Timeline for Noetherian module, over Noetherian ring, which is isomorphic to its double dual [duplicate]
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 5, 2018 at 18:06 | history | closed |
abx Ben McKay Chris Gerig მამუკა ჯიბლაძე András Bátkai |
Duplicate of Are there non-reflexive modules isomorphic to their bi-dual? | |
| Oct 1, 2018 at 22:48 | vote | accept | user521337 | ||
| Oct 1, 2018 at 15:25 | review | Close votes | |||
| Oct 5, 2018 at 18:06 | |||||
| Oct 1, 2018 at 14:29 | answer | added | Mare | timeline score: 1 | |
| Oct 1, 2018 at 12:51 | answer | added | A.G | timeline score: 1 | |
| Oct 1, 2018 at 6:16 | comment | added | YCor | Let me copy the relevant comment to Qiaochu's linked post: "For finitely generated modules over a Noetherian ring, no such examples exist. A student of Huneke proved this around 2004, but I don't think he ever published it. (It's possible it was already known at that time, but I never found a reference.) – Graham Leuschke Sep 21 '11 at 0:23" | |
| Oct 1, 2018 at 5:42 | comment | added | R. van Dobben de Bruyn | This is easy when $M^{**}$ is reflexive (e.g. if $R$ is a product of domains [Tag 0AV3]). Indeed, if $\phi \colon M \stackrel\sim\to M^{**}$ is any isomorphism, we get a commutative diagram $$\begin{array}{ccc}M & \stackrel{\operatorname{ev}}\to & M^{**}\\ \downarrow & & \downarrow \\ M^{**} & \stackrel{\operatorname{ev}}\to & M^{****},\!\end{array}$$ where the vertical arrows $\phi$ and $\phi^{**}$ are isomorphisms. Since the bottom arrow is also an isomorphism, we conclude that the top is as well. | |
| Oct 1, 2018 at 5:21 | comment | added | Qiaochu Yuan | mathoverflow.net/questions/76000/… | |
| Oct 1, 2018 at 4:12 | history | asked | user521337 | CC BY-SA 4.0 |