Timeline for Transversally intersecting divisors $C$ and $D$ in a Hartshorne's AG lemma
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 19, 2021 at 1:53 | comment | added | user267839 | i see it now, thank you. So this tensoring-by-structure-sheaf-of-the- divisor-trick keeping the involved sequence exact works only for surfaces and curves (+additional assumpion how they intersect) but hardly fails in higher dimensions without any hope to 'repair' it also if we require that the divisors intersect way 'nice'? | |
| Feb 17, 2021 at 20:14 | comment | added | Piotr Achinger | The ring $\mathcal{O}_{D, P}$ is one-dimensional ($D$ is a curve) and the maximal ideal is generated by one element. It is therefore regular, in particular a domain. | |
| Feb 17, 2021 at 20:05 | review | Close votes | |||
| Feb 19, 2021 at 0:43 | |||||
| Feb 17, 2021 at 19:51 | comment | added | user267839 | Why the fact that $f$ generates the maximal ideal of $\mathcal{O}_{X, P}/g = \mathcal{O}_{D, P}$ implies that $f$ is not a zerodivisor in this local ring? Clearly, if we eg know that the curve $D$ is regular in $P$, then $\mathcal{O}_{D, P}$ is dvr and we are done, but $D$ is not assumed to be non singular. Which property of $\mathcal{O}_{D, P}$ do you essentially exploit here to conclude that the generator of it's maximal ideal cannot be a zerodivisor? | |
| Feb 17, 2021 at 19:28 | comment | added | Piotr Achinger | If $f$ and $g$ generate the maximal ideal of $\mathcal{O}_{X, P}$, then $f$ generates the maximal ideal of $\mathcal{O}_{X, P}/g = \mathcal{O}_{D, P}$. In particular, $f$ is a nonzerodivisor in $\mathcal{O}_{D, P}$, which gives the injectivity of the map. | |
| Feb 17, 2021 at 19:08 | history | asked | user267839 | CC BY-SA 4.0 |