Timeline for On the stable model
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 26, 2016 at 5:35 | answer | added | Jason Starr | timeline score: 1 | |
| Dec 24, 2016 at 12:01 | comment | added | Jason Starr | You can prove this by induction on the number of unstable components of $C$, i.e., genus $0$ components $E$ with at most $2$ nodes. Let $f:C\to \widehat{C}$ be the contraction of $E$. If $E$ has $2$ nodes, then $\omega_C$ equals $f^*\omega_{\widehat{C}}$, so the canonical rings are equal. If $E$ has $1$ node, then $\omega_C$ equals $f^*\omega_{\widehat{C}}(\underline{E})$. Since $\omega_C$ has negative degree on $E$, every global section of $\omega_C^{\otimes n}$ comes from $f^*(\omega_{\widehat{C}}^{\otimes n})$. Again the canonical rings are equal. | |
| Dec 24, 2016 at 10:40 | comment | added | abx | Oops, my bad. I overlooked "semistable". | |
| Dec 24, 2016 at 10:29 | comment | added | Laurent Moret-Bailly | @abx: this proves that every nodal plane curve of degree $\geq4$ is stable. No contradiction there, I think. | |
| Dec 24, 2016 at 10:18 | comment | added | abx | "It is asserted", where? This is false. If you take any plane curve $C\subset \mathbb{P}^2$ of degree $d\geq 4$, then $\omega _C=\mathcal{O}_C(d-3)$ is very ample, so $C'=C$. | |
| Dec 24, 2016 at 10:07 | review | First posts | |||
| Dec 24, 2016 at 11:36 | |||||
| Dec 24, 2016 at 10:05 | history | asked | Gefeltig | CC BY-SA 3.0 |