Intersection Theory on a Surface

I have some problems to prove the exercise 20.2.A part (b) in Ravi Vakil's "Fondation of Algebraic Geometry". Here the excerpt: The setting is: We have a surface $$X$$ (therefore 2-dimensional, proper $$k$$-scheme) and two effective divisors $$C,D$$ (therefore curves) such that $$C$$ and $$D$$ don't have common irreducible components.

To show:

$$(D \cdot C) = h^0(D \cap C, \mathcal{O}_{C \cap D})$$

My efforts: Since by definition and exercise 20.2.A (a) we already know that $$(D \cdot C) = deg(\mathcal{O}_X(D) \vert _C) = \chi(\mathcal{O}_X(D) \vert _C) - \chi(\mathcal{O}_C)$$

here $$\chi$$ is the Euler characteristic and $$\mathcal{O}_X(D)$$ the corresponing invertible sheaf to divisor $$D$$.

The ideal is to show that following sequence is exact:

(*)$$0 \to \mathcal{O}_C(-D) \to \mathcal{O}_C \to \mathcal{O}_{D \cap C} \to 0$$

and then using the additivity of Euler charecteristic and the fact that $$\chi(\mathcal{O}_{C \cap D})= h^0(D \cap C, \mathcal{O}_{C \cap D})$$ since this intersection is zero dimensional.

Since this sequence arrises from the sequence $$0 \to \mathcal{O}_X(-D) \to \mathcal{O}_X \to \mathcal{O}_D \to 0$$ by tensoring with $$\mathcal{O}_C$$ the only cruical point is to prove that $$\mathcal{O}_C(-D) \to \mathcal{O}_C$$ is injective. By definition this can be done on level of stalks.

By definition the stalks are given by $$\mathcal{O}_{C,c} = \mathcal{O}_{X,c}/(g)$$ and $$\mathcal{O}(-D)_{X,c} = f\mathcal{O}_{X,c}$$ for regular $$f, g \in \mathcal{O}_{X,c}$$ since $$C,D$$ effective divisors.

Therefore we get $$\mathcal{O}_C(-D)_c = \mathcal{O}(-D)_{X,c} \otimes \mathcal{O}_{C,c}= f\mathcal{O}_{X,c} \otimes \mathcal{O}_{X,c}/(g) = f\mathcal{O}_{X,c}/(fg)$$

So I have to show that $$f\mathcal{O}_{X,c}/(fg) \to \mathcal{O}_{X,c}/(g)$$ is injective for every $$c \in C$$.

Since $$C$$ is a curve two cases could happen:

1. $$c$$ is generic point (of $$C$$)
2. $$c$$ is a closed

Could anybody help to show the desired injectivity for these two cases? I don't see where I can use the assumption that $$C$$ and $$D$$ don't have common irreducible components. Futhermore how to cope with the stalks where $$c$$ is an embedded component, so $$c \in Ass(\mathcal{O}_c)$$?

We want to show that if $$C,D$$ are two curves on a regular surface $$X$$ such that $$C$$ does not contain any associated point of $$D$$, then $$\mathcal{O}_C(-D) \rightarrow \mathcal{O}_C$$ is injective.

As you suggested it suffices to check this on stalks.

Let $$A=\mathcal{O}_{X,c}$$, and let $$f,g \in A$$ be local equations for $$C,D$$ respectively. Then we want to show that the map $$A/(f) \rightarrow A/(f)$$ given by multiplication by $$g$$ is injective. It suffices to show that $$g\in A/(f)$$ is not a zero-divisor.

Suppose otherwise, then there exists $$0\ne a\in A/(f)$$ such that $$ag \in (f)$$, i.e. $$ag = bf$$ in $$A$$ for some $$b\in A$$. Let's use the fact that $$A$$ is a UFD (as it is regular). If $$g$$ and $$f$$ do not share any common factors, then $$a=0 \in A/(f)$$. Therefore we may assume that $$g$$ and $$f$$ shares some common factor $$h\in A$$.

Now let's lift this result to an open set on $$X$$.

By a suitable localisation (i.e. take an affine open containing $$c$$, and then invert everything that shows up in the denominators in the above), we may assume that $$X$$ is locally given by $$\operatorname{Spec} A'$$, $$f,g,h \in A'$$ and that $$g = g'h, f=f'h$$ in $$A'$$ for some $$f,f'\in A'$$. But this implies that $$V(h) \subseteq V(g) \cap V(f)$$ in $$\operatorname{Spec} A'$$. Since $$A'$$ is an integral domain, $$V(h)$$ has dimension one. So this implies that $$V(g)$$ and $$V(f)$$ share an irreducible component on $$\operatorname{Spec} A'$$, which implies that they share an irreducible component on $$X$$. This gives a contradiction.

Note that we actually only used the fact that $$C,D$$ do not share irreducible components and nothing about embedded points - but as Vakil mentions right after this exercise - in fact $$D$$ is never going to have embedded points since $$X$$ is smooth.

I think you want to assume that $$S$$ is regular, so that $$(\mathcal{O}_{X,c})$$ is factorial. So we have a factorial ring $$R$$, and elements $$f$$, $$g$$ which cut out $$C$$, $$D$$. Since $$C$$, $$D$$ do not have common components, we $$f$$, $$g$$ are coprime---$$\text{gcd}(f,g) = 1$$. Now, we want to show that$$f: (gR)/f(gR) = \mathcal{O}_{C,c}(-D) \to \mathcal{O}_{C,c} = R/f\text{ is injective.}$$

To see this, take $$g a \in g R$$, with $$g a = 0$$ mod $$f$$. So $$g a = f b$$. So $$f \mid (g a)$$. Since $$f$$, $$g$$ are coprime, we have $$f \mid a$$. Hence $$a = c f$$, and $$g a = g f c$$, so $$g a \in f g R$$.

The best reference for this I know of is Beauville's book Complex Algebraic Surfaces, Chapter 1.

Update: Let's try a second explanation. Concerning your question, you are probably almost there. The issue is that the formula should hold whenever $$C$$ and $$D$$ "intersect properly". If say, $$X$$ is regular, $$C$$ and $$D$$ are reduced, this exactly means that $$C$$ and $$D$$ have no common irreducible components. In general, we need $$C$$ and $$D$$ to be Cartier divisors in $$X$$, and that $$C \cap D$$ remains a Cartier divisor in say, $$D$$. The last condition means that no associated point of $$C$$ is contained in $$D$$, or equivalently, that locally $$f$$ is a nonzero divisor in $$\mathcal{O}_{X,c}/(g)$$. So multiplication with $$f$$ is an injective morphism in $$\mathcal{O}_{X,c}/(g)$$, which looks like what we need.