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


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

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 $p \in \mathrm{Ass}(O_C)$ (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?


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.

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  • $\begingroup$ 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? $\endgroup$ – KarlPeter Jun 5 '19 at 11:03
  • $\begingroup$ No need to consider the embedded points, I think. Or I made any mistake? $\endgroup$ – Sheng Meng Jun 5 '19 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 $C= Spec(R/I)$. $\endgroup$ – KarlPeter Jun 5 '19 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? $\endgroup$ – KarlPeter Jun 5 '19 at 13:07
  • $\begingroup$ a typo: ...otherwise $R/I$ would have embedded points... $\endgroup$ – KarlPeter Jun 5 '19 at 13:20

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