocked to learn that, for $X$ a scheme of finite type, $\Gamma(X,\mathscr{O}_X)$ can fail to be of finite type. Now that I am no longer so young and bashful, I followed von Neumann's advice and “got used to it”. But what if, instead of getting used to it, we tried to generalize the “finite type” property so that the expected facts work?

Question: Let us define a property $\mathbf{F}$ of schemes over a field $k$, generated by the following conditions:

  • affine space $\mathbb{A}^n_k$ satisfies $\mathbf{F}$,

  • if $X,Y$ satisfy $\mathbf{F}$ then $X\times_{\operatorname{Spec}k} Y$ satisfies $\mathbf{F}$,

  • if $X$ satisfies $\mathbf{F}$, then any closed subscheme $Z \subseteq X$ satisfies $\mathbf{F}$,

  • if $X$ satisfies $\mathbf{F}$, then any open subscheme $U \subseteq X$ satisfies $\mathbf{F}$,

  • if $X$ is a $k$-scheme and $U_1,\ldots,U_n$ are finitely many open subschemes of $X$ which cover $X$ and satisfy $\mathbf{F}$, then $X$ satisfies $\mathbf{F}$,

  • if $X$ satisfies $\mathbf{F}$, then $\operatorname{Spec}\Gamma(X,\mathscr{O}_X)$ satisfies $\mathbf{F}$,

can we characterize the $k$-schemes which satisfy $\mathbf{F}$?

(Feel free to answer the question, instead, for the property $\mathbf{F}'$ adding any “reasonable” stability conditions which are satisfied by schemes of finite type, or maybe a generalization of the last condition, if it makes the answer easier or you think it's more natural.)

At the very least, can we give a decent necessary condition for satisfying $\mathbf{F}$? (Unless I made an embarrassing mistake, being of finite type is a sufficient condition.)

NB: this question is somewhat related.

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  • $\begingroup$ An example of a variety whose ring of global sections is not finitely-generated is given in Vakil's Foundations of Algebraic Geometry, Exercise 19.11.13. If I were to go out on a limb, I'd guess that $Spec k[x_1,x_2,\dots]$ is in $\mathbf F$, and therefore that $Spec R$ is in $\mathbf F$ for any countably-generated $k$-algebra $R$, so that any scheme admitting a finite open cover by $Spec R$'s where $R$ is countably generated, is in $\mathbf F$, and maybe that's exactly what $\mathbf F$ is? $\endgroup$ Commented Feb 19, 2021 at 19:36
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    $\begingroup$ For every scheme with this property, the total ring of fractions is essentially of finite type. $\endgroup$ Commented Feb 19, 2021 at 20:37

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