Timeline for Restriction of a Cartier divisor
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 5, 2019 at 16:07 | comment | added | Sheng Meng | Let us continue this discussion in chat. | |
| Jun 5, 2019 at 15:58 | comment | added | Sheng Meng | You may apply Chow's lemma and minimal resolution to reduce to the case when $X$ is a smooth projective surface, no need to get so involved with such things. On the other hand, your $f$ is not a zero divisor of $R/I$, for otherwise, $C$ is a component of $D$ (your assumption). | |
| Jun 5, 2019 at 15:54 | comment | added | Sheng Meng | But restriction $D|_C$ is just pullback of line bundle. So it is always Cartier? | |
| Jun 5, 2019 at 13:20 | comment | added | user267839 | a typo: ...otherwise $R/I$ would have embedded points... | |
| Jun 5, 2019 at 13:07 | comment | added | user267839 | The problem is to verify that $f$ is a non zero divisor in $R/I$,right? If $f$ would be a zero divisor then it would be contained in a associated prime ideal $p$ of $R/I$. This prime ideal must be generic/minimal ideal of $R/I$ since otherwise $R/I$ would have generic points. But if $f$ is contained in a minimal ideal then this would be also a minimal ideal of $R/(f)$ by dimension argument. Then $C,D$ would share a component. Does this argumentation work? | |
| Jun 5, 2019 at 13:07 | comment | added | user267839 | meanwhile I think this condition regarding embedded points sits in the conclusion that $D \vert _C$ is Cartier. Indeed there is a statement that if $C,D$ are already Cartier divisors then $D \vert _C$ is also a CD. And the fact that $C$ hasn't embedded points and has codimension $1$ implied that $C$ is also a CD in our situation. Maybe more explicitely & elementary these two conditions can be exploited in following way: Essentially if we think locally then assume $X=Spec(R), D=Spec(R/(f))$ for non zero divisor $f \in R$ and $C= Spec(R/I)$. | |
| Jun 5, 2019 at 11:13 | comment | added | Sheng Meng | No need to consider the embedded points, I think. Or I made any mistake? | |
| Jun 5, 2019 at 11:03 | comment | added | user267839 | hmmm this argument seems not using that $C$ hasn't embedded points,right? Or do I oversee a point ...literally :)? As soon as we know that $D \vert_C$ is effective Cartier then $\mathcal{O}_C(-D)\to \mathcal{O}_C$ is already injective as "ideal sheaf". Where I miss the detail with absence of embedding points? | |
| Jun 5, 2019 at 10:30 | history | answered | Sheng Meng | CC BY-SA 4.0 |