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I have recently learnt Riemann-Roch formula for surfaces. Roughly speaking, the theorem says that on a reasonably nice surface we have the relation: $$ \chi(D) = \frac{1}{2}(D.D - D.K) + p_a + 1 $$ where $D$ is a divisor, $\chi$ is the Euler characteristic, $K$ is the canonical divisor, and $p_a$ is the arithmetic genus. I am trying to use this to figure out $\chi(D)$ (or some components in the sum defining it), and am naturally led to the question of what can be said about the arithmetic genus $p_a$. In the case of curves, there is the degree-genus formula which allows one to compute the genus of a curve on a plane. Is there an analogous formula for the genus of a surface given by an equation in a three dimensional space? If not, what methods are there to compute $p_a$?

Note: This is related, I hope not too closely, to an assignment. Hopefully, using SE is no more morally objectionable than using a book.

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up vote 4 down vote accepted

In the general case the arithmetic genus of a hypersurface $H\subset \mathbb P^n$ (smooth or not) of degree $d$ is $$ p_a(H)=\binom {d-1}{n} $$ so that $ p_a(S)=\binom {d-1}{3} $ for a surface $S\subset \mathbb P^3$, the special case that interests you.

Proof of the formula in the general case :
The Hilbert polynomial of $H$ is $P_H(z)= \binom {z+n}{n}- \binom {z-d+n}{n} \quad$
[cf. Hartshorne's proof of Prop 7.6, Chap. I, page 52]
By definition $p_a(H)=(-1)^{n-1}(P_H(0)-1)$.
Since $P_H(0)= \binom {n}{n}- \binom {-d+n}{n}=1-(-1)^n \binom {d-1}{n} $, we get $$p_a(H)=(-1)^{n-1}(P_H(0)-1)=(-1)^{n-1}[1-(-1)^n \binom {d-1}{n}-1]=\binom {d-1}{n}$$ as announced.

Edit: a cultural note.
The arithmetic genus of a smooth projective surface over an algebraically closed field $k$ is a birational invariant.
The same is true for higher dimensional projective manifolds if $char.(k)=0$: this is highly non trivial since it requires Hironaka's desingularization theorem.
Turning back to the case of a smooth surface $S\subset \mathbb P^3_k$ of degree $d$, it is well known that for $d=1$ (a plane) or $d=2$ ( a quadric) $S$ is rational.
However the formula $ p_a(S)=\binom {d-1}{3} $ shows that for degree $d \geq 4$ one has $p_a(S)\gt 0$ and that $S$ is thus not rational, since a rational smooth projective surface has $p_a=0$.
The only remaining case is $d=3$, for which $p_a(S)=0$ . The result in that case is that such a cubic surface is rational.
To sum up, for a smooth surface $S\subset \mathbb P^3_k$ of degree $d$ : $$S \:\text{is rational} \iff p_a(S)=0\iff 1\leq d\leq 3 .$$

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