Timeline for Injective indecomposable modules over Laurent polynomial rings
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Oct 24, 2018 at 18:46 | vote | accept | Benjamin Steinberg | ||
| Oct 24, 2018 at 17:27 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Oct 24, 2018 at 16:20 | answer | added | Uriya First | timeline score: 4 | |
| Oct 24, 2018 at 15:03 | comment | added | Benjamin Steinberg | Probably they do. Maybe you just invert the powers of the prime $p$. | |
| Oct 24, 2018 at 15:01 | comment | added | rschwieb | I would have thought the Prüfer groups would have an analogue for any prime $p$ in a PID, but I don't understand them well :) Sorry I can only make comments like these instead of resolving the problem... | |
| Oct 24, 2018 at 14:42 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Oct 24, 2018 at 14:41 | comment | added | Benjamin Steinberg | @rschwieb, If you think about $\mathbb Z$ where you get $\mathbb Q$ and the Prufer p-groups, then it seems not so easy. That is why I want $\mathbb C$, so we have fewer irreducibles., which I realize I forgot to say. | |
| Oct 24, 2018 at 14:32 | comment | added | rschwieb | It does seem that the injective envelope of arbitrary cyclic modules over a PID should be well-known! | |
| Oct 24, 2018 at 14:29 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Oct 24, 2018 at 14:29 | comment | added | Benjamin Steinberg | k should be C. Sorry. I'll fix. | |
| Oct 24, 2018 at 14:11 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Oct 24, 2018 at 14:00 | history | asked | Benjamin Steinberg | CC BY-SA 4.0 |