# Is the global section ring of a Noetherian Scheme Noetherian as well?

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.

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.
• For algebraists $\Gamma(U)=k+yk[x,y]$. – user26857 Sep 9 '19 at 22:24