Del Pezzo surface of degree $4$ I'd like to show that the del Pezzo surface $S_4\subset\mathbb{P}^4$ (i.e. the complete intersection of two quadrics) is rational.
I've got two possibilities:
1- I show that is the blow-up of $\mathbb{P}^2$ at 5 points.
2- I use Castelnuovo's rationality criterion, so i prove that both $q(S_4)=h^1(\mathcal{O}_{S_4})=0$ and $P_2=h^0(\mathcal{O}_{S_4}(2K))=0$, where $K$ is the canonical divisor on $S_4$.
I'd like to do it "by hand" (for example i don't want to use theorems like the Kodaira vanishing theorem in the second case).
Is there anyone that could give me any hints?
 A: Let me give some thoughts on this. 
First, as I mentioned in the comments, Castelnuovo's criterion certainly works. By the Lefschetz theorem, $S_4$ is simply connected, so $q=0$. Moreover, adjunction tells you that $K_{S_4}={K_{{\mathbf P}^4}}_{|S_4} \otimes \bigwedge^2 N_S$ where $N_S$ is the normal bundle of the surface; unpacking this gives you $-K_{S_4}=\mathcal{O}_{\mathbf P^4}(1)_{|S_4}$. So $-K$ of your surface is very ample: this implies (as discussed in earlier questions) that the bundle $mK$ cannot have sections for any positive $m$. In particular $P_2=0$, as you want.
(I don't know how to show $q=0$ without Lefschetz; however, since we are already using Castelnuovo's criterion, we're far from proving rationality "by hand". So it seems a little arbitrary to disallow other theorems like Lefschetz and Kodaira.)
OK, but this all seems a little unsatisfactory: there must be a more direct way. (How would del Pezzo have proved this, for instance?) Here's what I came up with so far. 
First, I claim that $S_4$ contains a conic curve $C$. (I wasn't able to come up with an elementary argument for this. There's a pretty standard argument using Chern classes, but I would like to know a simpler one.) Let's take this as given. Then $C$ is contained in a plane $\Pi \subset \mathbf P^4$. If $T$ is any 3-dimensional linear space containing $\Pi$, then $T \cap S_4$ is a degree-4 curve containing $C$ as a component: let's write it as $C \cup \Gamma_T$. Note that for each $T$, the curve $\Gamma_T$ is again a conic.
Now we can apply adjunction again to calculate the self-intersection of the curves $\Gamma_T$ on $S$: we obtain $\Gamma_T^2=0$. This means that the family of curves has no basepoints on $S_4$, so it gives a map $S_4 \rightarrow \mathbf P^1$ whose fibres are exactly the conics $\Gamma_T$. So $S_4$ is a conic bundle over $\mathbf P^1$, and (assuming the base field is algebraically closed) such a surface is always rational. 
(Extra credit problem: show that the map $S_4 \rightarrow \mathbf{P}^1$ has exactly 4 fibres that consist of a pair of distinct lines. Each line in such a pair is a $(-1)$-curve, so one of each can be blown down; conclude that $S_4$ is the blowup of $\mathbf P^2$ in 5 points.)
I've skipped some details, but I hope the answer is still helpful.
