As the title suggests, I am asked to prove that, given a Noetherian scheme $(X,\ \mathcal{O}_{X})$ and any open subset $U\subseteq X$, $\Gamma(U,\ \mathcal{O}_{X}):=\mathcal{O}_{X}(U)$ is a Noetherian ring.

Up to now, I have been able to show that the result is true if $U$ is an affine open subset of $X$, i.e. $U\simeq\mathrm{Spec}(A)$ for some ring $A$ (and this is actually true when $X$ is just locally Noetherian). I have also shown that, given $U$ as above, $(U, \mathcal{O}_{X\vert U})$ is a Noetherian scheme as well, which should then allow me to reduce the problem to the case $U=X$. So, when all is said and done, I should try to prove that the ring $\mathcal{O}_{X}(X)$ is Noetherian. However, I can not go any further and I am stuck here.

Any help or suggestion would be grately appreciated.

Thank you.


The answer is No. See the note Un ouvert bizarre by Manuel Ojanguren.

Here is an outline of the construction: Let $A,B \subseteq \mathbb{P}^3_k$ be two projective planes which intersect in a projective line $L$. Let $X = A \cup B$. Let $D \neq L$ be a projective line on $A$ with $D \cap L = \{P\}$. Let $U = X \setminus D$. Then $U$ is noetherian, but $\Gamma(U) \cong \{f \in k[x,y] : f(x,0)=f(0,0)\}$ is not noetherian.

  • $\begingroup$ Ah, that's consolatory: I was asked to prove something false! However, Mr. Brandenburg, since I do not understand French too much, would you be so kind to add a translation/explanation of the linked counterexample in your answer, so that I will accept it? Thank you! $\endgroup$ – Marco Vergura Jun 2 '14 at 21:45
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    $\begingroup$ For algebraists $\Gamma(U)=k+yk[x,y]$. $\endgroup$ – user26857 Sep 9 '19 at 22:24
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    $\begingroup$ @user26857 Thanks! $\endgroup$ – Martin Brandenburg Dec 22 '19 at 14:27

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