Hartshorne Exercise V.1.3.
Recall that the arithmetic genus of a projective scheme D of dimension 1 is defined as $p_a=1-\chi(\OO_D)$ (III, Ex. 5.3).
(a) If $D$ is an effective divisor on the surface $X$, use (1.6) to show that $2p_a – 2 =
D.(D + K).$
(b) $p_a(D)$ depends only on the linear equivalence class of $D$ on $X$.
(c) More generally, for any divisor $D$ on $X$, we define the virtual arithmetic genus (which is equal to the ordinary arithmetic genus if $D$ is effective) by the same formula: $2p_a – 2 = D.(D + K)$. Show that for any two divisors $C,D$ we have
$$p_a(-D)=D^2 – p_a(D)+2$$
and
$$p_a(C +D) = p_a(C) +p_a(D) +C.D – 1.$$
(a) Using the SES,
$$0\ra \OO_X(-D)\ra\OO_X\ra\OO_D\ra 0$$
we find that $p_a-1=\chi(\OO_X(-D))-\chi(\OO_X)$. Then by Riemann Roch for surfaces with the divisor $-D$, we find that
$$\chi(\OO_X(-D))-\chi(\OO_X)=\frac 12 D.(D+K).$$
(b) Indeed, the intersection product on a surface $X$ is defined by the degree homomorphism on isomorphism classes of line bundles, and linearly equivalent divisors give rise to isomorphic line bundles on $X$.
(c) Observe that
$$ p_a(D)+p_a(-D)=\frac 12 (D^2+D.K+D^2-D.K)+2=D^2+2.$$
Also,
$$p_a(C+D)-p_a(C)-p_a(D)=\frac 12 ((C+D)^2+C.K+D.K-C^2-C.K-D^2-D.K)-1=C.D-1.$$
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