Counting the dimension of a component of $\mathsf{hilb}^{2t+1}_{3}$ Consider the Hilbert scheme $\mathsf{hilb}^{2t+1}_{3}$, parametrizing varieties of degree $2$ and genus $0$ in $\mathbb{P}^{3}_{k}$, with $k$ an algebraically closed field. Consider the component
$
\quad \quad \mathsf{hilb}_{\text{conic}}=\{ p\in \mathsf{hilb}^{2t+1}_{3}\ |\ \text{the corresponding variety is a curve in $\mathbb{P}^{3}_{k}$}\}
$
My goal is to prove that such a component is irreducible of dimension $8$. I am not sure how to prove the irreducibility. To prove that $\dim (\mathsf{hilb}_{\text{conic}})=8$, one way would be to show that any conic can be obtained as the complete intersectio of a quartic and a hyperplane. However, in the context that I am working in, I would rather proceed in another way, and use the following
$\mathbf{Lemma.}$ Any conic $C\in \mathbb{P}^{3}_{k}$ is a complete intersection of a plane $L$ and a quadric surface $S$.
Does anyone know how to compute the dimension of $\mathsf{hilb}_{\text{conic}}$, using only the lemma above, and not other classical results concerning the nature of quadrics in projective space?
 A: A standard fact is that the dimension of the Zariski tangent space to the Hilbert scheme at a point $[C]$ is $h^0(\mathcal N_{C})$, where $C\subseteq\mathbb P^3$ is the corresponding curve and $\mathcal N_C$ is its normal sheaf in $\mathbb P^3$. Another fact is that the Hilbert scheme is nonsingular at a local complete intersection. Thus, one way to compute the dimension of the component containing such a curve is to compute $h^0(\mathcal N_C)$. To do this, one can use the following "elementary short exact sequence of normal sheaves":
$$ 0\to \mathcal N_{H/\mathbb P^3}|_C\to \mathcal N_C\to \mathcal N_{C/H}\to 0$$
Here we have chosen $H$ to be the hyperplane cutting the quadric to form $C$, so that $C\subseteq H\subseteq \mathbb P^3$. This is useful because $H$ is a divisor on $\mathbb P^3$ and $C$ is a divisor on $H$. Thus, $\mathcal N_{H/\mathbb P^3}\cong\mathcal O(1)|_H$ and $\mathcal N_{C/H}\cong\mathcal O_H(2)|_C = \mathcal O_C(2)$. The former gives $\mathcal N_{H/\mathbb P^3}|_C\cong \mathcal O(1)|_H|_C = \mathcal O_C(1)$.
Assuming that $C$ is nonsingular, we can pull back these sheaves to $\mathbb P^1$ to compute cohomology. Thus, the first term in the associated long exact sequence in cohomology has dimension ${1 + 2\choose 1} = 3$, the third term has dimension ${1 + 4\choose 1} = 5$, the fourth term vanishes, and thus the normal module has dimension $3+5=8$.
Another, probably easier way is to do the dimension counting the following. A (smooth irreducible) conic curve in $\mathbb P^3$ must lie in some hyperplane $H\cong \mathbb P^2$ (e.g., see the exercises to Hartshorne section IV.3). The Hilbert scheme of conics in $\mathbb P^2$ is ${2+2\choose 2}-1 = 5$ dimensional, and the hyperplane moves in the Grassmannian $\mathbb G(2,3) = G(3,4)$ which is $3(4-3) = 3$ dimensional, so the total family is $8$ dimensional.
