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?

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    $\begingroup$ $O_X(U)$ is also a localization of $k[X]$. $\endgroup$
    – YCor
    Mar 17, 2016 at 18:23
  • $\begingroup$ @DenisNardin Are you saying $\mathcal{O}_X(U)=\{\frac{f}{g}\mid f,g\in k[X], g(p)\neq 0\forall p\in U\}$?. Because this is not true. Consider $X=V(xy-zw)\subseteq\mathbb{A}^4$, and $U=U_y\cup U_w$, where $U_y=\{(x,y,z,w)\mid y\neq 0\}$ and $U_w=\{(x,y,z,w)\mid w\neq 0\}$. Then there is $h\in\mathcal{O}_X(U)$ such that $h=\frac{z}{y}$ in $U_y$ and $h=\frac{x}{w}$ in $U_w$, but there is no global expression for $h$. $\endgroup$ Mar 17, 2016 at 18:59
  • $\begingroup$ Why the downvote? $\endgroup$
    – Fan Zheng
    Mar 18, 2016 at 1:04
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    $\begingroup$ @Fan Zheng is absolutely right: this is a very interesting and highly non-trivial question. I'm upvoting it right now. $\endgroup$ Mar 27, 2016 at 10:47
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    $\begingroup$ I started a bounty for my original question: math.stackexchange.com/questions/1696350/… I am going to delete this version of the question in a couple of days. $\endgroup$ Mar 27, 2016 at 16:41

1 Answer 1


For the sake of completeness, let me just note that this question has been re-crossposted on Mathoverflow and got an answer there.


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