Let $Q\subset\mathbb{P}^3\subset\mathbb{P}^4$ be a smooth quadric surface, and let $X = Bl\_Q\mathbb{P}^4$ the blow-up of $\mathbb{P}^4$ along $Q$. Let $H$ be the pull-back of the hyperplane section of $\mathbb{P}^4$ and $E$ the exceptional divisor. I would like to compute te intersection numbers of these divisors. 


For instance $H^4 = 1$. Then we should have $H^3\cdot E = 0$ because $H^3$ is the pull-back of the class of a line and $Q$ is in codimension two. Then $H^2\cdot E^2 = -2$ because $H^2$ is the pull-back of the class of a plane and $deg(Q) = 2$. Finally the first chern class of $N\_{Q/\mathbb{P}^4}$ is $3$. Therefore we should have $H\cdot E^3 = H\cdot E^2\cdot E = -2\cdot 3$, and $E^3 = -2\cdot 3^2$. Summing up we should have


$$H^4 = 1, \quad H^3\cdot E = 0, \quad H^{4-i}E^i = -2\cdot 3^{i-2},\: for \: i\geq 2.$$


Is this computation correct or am I missing something?



Blow-up of $\mathbb{P}^4$ along a quadric surface
The intersection numbers $H^4 =1,\; H^3\cdot E = 0,\; H^{2}\cdot E^2 = -2,\; H\cdot E^3 = -6$ are correct. The last one $E^4 = H^0\cdot E^4 = -18$ is wrong.


A way to compute $E^4$ is the following. The blow-up $X$ of a smooth quadric surface $Q\subset\mathbb{P}^4$ is isomorphic to the blow-up of a smooth $4$-dimensional quadric in a point. Let us call $Y$ this blow-up, and let $E\_p\subset Y$ be the exceptional divisor. Then $E\_p^4 = -1$. On the other hand $E\_p$ is the strict transform through the blow-up map $\epsilon:X\rightarrow \mathbb{P}^4$ of the $3$-plane spanned by $Q$. Then $E\_p = H-E$, and
$$E\_p^4 = H^4-4H^3\cdot E+6H^2\cdot E^2-4H\cdot E^3+E^4.$$
Using the first four intersection numbers we get
$$-1 = E\_p^4 = 1-12+24+E^4,$$
and finally $E^4 = -14$. Summing up the intersection numbers you are looking for are:
$$H^4 =1,\; H^3\cdot E = 0,\; H^{2}\cdot E^2 = -2,\; H\cdot E^3 = -6,\; E^4 = -14.$$
Another way to see this is the following:
$$E^4 = -deg(Q)s\_2 = -2s\_2,$$
where $s\_2$ is the second Segre class of the conormal bundle $N\_{Q/\mathbb{P}^4}^{\vee} = \mathcal{O}\_{Q}(-1)\oplus\mathcal{O}\_{Q}(-2)$. The chern classes of this bundle are $c\_1 = -3$ and $c\_2 = 2$. Therefore $s\_1 = 3$ and $s\_2 = s\_1^2-c\_2 = 7$. Finally you get again $E^4 = -2s\_2 = -14$.


