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Let $X\subseteq \mathbb{A}^n$ be an affine variety.

The local ring of $X$ at $p\in X$, given by $\mathcal{O}_{X,p}=\{f\in k(X):f \text{ regular at } p\}$ is noetherian because it is a localization of $k[X]$.

If $U\subseteq X$ is open, let $\mathcal{O}_X(U)=\bigcap_{p\in U}\mathcal{O}_{X,p}$. Is this ring noetherian as well?

Note This question has already been asked here before, but was migrated to stackexchange. After the migration, we seem to realize that it is not as trivial as we have thought. Hence I take the risk to crosspost it back here.

Arguments against the claim comes from Section 19.11.13 of Ravi Vakil's notes, where he produced a variety whose ring of global section is not Noetherian. His example, though, is not affine, and it remains a question if we can produce an affine version of that example. This argument is suggested by Marco Flores et al on stackexchenge.

Added later: here is the link to the questin on stackexchange.

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  • $\begingroup$ $X$ here is an affine spectrum of some finitely generated $k$-algebra so it's noetherian in particular. Any localization of this algebra will be noetherian. The structure sheaf is just the algebra here. Any $U \subset X$ open is complementary to a closed subset defined by a finitely generated ideal. localize away from the product of all the generators and you're done. $\endgroup$
    – Saal Hardali
    Commented Apr 7, 2016 at 21:17
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    $\begingroup$ I posted this on Marco Flores's answer, since I was confused why the example was supposed to be affine as well. Vakil shows in this note, §3 that you can modify the construction you refer to to get an affine example. $\endgroup$
    – Takumi Murayama
    Commented Apr 8, 2016 at 2:51
  • $\begingroup$ @SaalHardali The global sections of $U$ are not simply localizations of the global sections on $X$, see the corresponding question on stackexchange. $\endgroup$
    – Fan Zheng
    Commented Apr 8, 2016 at 21:41
  • $\begingroup$ @FanZheng Ah thanks I see why. By definition If $X= Spec A$ we have $\mathcal{O}_X(U)=lim_{supp(f) \subset U} \mathcal{O}_X (X_f)=lim_{supp(f) \subset U} A_f$. Such a limit of noetherian rings doesn't have to be noetherian. $\endgroup$
    – Saal Hardali
    Commented Apr 8, 2016 at 22:33
  • $\begingroup$ Fan Zheng, did you realize that the link given by @Takumi Murayama answers your question? $\endgroup$
    – abx
    Commented Apr 9, 2016 at 3:55

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I am just reposting my comment as an answer.

Vakil shows in this note, §3 that you can modify the construction you refer to to get an affine example. For completeness, I'll write down the example here.

Let $E$ be an elliptic curve over a field $k$, let $N$ be a degree $0$ non-torsion invertible sheaf, and $P$ an invertible sheaf of degree $\ge3$. Then, the total space $Y$ of the bundle $N \oplus P^*$ (where $P^*$ is the $k^*$ bundle associated to $P$) is quasi-affine, and has a ring of global sections that is non-noetherian.

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