Let $X$ be a Noetherian scheme (in particular, we assume that it has only finitely many irreducible components). Is it true that for any open set $U$, the ring $\Gamma(U, \mathscr{O}_X)$ is a Noetherian ring. Let $\mathscr{F}$ be a coherent sheaf on $X$. Is it true that for any open set $U$, $\Gamma(U,\mathscr{F})$ is a finitely generated $\Gamma(U,\mathscr{O}_X)$ module.
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$\begingroup$ Since every open subset of a noetherian topological space is quasi-compact, the scheme $U$ is also noetherian. In other words you can assume that $U=X$ in your question. $\endgroup$– Georges ElencwajgCommented Jan 15, 2012 at 19:15
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2$\begingroup$ Georges, I think you are correct in your statement, but not in the proof. A scheme whose underlying topological space is noetherian is not necessarily a noetherian scheme. A (locally) noetherian scheme is covered by open sets that are Spec's of noetherian rings. Here is an example:let $A=k[x_n\vert n\in\mathbb N]/\mathfrak m^2$ where $\mathfrak m=(x_n\vert n\in\mathbb N)$. The topological space $\mathrm{Spec}A$ is just a point and hence a noetherian topological space, but $\mathfrak m$ is not finitely generated, so $A$ is not noetherian and hence $\mathrm{Spec}A$ is a non-noetherian scheme. $\endgroup$– Sándor KovácsCommented Jan 15, 2012 at 22:52
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$\begingroup$ Rex, if you added that $U$ were an affine scheme, then the statement is actually true.... $\endgroup$– Sándor KovácsCommented Jan 15, 2012 at 22:53
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$\begingroup$ Dear @Sándor what you say is correct, but I knew the distinction between noetherian scheme and noetherian topological space. Actually I didn't give a proof of my assertionj because I thought it was sufficiently straightforward! To spell it out: we can cover $U$ by affine spectra $U_i=spec(A_i) $ of noetherian rings $A_i$, because $X$ has a basis of such affines ( $X$ being a noetherian scheme is a fortiori a locally noetherian scheme) . Then by quasi-compactness we can extract a finite covering of $U$ by these $U_i$'s, proving that $U$ is indeed a noetherian scheme. $\endgroup$– Georges ElencwajgCommented Jan 15, 2012 at 23:46
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1$\begingroup$ Maybe the clearest way to sum up the above is: A scheme is noetherian iff it is locally noetherian and quasi-compact $\endgroup$– Georges ElencwajgCommented Jan 15, 2012 at 23:57
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There exists a noetherian scheme , which is even a variety over a field $k$, such that
$\Gamma(X, \mathscr{O}_X)$ is not a Noetherian ring.
It is given as Exercise 21.9. D. in Ravi Vakil's wonderful online book.
Ravi takes for $X$ the total space of the vector bundle associated to a locally free sheaf $\mathcal E$ of rank 2 on an elliptic curve $E$.
The locally free sheaf is the direct sum $\mathcal E=\mathcal P\oplus \mathcal N$ of a an invertible sheaf of positive degree $\mathcal P$ and of a non-torsion invertible sheaf $\mathcal N$ of degree $0$ on $E$.
(Full disclosure: I have just discovered that exercise and I have not yet checked it and the ones leading to it in detail. But to say that I trust Ravi is a vast understatement...)
answered Jan 15, 2012 at 20:08
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3$\begingroup$ To add a few more details: $\Gamma(X, \mathcal O_X)$ is isomorphic to the graded ring $\oplus_{n,m} \Gamma(E, \mathcal P^n \otimes \mathcal N^m)$. A graded component of this ring is non-zero if and only if either $n=m=0$ or $n > 0$. If we consider the ideal $I$ consisting of all graded pieces with $n > 0$, then any homogeneous generating set for $I$ must included at least one element from the graded piece with $n=1$ and for each $m$. Thus, $I$ is not finitely generated, so $\Gamma(U, \mathcal O_X)$ is not Noetherian. $\endgroup$ Commented Jan 15, 2012 at 22:29
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$\begingroup$ Dear @Dustin, thank you for your interest and your comment. $\endgroup$ Commented Jan 15, 2012 at 23:25