Timeline for Vanishing of Tor
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 12, 2021 at 9:13 | comment | added | user521337 | I wonder if this is true even in the $n=1$ case? I know it is true if $t=1$, but I wonder if it would be true when $n=1$ and $t>1$? | |
| Jul 25, 2013 at 5:41 | comment | added | Angelo | $I^{n-1}/I^n$ is a free $R/I$-module, so the statement follows from a simple induction. | |
| Jul 24, 2013 at 22:06 | history | edited | Ricardo Andrade |
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| Jan 12, 2013 at 15:43 | vote | accept | Pham Hung Quy | ||
| Jan 11, 2013 at 20:26 | answer | added | user26857 | timeline score: 2 | |
| Jan 10, 2013 at 4:22 | comment | added | Mahdi Majidi-Zolbanin | So Lemma 18.2 applies at least when $M$ has finite projective dimension. | |
| Jan 10, 2013 at 2:41 | comment | added | Pham Hung Quy | Lemma 18.2 in Matsumura need $x$ is both $R$-regular and $M$-regular. | |
| Jan 9, 2013 at 19:11 | comment | added | David Hansen | It's true for $n=1$ if the sequence is also $R$-regular, since then $\mathrm{Tor}_{1}^{R}(R/I,M) \simeq \mathrm{Tor}_{1}^{R/I}(R/I,M/I)$. (see Lemma 18.2.iii in Matsumura's CRT.) | |
| Jan 9, 2013 at 18:56 | history | asked | Pham Hung Quy | CC BY-SA 3.0 |