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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.

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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.

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  • $\begingroup$ 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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