Hartshorne, *Algebraic Geometry*, Chapter V, Lemma 1.3, reads (in part):

> 
> Throughout this chapter, a *surface* will mean a nonsingular projective surface over an algebraically closed field $k$. [...] Let $X$ be a surface. 
> 
> 
> [...]
> 
> 
> **Lemma 1.3.** Let $C$ be an irreducible nonsingular curve on $X$ , and let $D$ be any curve meeting $C$ transversally. Then
>  $$\#(C \cap D) = \deg\_C(\mathscr{L}(D) \otimes \mathcal{O}\_C).$$
>  Proof. [...] We use the fact that $\mathscr{L}(D)$ is the ideal sheaf of $D$ on $X$. Therefore, tensoring with $\mathcal{O}\_X$, we have an exact sequence
>  $$0 \to \mathscr{L}(-D) \otimes \mathcal{O}\_C \to \mathcal{O}\_C \to \mathcal{O}\_{C \cap D} \to 0$$
>  where now $C \cap D$ denotes the scheme-theoretic intersection. [...]
> 
> 
> 

So apparently we've taken the exact sequence
$$0 \to \mathcal{O}\_X(-D) \to \mathcal{O}\_X \to \mathcal{O}\_D \to 0$$
of coherent sheaves on $X$, tensored by the coherent sheaf $\mathcal{O}\_C$, and the resulting sequence is still exact. 

Now, as I understand it, $\mathcal{O}\_C$ is not flat: Proposition III.9.2(e) says that a coherent sheaf on a noetherian scheme is flat iff it is locally free, and $\mathcal{O}\_C$ is clearly not locally free since it's supported on a curve. (Of course I guess I don't actually *need* flatness here.)