Timeline for Homomorphism between ideal sheaves of codimension $2$?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 14, 2023 at 22:01 | comment | added | Sasha | One way to see that $\mathcal{E}xt^1(\mathcal{O}_Y, \mathcal{O}_X) = 0$ is by using the local-to-global spectral sequence and Serre duality on $X$. | |
| Nov 14, 2023 at 4:23 | comment | added | DVL-WakeUp | @Sasha Oh I'm sorry, this may follows from $\mathscr{depth}_{I_Y}(\mathscr{O}_X)=\inf\{\mathrm{depth}(\mathscr{O}_{X,x}):x\in V(I_Y)\}$ and the Serre's criterion for normality. | |
| Nov 14, 2023 at 4:15 | comment | added | DVL-WakeUp | @Sasha I think this is right if $\mathrm{depth}_{I_Y}O_X\geq2$，maybe we need use some kind of Cohen-Macaulay? | |
| Nov 14, 2023 at 3:46 | comment | added | DVL-WakeUp | @Sasha Sorry for my stupid questions, why $\mathcal{E}xt^1(\mathscr{O}_Y,\mathscr{O}_X)=0$? Is this just a direct calculation? | |
| Nov 14, 2023 at 3:20 | comment | added | DVL-WakeUp | @Sasha Thank you for your nice solution! | |
| Nov 14, 2023 at 2:37 | vote | accept | DVL-WakeUp | ||
| Nov 13, 2023 at 14:04 | comment | added | Sasha | @CraniumClamp: Start with the standard exact sequence $$0 \to I_Y \to \mathcal{O}_X \to \mathcal{O}_Y \to 0$$, apply the dualization functor, and use the fact that $$\mathcal{H}om(\mathcal{O}_Y, \mathcal{O}_X) = \mathcal{E}xt^1(\mathcal{O}_Y, \mathcal{O}_X) = 0$$ since the codimension of $Y$ is at least 2. | |
| Nov 13, 2023 at 9:04 | comment | added | Cranium Clamp | Is the assertion that $ (I_Y)^{\vee} = \mathcal{O}_X $ a consequence of the following steps: (1) $ (I_Y)^{\vee} $ is reflexive by a short lemma (2) $ (I_Y)^{\vee \vee} = \mathcal{O}_X $ because the double dual is a line bundle which is trivial outside a codimension $ 2 $ closed subset, hence is trivial by Hartog's lemma. I don't immediately see how you get the first dual trivial and then the second. | |
| Nov 12, 2023 at 19:38 | history | answered | Sasha | CC BY-SA 4.0 |