Hartshorne Example V.1.4.1 - Why $C^2=\deg_C(\mathcal{L}(C)\otimes\mathcal{O}_C)$?

Question

In the book Algebraic Geometry by Hartshorne, Example V.1.4.1 says $C^2=\deg_C(\mathcal{L}(C)\otimes\mathcal{O}_C)$ holds due to Lemma V.1.3. Here, $C$ is a nonsingular curve on a nonsingular projective surface $X$. The statement of Lemma V.1.3 is

Lemma 1.3. Let C be an irreducible nonsingular curve on $X$, and let $D$ be any curve meeting $C$ transversally. Then $$\#(C\cap D)=\deg(\mathcal{L}(D)\otimes\mathcal{O}_C).$$

However, since $C$ and $C$ itself does not meet transversally and also $\#(C\cap C)$ does not make a sense, we have to find a curve $D\in|C|$ that is transversal to C. If we find such $D$, then we may conclude $C^2=C\cdot D=\#(C\cap D)=\deg(\mathcal{L}(D)\otimes\mathcal{O}_C)=\deg(\mathcal{L}(C)\otimes\mathcal{O}_C)$. So I'm trying to find such $D\in|C|$.

One sufficient condition is to find a very ample divisor $E\in|C|$. In that case, almost curves $D\in|C|$ are transversal to $C$ by Lemma V.1.2.

So my question is,

1. If $C$ is an irreducible nonsingular curve on a nonsingular projective surface $X$, is there always a very ample divisor $E\in|C|$?
2. If the question 1 is not the case, why is there a curve $D\in|C|$ that meets $C$ transversally?

Answer 1

$\text{}$1. If $C$ is an irreducible nonsingular curve on a nonsingular projective surface $X$, is there always a very ample divisor $E\in|C|$?

No. For example, a line on a nonsingular quadric surface has $L^2 = 0$. Your question is not grammatical, since very ample is a property of the sheaf $\mathcal{O}_S(C)$, not of the particular curve. In any case,$$C.D = \deg_C \mathcal{O}_S(D)$$rest $C$ for any $C$, $D$.

$\text{}$2. If the question 1 is not the case, why is there a curve $D\in|C|$ that meets $C$ transversally?

It is just not true, and you do not need it. If $C$ and $D$ are effective curves and intersect in dimension $0$ only (i.e. only finite set of points $O_i$, so no common components), there is a definition of local multiplicity $(C.D)_Pi$ so that $C.D =$ sum of those. However, even if $C$ or $D$ is not effective, or they have common components, the intersection $C.D$ is still perfectly well-defined. It is described in detail in Shafarevich's book on algebraic geometry and in excruciating detail in Fulton's book on algebraic curves.

Answer 2

As the surface $X$ is supposed to be projective there exist an ample divisor $H$ on $X$. Following the same reasoning as the begining of the proof of V.1.1 the exist an integer $n>0$ such that $nH$ and $C+nH$ are very ample.

For all very ample divisor $D$ and irreducible nonsingular curve $C$ we can apply lemma V.1.2. to get some $D'\in|D|$ transversal to $C$. Then we can apply V.1.3 and V.1.1 to get that $$ C.D=C.D'=\#(C\cap D')=\deg(\mathcal{L}(D')\otimes\mathcal{O}_C)=\deg(\mathcal{L}(D)\otimes\mathcal{O}_C) $$

As Mohan says above we then have that $$ \begin{align} C^2 & = (C+nH).C-nH.C \\ & = \deg(\mathcal{L}(C+nH)\otimes\mathcal{O}_C)-\deg(\mathcal{L}(nH)\otimes\mathcal{O}_C) \\ &= \deg(\mathcal{L}(C+nH)\otimes\mathcal{O}_C\otimes_{\mathcal{O}_C}\mathcal{O}_C\otimes\mathcal{L}(-nH)) \\ &= \deg(\mathcal{L}(C+nH-nH)\otimes\mathcal{O}_C) \\ &= \deg(\mathcal{L}(C)\otimes\mathcal{O}_C) \end{align} $$