TITLE: Hartshorne Example V.1.4.1 - Why $C^2=\deg_C(\mathcal{L}(C)\otimes\mathcal{O}_C)$?
QUESTION [5 upvotes]: 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,

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

REPLY [2 votes]: 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
$$ \begin{align} 
C^2 & = (C+nH).C-nH.C \\ 
 & = \deg(\mathcal{L}(C+nH)\otimes\mathcal{O}_C)-\deg(\mathcal{L}(nH)\otimes\mathcal{O}_C) \\
 &= \deg(\mathcal{L}(C+nH)\otimes\mathcal{O}_C\otimes_{\mathcal{O}_C}\mathcal{O}_C\otimes\mathcal{L}(-nH)) \\
 &= \deg(\mathcal{L}(C+nH-nH)\otimes\mathcal{O}_C) \\
 &= \deg(\mathcal{L}(C)\otimes\mathcal{O}_C)
\end{align} $$