# Restriction of a Cartier divisor

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.

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 $$p \in \mathrm{Ass}(O_C)$$ (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.
• 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? – KarlPeter Jun 5 '19 at 11:03
• 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 $C= Spec(R/I)$. – KarlPeter Jun 5 '19 at 13:07
• 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? – KarlPeter Jun 5 '19 at 13:07
• a typo: ...otherwise $R/I$ would have embedded points... – KarlPeter Jun 5 '19 at 13:20