Hartshorne Exercise V.1.2
Let $H$ be a very ample divisor on a surface $X$, corresponding to a projective embedding $X\subseteq\PP{k}{N}$. If we write the Hilbert polynomial of $X$ (III, Ex. 5.2) as
$$P(z)=\frac 12 az^2+bz+c,$$
show that $a=H^2$, $b=\frac 12 H^2+1-\pi$, where $\pi$ is the genus of a nonsingular curve representing $H$, and $c=1+p_a$. Thus the degree of $X$ in $\PP{k}{N}$, as defined in (I.7), is just $H^2$. Show also that if $C$ is any curve in $X$, then the degree of $C$ in $\PP{k}{N}$ is just $C.H$.
Write $\iota:X\ra \PP{k}{N}$ for the embedding of $X$, so In the language of exercise III.5.2, $\OO_X(H)=\iota^*\OO_{\PP{k}{N}}(1)=\OO_X(1)$ and $\mathcal{F}=\OO_X(1)$. Note that $\OO_X(n)=\OO_X(nH)$, using that $\iota^*:\op{Pic}\PP{k}{N}\ra \op{Pic}X$ is a homomorphism of groups. If $P(z)\in\Q[z]$ is the Hilbert polynomial of $\OO_X(1)$ relative to $\OO_X(1)$, then we see by Riemann Roch for surfaces that,
$$P(n)=\frac 12 n^2 H^2-\frac 12 nH.K_X+\chi (\OO_X),$$
confirming that $\deg P(z)=2$. It is clear that $a=\frac 12 H^2$. Note that the divisor $H$ on $X$ corresponds to a hyperplane section of $X$, which we take WLOG to be a general one; by Bertini’s theorem, $H$ is a nonsingular and irreducible curve. We find that,
$$\frac 12 H^2+1-\pi = \frac 12H^2+1 -\big (\frac{H.(H+K_X)+2}{2}\big )=-\frac 12 H.K_X=b.$$
By definition,
$$1+p_a=\chi(\OO_X)=c.$$
Now suppose $C$ is any curve in $X$, by which Hartshorne means any effective (Cartier) divisor on $X$, i.e., $C$ need not be irreducible. The projective embedding of $C$ is given by $\iota |_C$ which we take to mean $\iota \cdot j$ where $j:C\ra X$ is inclusion. By definition,
$$\deg C := \deg (\iota\cdot j)^*\OO_{\PP{k}{N}}(1)=\deg j^*\iota^*\OO_{\PP{k}{N}}(1)=\deg j^*\OO_X(H)=:C.H.$$
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