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Let $X\subseteq \mathbb{A}^n$ be an affine variety. The ring $k[x_1,\ldots,x_n]$ is noetherian because of Hilbert's basis theorem.

The coordinate ring $k[X]=k[x_1,\ldots,x_n]/I(X)$ is noetherian because ideals of $k[X]$ are of the form $J/I(X)$, where $J\supseteq I(X)$ is an ideal of $k[x_1,\ldots,x_n]$.

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 $k[X]$, and the ideals of a ring of fractions $S^{-1}A$ are of the form $S^{-1}J$, where $J$ is an ideal of $A$.

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?

Marco Flores
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  • Are you requiring $X$ to be irreducible? – D_S Mar 13 '16 at 23:48
  • Yes, by affine variety I mean irreducible affine algebraic set. – Marco Flores Mar 13 '16 at 23:51
  • A direct approach is not immediate to me. However, if you're familiar with schemes, you can try identifying such an affine variety with an integral separated scheme of finite type over $k$ whose structure sheaf is induced by the sheaf of regular functions. – Future Mar 14 '16 at 04:43
  • Are you sure that it holds for any open set? It certainly holds for affine opens. – MooS Mar 14 '16 at 09:51
  • I changed the question because I am not sure anymore. But, how would you prove it for affine opens? – Marco Flores Mar 14 '16 at 14:21
  • It is true for principal affine opens, since $R_{f}$ is noetherian if $R$ is noetherian. If $U=\operatorname{Spec} A \subset X = \operatorname{Spec} R$ is affine open, we can cover $U$ by finitely many principal opens $D(f_i)$ with $f_i \in A$. We get that $A_{f_i}$ is noetherian for all $i$ and $(f_1, \dotsc, f_n) = A$. It is a standard exercise in commutative algebra, that this implies that $A$ is noetherian. – MooS Mar 14 '16 at 16:22

3 Answers3

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There is a counterexample in section 19.11.13 of Ravi Vakil's Foundations of Algebraic Geometry https://math216.wordpress.com/

Marco Flores
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Since any open subscheme of a noetherian scheme is noetherian (Corollary 3.22 in Görtz/Wedhorn), we can reduce to case of global sections.

If $X = \operatorname{Spec} R$ is affine, then $X$ is noetherian if and only if $R$ is noetherian (Prop. 3.19 in Görtz/Wedhorn), hence the section ring of any affine open will be noetherian.

But in general, the answer is negative, see this answer:

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

Community
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MooS
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    The thing you linked to is for open sets of arbitrary noetherian schemes. But OP is asking about the global section of a quasi-affine variety. – D_S Mar 14 '16 at 18:52
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    Nevertheless it gives some insight, that the problem is non-trivial. – MooS Mar 15 '16 at 06:30
  • Do you have an answer in the quasi-affine integral case? I am not sure if one can realize the ring $$k[x,y] \subset k[x,y,\frac{y^2}{x}, \frac{y^3}{x^2}, \frac{y^4}{x^3}, \dotsc] \subset k(x,y)$$ (the middle term is the non-noetherian ring from the counterexample in the link) as a section ring of some quasi affine variety in the affine plane. I guess the answer is no. Some other non-noetherian ring between the polynomial ring and the function field? More variables? – MooS Mar 15 '16 at 06:30
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We can do the following in the local case, but the details of it are left to the reader.

Let $X$ be an irreducible affine variety over the complex numbers. The answer to your question is affirmative because by taking the intersection in the field of fractions $\mathbb{C}(X)$, we have that $$\mathbb{C}[X]=\bigcap_{x\in X}\mathcal{O}_{x,X}.$$ This says that the local rings of $X$ determine the coordinate ring $\mathbb{C}[X]$. Now, to see why this is true, one inclusion is clear. The other one goes as follows. Say $f\in \mathcal{O}_{x,X}$ for all $x\in X$. Let us analyze the ideal which lifts $f$ to the coordinate ring, that is $$Z=\{g\in \mathbb{C}[x_0,\ldots , x_n]| \mbox{ if } \overline{g}\in \mathbb{C}[X],\overline{g}.f\in\mathbb{C}[X]\}.$$ Observe that $f$ regular at $x$ means that $f=h_x/g_x$, where $g_x\ne 0$, which tells us that $g_x\in Z$, and therefore $x\notin V(Z)$ for all $x$ (this is an open condition). Note the ideal defining $X$, $I(X)$, is contained in $Z$, and openness tells us $V(Z)=\emptyset$. Now, Nullstellensatz says that $1\in I(V(Z))$, which implies that $f\in \mathbb{C}[X]$.

Csar
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  • I think everyone involved here knows that you can compute section rings of integral varieties as the intersection of the stalks...But then, is the intersection of noetherian rings again noetherian? In general, definitely not! – MooS Mar 16 '16 at 06:07