Why do intersection of two quadratic forms implies elliptic curve? Let $k$ be a field and $S=k[T_0,T_1,T_2,T_3]$ and $f,g\in S_2$ two relatively prime quadratic forms. How can I show that the intersection $X\subset \mathbb P_k^3$ of second degree surfaces $V_+(f)$ and $V_+(g)$  has genus 1?
Does this follows from syzygy theorem or what should I use here?
 A: Yes, one way is indeed to use syzygies. 
Denote the curve by $C$. Then there is a resolution of the form 
$$
0 \to \mathscr O_{\mathbb P^3}(-4) \to \mathscr O_{\mathbb P^3}(-2)^{\oplus 2} \xrightarrow{\alpha} \mathscr O_{\mathbb P^3} \to \mathscr O_C \to 0.
$$
This is not a short exact sequence, so we have to break it into two short exact sequences:
$$
0 \to \ker \alpha \to \mathscr O_{\mathbb P^3} \to \mathscr O_C \to 0.
$$
and 
$$
0 \to \mathscr O_{\mathbb P^3}(-4) \to \mathscr O_{\mathbb P^3}(-2)^{\oplus 2}  \to \ker \alpha \to 0.
$$
We are interested in $H^1(\mathscr O_C)$. By the long exact sequence of cohomology, we have 
$$
0 \to H^1(\mathscr O_C) \to H^2(\ker \alpha) \to 0
$$
since $H^1(\mathscr O_{\mathbb P^3})=H^2(\mathscr O_{\mathbb P^3})=0$. Then from the second exact sequence we get:
$$
0 \to H^2(\ker \alpha) \to H^3(\mathscr O_{\mathbb P^3}(-4)) \to H^3(\mathscr O_{\mathbb P^3}(-2))^{\oplus 2}=0
$$
Hence $h^2(\ker \alpha)=h^3(\mathscr O_{\mathbb P^3}(-4))=h^0(\mathscr O_{\mathbb P^4})=1$.
