Timeline for Splitting a nilpotent into square-zeros by ring extension
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 27, 2022 at 2:50 | history | became hot network question | |||
| Jun 26, 2022 at 22:14 | history | edited | darij grinberg | CC BY-SA 4.0 |
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| Jun 26, 2022 at 22:13 | vote | accept | darij grinberg | ||
| Jun 26, 2022 at 21:28 | answer | added | Will Sawin | timeline score: 7 | |
| Jun 26, 2022 at 21:20 | comment | added | Z. M | Base change this map along $\mathbb Z[T]/T^{k+\ell-1}\to R,T\mapsto a$, you get a universal candidate of $S$ (without assuming that $R\to S$ being injective), in the sense that every other $S$ factors uniquely through that. The faithful flatness is stronger than the injectivity. | |
| Jun 26, 2022 at 19:40 | comment | added | darij grinberg | @Z.M: Can you walk me through the argument for why this would help? | |
| Jun 26, 2022 at 19:38 | comment | added | Z. M | Seemingly it suffices to study the universal case. For example, is the map $\mathbb Z[T]/T^{k+\ell-1}\to\mathbb Z[X,Y]/(X^k,Y^\ell),T\mapsto X+Y$ faithfully flat? | |
| Jun 26, 2022 at 18:44 | history | asked | darij grinberg | CC BY-SA 4.0 |