2
$\begingroup$

Let $X$ be a surface (so $2$-dimensional proper $k$-scheme)

$D \subset X$ an effective Cartier divisor of $X$ which corresponding to an invertible sheaf $\mathcal{L}=O_X(D)$ and

$C \subset X$ a closed curve which hasn't embedded components.

Assume futhermore that $C$ and $D$ haven't common irreducible components.

Let

$$0 \to O_X(-D) \to O_X \to O_D \to 0$$

the short exact sequence defining the Cartier divisor $D$. Restricting this sequence to $C$ we obtain exact sequence

$$O_C(-D) \to O_C \to O_{D \cap C} \to 0$$.

My question is how to verify that under given assumptions of $C$ and $D$ the arrow $O_C(-D) \to O_C$ is injective? By definition the whole story is told on stalks so we have to investigate $O_{C,q}(-D) \to O_{C,q}$ for diverse primes $q$:

More concretely here I encountered following problem: Firsty I reduced the problem to stalks $O_{C,p}$ with $p \in Ass(O_C)$ (so $p$ associated point)

$C$ has as a curve only associated (generic & embedded) and closed points(=maximal ideals). Since by assumption $C$ hasn't embedded points all associated points are already generic.

Let $c$ be a closed point with corresponding prime $p_c$. (for sake of simplicity identify $c=p_c$). Then $p_c$ contains a generic (=minimal) ideal $\eta:=p_{\eta}$.

Then by universal property of localization we obtain the commutative diagram

$$ \require{AMScd} \begin{CD} O_{C,c}(-D) @> >> O_{C,c} \\ @VVV @VVV \\ O_{C, \eta}(-D) @> >> O_{C, \eta} \end{CD} $$

If we assume that the lower map is injective (since $\eta$ generic) (*) and can show that the canonical vertical maps are injective (**) then the upper map is also injective.

To (**): (only $O_{C,c} \to O_{C,\eta}$): Let $r \in O_{C,c}$ with $r=0$ in $O_{C,\eta}$. Then there exist a $s \in O_{C,c} \backslash p_{\eta}$ with $rs=0$. So $s$ is a zero divisor not containing in a minimal prime ideal so it must be contained in an embedded ideal. But by assumtion $C$ has't embedded components. Contradiction, so the vertical maps are also injective.

Now my PROBLEM is to verify (*) that for minimal primes $p_{\eta}$ the map $O_{C, \eta}(-D) \to O_{C, \eta}$ is injective using given assumptions on $C$ and $D$.

Does anybody see an argument?

$\endgroup$
0
$\begingroup$

Since $C$ and $D$ have no common component, $D|_C$ is an effective (nef) Cartier divisor (just by the definition of effective Cartier divisor, the restriction of the rational function of $D|_{U_i}$ on $C$ is a non-zero rational function of $C$, and ...). So $\mathcal{O}_C(-D)\to \mathcal{O}_C$ is injective.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Karl_Peter Jun 5 at 11:03
  • $\begingroup$ No need to consider the embedded points, I think. Or I made any mistake? $\endgroup$ – Sheng Meng Jun 5 at 11:13
  • $\begingroup$ 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 $C= Spec(R/I)$. $\endgroup$ – Karl_Peter Jun 5 at 13:07
  • $\begingroup$ 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? $\endgroup$ – Karl_Peter Jun 5 at 13:07
  • $\begingroup$ a typo: ...otherwise $R/I$ would have embedded points... $\endgroup$ – Karl_Peter Jun 5 at 13:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.