I don't know how to solve part (a) of exercise 1.5.8 in Robin Hartshorne's book Deformation Theory 1.5.8 (page 42):



> 
> Consider the Hilbert scheme of zero-dimensional closed subschemes
> of $\mathbb{P}^4\_k$ of length $8$, the field $k$ is algebraically closed. There is one component of dimension $32$
> that has a nonsingular open subset corresponding to sets of eight
> distinct points. We will exhibit another component containing a nonsingular open
> subset of dimension $25$.
> 
> 
> 



> 
> (a) Let $R=k[x,y,z,w]$, let $\mathfrak{m}$ be a maximal ideal in
> this ring, and let $I=V+\mathfrak{m}^3$, where
> $V$ is a $7$-dimensional subvector space of
> $\mathfrak{m}^2/\mathfrak{m}^3$. Let $B=R/I$, and
> let $Z= \operatorname{Spec}(B)$ be the associated closed subscheme of
> $\mathbb{A}^4 \subset \mathbb{P}^4$.
> Show that the set of all such $Z$, as the point of its support
> ranges over $\mathbb{P}^4$, forms an irreducible $25$-dimensional
> subset of the
> Hilbert scheme $H=\operatorname{Hilb}^8(\mathbb{P}^4)$.
> 
> 
> 


How can I show that this subscheme of the Hilbert scheme is irreducible? I have no idea even how to start. Is there a general strategy how to deal with that kind of questions to show that certain subscheme of a moduli space is irreducible? Clearly it suffice to construct an irreducible open dense subscheme sitting inside it, but I not see how to manage it in this exercise.



Exercise 1.5.8 from Robin Hartshorne's Deformation Theory
As the comment also indicates, consider the Grassmannian $ Gr\_3(\operatorname{Sym}^2 \Omega\_{\mathbb{P}^4} ) $ where the vector bundle $ \operatorname{Sym}^2 \Omega\_{\mathbb{P}^4} $ has fiber $ m\_p^2/m\_p^3 $ over a point $ p \in \mathbb{P}^4 $. Then there is a morphism $ Gr\_3(\operatorname{Sym}^2 \Omega\_{\mathbb{P}^4} ) \rightarrow Hilb^8(\mathbb{P}^4) $ and it suffices to give it by specifying it on $ \mathbb{C} $-valued points. By the universal property, a point in $ Gr\_3(\operatorname{Sym}^2 \Omega\_{\mathbb{P}^4} ) ( \mathbb{C} ) $ is given by a point in $ \mathbb{P}^4 $ along with a $ 3 $-dimensional quotient $ \operatorname{Sym}^2 \Omega\_{\mathbb{P}^4}|\_p = m\_p^2 / m\_p^3 \rightarrow Q \rightarrow 0 $, or equivalently a $ 7 $-dimensional subspace, which in turn gives the appropriate length $ 8 $ subscheme of $ \mathbb{P}^4 $.


It only remains to see that $Gr\_3(\operatorname{Sym}^2 \Omega\_{\mathbb{P}^4} ) $ is $ 25 $-dimensional and irreducible.


