Reducibility of a Hilbert scheme in projective space My question concerns the computation of the Hilbert scheme $\mathsf{Hilb}_{3}^{2x+1}$, which parametrizes all curves of degree $2$ and genus $0$ in $\mathbb{P}^{3}_{k}$, with $k$ algebraically closed. I would like to prove that $\mathsf{hilb}_{3}^{2x+1}$ is reduced. To do that, I proceed in the following way:
(1) I want to show that $\text{dim}(\mathsf{Hilb}^{2x+1}_{3})=8$;
(2) I want to show that $\text{dim}\ \text{T}_{p}(\mathsf{Hilb}^{2x+1}_{3})=8$, for any closed point $p\in \mathsf{Hilb}^{2x+1}_{3}$.
I cannot seem to be able to compute part (1). I thought about simply counting the parameters of the generic quadric in $\mathbb{P}^{3}$, but it clearly doesn't work. With part (2) I made a little progress. I found on "Curves in Projective Space", Joe Harris, Les Presses de L'université de Montréal, 1982, the following equivalence
$\qquad \qquad \qquad \qquad \qquad$ $h^{0}(C,\mathcal{N}_{C/\mathbb{P}^{3}})=h^{0}(C,\mathcal{O}(1)\oplus \mathcal{O}(2))=3+5$
I thought this could be traced back to some split exact sequence, but I don't seem to be able to find which one. I also don't understand why $h^{0}(C,\mathcal{O}(1))=3$ and $h^{0}(C,\mathcal{O}(2))=5$.
 A: (1) A naive count of parameters could go as follows: we have that the space of polynomials of degree $2$ in two variables (the homogeneous coordinates of $\mathbb P^1$) has dimension $h^0(\mathbb P^1,\mathscr O_{\mathbb P^1}(2))=3$. You need $4$ such polynomials to get a degree $2$ curve in $\mathbb P^3$. But you certainly want to identify conics given by polynomials who are multiples of each other (so we remove $1$ parameter), and you might also want to forget automorphisms of $\mathbb P^1$ (which form a $3$-dimensional group).
Putting all this together we get 
$$\#\,\textrm{parameters}=(4\cdot 3)-1-3=8.$$
This is not quite a proof that the Hilbert scheme has dimension $8$, because there might be extraneous components in the Hilbert scheme, like for twisted cubics for instance. But this is not the case, as (roughly) conics are much less rigid than twisted cubics.
(2) Let us embed $C$ as a conic hyperplane section inside a smooth quadric surface $S\subset \mathbb P^3$. Then we have the exact sequence
$$0\to \mathcal N_{C/S}\to \mathcal N_{C/\mathbb P^3}\to \mathcal N_{S/\mathbb P^3}|_C\to 0.$$
This sequence splits because $C$ is a complete intersection. We have 
$$\mathcal N_{C/S}\cong\mathscr O_C(1)\cong\mathscr O_{\mathbb P^1}(2) $$
and $$\mathcal N_{S/\mathbb P^3}|_C\cong\mathscr O_C(2)\cong \mathscr O_{\mathbb P^1}(4).$$
Thus, the splitting condition says $\mathcal N_{C/\mathbb P^3}\cong\mathscr O_{\mathbb P^1}(2)\oplus\mathscr O_{\mathbb P^1}(4)$, and we have
$$h^0(C,\mathcal N_{C/\mathbb P^3})=h^0(\mathscr O_{\mathbb P^1}(2))+h^0(\mathscr O_{\mathbb P^1}(4))=3+5=8.$$
