I try to understand an argument on page 358 of Hartshorne, proof of Lemma V.1.2. Consider a surface X, irreducible curves $C_i $on X and a very ample divisor D. Hartshorne says that since the intersections $C_i \cap D$ are nonsingular, the $C_i$ and $D\,'$ meet transversally.
My questions:
Thanks in advance!
First of all, he picks $D'$ to be some nonsingular curve in the linear system $|D|$. The intersection $C_i\cap D'$ is ment scheme-theoretically. If the intersection is nonsingular, this means that the local rings of each point in that dimension zero scheme are regular points. A regular local ring of dimension zero is a field, so the scheme $C_i\cap D'$ has no embedded points. Differently put, let $U=\mathrm{Spec}(A)$ be an affine neighbourhood of the intersection point containing none of the others. By going more local, assume that $C_i$ is defined on $U$ by $f\in A$ and $D'$ is defined on $U$ by $g\in A$. Then, the local ring at the scheme-theoretic intersection point is $A/(f,g)$, which we have seen to be a field, so $f$ and $g$ generate the maximal ideal of the (reduced) intersection point - that's Hartshorne's definition of transversal intersection.
