ag.algebraic geometry - Component of Hilbert Scheme

The short (and unfriendly!) answer is that it is an irreducible component of the Hilbert scheme. Here is a slightly more detailed answer.

Given a subvariety $Xsubset mathbb P^N$, you can attach to it a polynomial, the Hilbert polynomial $P_X(t) in mathbb Q[t]$. It is a rough invariant, which nevertheless contains much qualitative information about $X$. Its degree gives the dimension $d$ of the variety: $ degree P_X(t)=dim(X)$. The leading coefficient of $P_X(t)$ calculates the degree of the variety $lead(P(t))=(1/d!): deg(X)$, the arithmetic genus of the variety is $; p_a(X)=(-1)^d(P_X(o)-1) $, etc.

Now, the Hilbert scheme (introduced by Grothendieck, as the name does not say...) parametrizes all subschemes of $mathbb P^N$ and Hartshorne has proved the wonderful theorem that if you fix the Hilbert polynomial $P(t)$, the corresponding Hilbert scheme will be connected.

Since then people have investigated these connected schemes and, in particular, their irreducible components which your question is about. The best introduction to these questions might still be Mumford's 1966 notes notes:

References
Mumford, D. Lectures on Curves on Algebraic Surfaces (with George Bergman), Princeton University Press, 1964.

Hartshorne R. Connectedness of the Hilbert scheme. Publications Mathématiques de l'IHÉS, 29 (1966), p. 5-48

Comment for functor aficionados Here are two sentences from Hartshorne's
article that will warm your heart

"It also appears that the Hilbert scheme is never actually needed in the proof. ... we define the notion of a connected functor, and prove that the functor $Hilb^p$ is connected" [From the introduction]

"This section is a variation on the theme " anything you can do with preschemes, you can do with the functors they represent " " [page 12]

Finally, you can't even define Hilbert schemes rigorously if you don't first define the functor it represents...