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?
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.