Timeline for Non-isomorphic rings that are localizations of each other
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 18, 2021 at 2:46 | comment | added | jdc | @user111524: An inverse is given by $a/s \mapsto a/s \otimes 1$. To see this is an inverse, note that $a/s \otimes b/t$ is equal to $ab/s \otimes 1/t$ and that this is equal to $(ab/st)t \otimes 1/t = ab/st \otimes 1$. | |
| May 19, 2018 at 19:33 | comment | added | user111524 | Could you please elaborate on why the natural map $A' \otimes_A A' \to A'$ is an isomorphism (or give a reference) ? | |
| May 7, 2017 at 6:50 | history | edited | R. van Dobben de Bruyn | CC BY-SA 3.0 |
Numdam links have been reindexed.
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| Oct 31, 2016 at 8:38 | comment | added | HeinrichD | Are there finitely generated examples? (Not finitely presented.) | |
| Oct 31, 2016 at 8:10 | history | edited | R. van Dobben de Bruyn | CC BY-SA 3.0 |
The argument erroneously said 'endomorphism' where 'automorphism' was meant.
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| Jun 23, 2016 at 22:09 | history | edited | R. van Dobben de Bruyn | CC BY-SA 3.0 |
A remark.
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| Jun 20, 2016 at 7:21 | vote | accept | Yuzhou Gu | ||
| Jun 20, 2016 at 7:10 | comment | added | R. van Dobben de Bruyn | Maybe I should use the suggestive notation $A = K[x_0, x_{-1},x_{-2}, \ldots]$. | |
| Jun 20, 2016 at 7:04 | history | answered | R. van Dobben de Bruyn | CC BY-SA 3.0 |