Timeline for Coherent Sheaves on Noetherian schemes
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 20, 2012 at 5:08 | vote | accept | Rex | ||
| Jan 16, 2012 at 9:12 | vote | accept | Rex | ||
| Jan 16, 2012 at 9:12 | |||||
| Jan 16, 2012 at 8:28 | comment | added | Martin Brandenburg | Good question, +1: I always thought that $\Gamma(X)$ is noetherian when $X$ is noetherian. | |
| Jan 15, 2012 at 23:57 | comment | added | Georges Elencwajg | Maybe the clearest way to sum up the above is: A scheme is noetherian iff it is locally noetherian and quasi-compact | |
| Jan 15, 2012 at 23:46 | comment | added | Georges Elencwajg | 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. | |
| Jan 15, 2012 at 22:53 | comment | added | Sándor Kovács | Rex, if you added that $U$ were an affine scheme, then the statement is actually true.... | |
| Jan 15, 2012 at 22:52 | comment | added | Sándor Kovács | 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. | |
| Jan 15, 2012 at 20:08 | answer | added | Georges Elencwajg | timeline score: 10 | |
| Jan 15, 2012 at 19:15 | comment | added | Georges Elencwajg | 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. | |
| Jan 15, 2012 at 15:46 | history | asked | Rex | CC BY-SA 3.0 |