Timeline for Relative version of Hilbert syzygy theorem
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| May 21, 2021 at 18:33 | history | edited | Leonid Positselski | CC BY-SA 4.0 |
corrected a typo
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| May 21, 2021 at 17:27 | comment | added | A.G | Thank you very much for your time. | |
| May 21, 2021 at 17:21 | comment | added | Leonid Positselski | When $G$ is of finite flat dimension $m$ over $R$, resolve it by flat $S$-modules. The $m$-th syzygy module $\Omega^mG$ will be an $R$-flat $S$-module. Then the proof in my answer shows that the $(n+m)$-th syzygy module $\Omega^{n+m}G$ is a flat $S$-module. | |
| May 21, 2021 at 17:18 | comment | added | Leonid Positselski | @A.G I added an EDIT at the end of the answer with a lemma purporting to answer your first question (existence of the distinguished triangle). | |
| May 21, 2021 at 17:17 | history | edited | Leonid Positselski | CC BY-SA 4.0 |
EDIT with an explanatory Lemma added
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| May 21, 2021 at 17:05 | comment | added | A.G | It is beacuse I was trying to re-write your proof when $G$ is of finite flat dimension over $R$ and we want it to be of finite flat dimension over $S$· | |
| May 21, 2021 at 17:01 | comment | added | A.G | Thank you, and sorry for making waste your time on such easy question. I din't notice the flatness of G over R. | |
| May 21, 2021 at 16:58 | comment | added | Leonid Positselski | @A.G Furthermore, tensoring an $S$-module with $R/aR$ over $R$ is the same thing as tensoring it with $S/aS$ over $S$ (both the functors amount to $L\longmapsto L/aL$). Tensoring a flat $S$-module with any ring $T$ over $S$ produces a flat $T$-module. In particular, tensoring a flat $S$-module with $S/aS$ produces a flat $S/aS$-module. Does this answer your second question? | |
| May 21, 2021 at 16:56 | comment | added | Leonid Positselski | @A.G Answering your second question: a flat resolution of the $S$-module $G$ is a resolution of a flat $R$-module by flat $R$-modules (because any flat $S$-module is a flat $R$-module, since $S$ is a flat $R$-module). Tensoring a flat $R$-module resolution of a flat $R$-module with any $R$-module, we obtain an exact complex of $R$-modules. | |
| May 21, 2021 at 16:46 | comment | added | A.G | Essentially the same question for the case of zero-divisors: why "Tensoring a flat resolution of the $S$-module $G$ with $R/aR$ over $R$, we obtain a flat resolution of the $S/aS$-module $G/aG$"? | |
| May 21, 2021 at 16:40 | comment | added | A.G | Why there exists such distinguished triangle? | |
| May 21, 2021 at 16:01 | history | edited | Leonid Positselski | CC BY-SA 4.0 |
"Let $M$ be an $S$-module" inserted
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| May 21, 2021 at 15:54 | history | answered | Leonid Positselski | CC BY-SA 4.0 |