Let $F_1$ and $F_2$ be two (smooth) surfaces in $\mathbb{P}^3$, of degrees $d_1$ and $d_2$ respectively. Let $C$ denote curve given as their intersection. How one can compute arithmetical genus of the curve $C$?
Perhaps I need to add some assumptions on $F$ and $G$ to exclude degenerate cases.
You can get the answer by using the adjunction formula.
First assume that $C$ is smooth. Then its normal bundle $N_C$ is the restriction of $\mathscr O (d_1) \oplus \mathscr O(d_2)$ to $C$. Then the adjunction formula (Hartshorne II.8.20) says that the canonical bundle of $C$ is $$K_C = (K_{\mathbf P^3})_{|C} \otimes \bigwedge^2 N_C = (\mathscr O(-4) \otimes \mathscr O(d_1+d_2))_{|C}.$$
Since $C$ is a curve of degree $d_1d_2$, that means $K_C$ is a line bundle of degree $d_1d_2(d_1+d_2-4)$.
On the other hand, the genus of $C$ is related to this number by
$$2g-2= \operatorname{deg} K_C = d_1d_2(d_1+d_2-4)$$
so one gets $g= \frac12 (d_1d_2(d_1+d_2-4)+2)$.
As a sanity check, if say $d_1=1$, then $C$ is actually a plane curve of degree $d_2$, and the formula above gives its genus as $\frac12(d_2(d_2-3)+2)=\frac12(d_2-1)(d_2-2)$ as expected.
Now, what if $C$ is not smooth? I claim it doesn't matter, as long as $C$ is viewed the scheme-theoretic intersection of the surfaces and we calculate arithmetic genus accordingly. By deforming the coefficients of the defining equations of $F_1$ and $F_2$ we get a flat family of curves in $\mathbf P^3$ which has $C$ as one member and whose general member is a smooth intersection of surfaces of degrees $d_1$ and $d_2$. Hartshorn III.9.10 shows that arithmetic genus is constant in such a family.
Edit: The OP asked me to justify the claim that $N_C$ is $\mathscr O (d_1) \oplus \mathscr O(d_2)$ restricted to $C$, so let's do that.
Here $\mathscr O(d_1)_{|F_i}$ is the normal bundle to $F_i$ for each $i$.
(From now on, I won't write the restriction everywhere. It should be clear from context where the bundles in question live.)
Now $N_C = T_X/T_C$ and $N_{|F_i} = T_X/T_{F_i}$ so there is a canonical quotient map $$N_C \rightarrow N_{F_i}$$ for each $i$. This gives a map $$\varphi: N_C \rightarrow N_{F_1} \oplus N_{F_2}$$ which I claim is an isomorphism. To see this, observe that both bundles have the same rank at each point, so it is enough to prove it is injective. And now the key point is that since $C$ is smooth, the two surfaces intersect transversely at each point of $C$: that is
$$ T_{F_1} \cap T_{F_2} = T_C $$
at each point of $C$. Therefore the map $\varphi$ above has trivial kernel.
The cohomology of the intersection of two surfaces can also be computed by writing resolutions. First of all, we have the ideal sheaf sequence $$ 0 \to \mathcal I \to \mathcal O_\mathbb P \to \mathcal O_C \to 0.$$
We also have the exact sequence $$ 0 \to \mathcal O(-d-e) \to \mathcal O(-d) \oplus \mathcal O(-e) \to \mathcal O \to \mathcal I \to 0. $$
Then you can use arguments with long exact sequences to compute the cohomology of $C$. First of all, the cohomology of the ideal sheaf is completely determined by the second exact sequence together with results on the cohomology of projective space. Then combine this with the first sequence to determine the cohomology of $\mathcal O_C$.
(this is more bothersome than Asal Beag Dubh's method however...)
