Del Pezzo surface of degree 4 is intersection of two quadrics? Let $S$ be a del Pezzo surface $S$ of degree $4$. There is an exact sequence
$$ 0\to H^0(\mathbb{P}^4,I_S(2)) \to H^0(\mathbb{P}^4,\mathcal{O}(2))\to H^0(S,\mathcal{O}_S(2))\to0$$ where $I_S$ is the ideal sheaf of $S$. Let $K$ be a canonical divisor. We have $$H^0(S,\mathcal{O}(-2K_S))=13.$$
The claim is $$H^0(S,\mathcal{O}(-2K_S))=H^0(S,\mathcal{O}(2))$$ and from this it follows that $S$ is the base locus of a pencil of quadrics. I neither understand the isomorphism nor how the statement about the base locus follows, could anyone elaborate? 
This is from Dolgachev:" Classical Algebraic Geometry: a Modern View" (Thm 8.2.6)
Thanks a lot.
PS: a base locus of a pencil of quadrics is the intersection of two quadrics. 
 A: Your Del Pezzo surface $S$ is embedded into $\mathbb P^4$ anticanonically, i.e. $\mathcal O_S(1) = - K_S$.  (For this, recall that $-K_{\mathbb P^2} = \mathcal O_{\mathbb P^3}(3)$.  Remember also that blowing up adds a copy of the exceptional divisor to the canonical bundle, and that $S$ is obtained by blowing up $\mathbb P^2$ at five points, and is embedded into $\mathbb P^4$ via the linear system of cubics passing through these five points.  These remarks taken together will let you verify the claim.)  Consequently  $\mathcal O_S(-2K_S) = \mathcal O_S(2)$, and the stated isomorphism follows.
Now the dimension of $H^0(\mathbb P^4, \mathcal O(2))$ is $15$, and so, from the exact sequence
$$0 \to H^0(\mathbb P^4,\mathcal I_S(2)) \to H^0(\mathbb P^4, \mathcal O(2)) \to H^0(S, \mathcal O_S(2) ) \to 0$$
we find
that $H^0(\mathbb P^4, \mathcal I_S(2))$ is $2$-dimensional.  This precisely means that there is a two-dimensional space of quadratic forms that vanish on $S$, and hence that $S$ is the base locus of the corresponding pencil of quadrics. 
