$\operatorname{Hilb}^8(\mathbb{P}^4_k)$ not irreducible (Ex. in Hartshorne's Deformation Theory book) Exercise 1.5.8 from Robin Hartshorne's Deformation Theory:
5.8. $\operatorname{Hilb}^8(\mathbb{P}^4_k)$ is not irreducible.
Consider the Hilbert scheme of zero-dimensional
closed subschemes of $\mathbb{P}^4_k$
of length $8$, the ground field $k$ is assumed to be algebraically closed. There is one component of dimension $32$ that
has a nonsingular open subset corresponding to sets of eight distinct points. (I suppose that the author uses it as nontrivial fact)
We will
exhibit another component containing a nonsingular open subset of dimension $25$.
The Exercise comprises of four parts and I have problems with the first part:
(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 $Z$ 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 to show that the "set" of the $Z$'s as defined in (a) is irreducible?
Let call it $S \subset H$. The Hilbert scheme $H$ is constructed as closed subscheme of the Grassmanian defined by the vanishing of various determinants and is therefore we can endow the "set" $S$ as subscheme of $H$ with unique reduced scheme structure.
On the set level / on $k$-valued points $S(k)$ we can define canonically the map $p(k): S(k) \to \mathbb{P}^4(k)$ sending $Z$ the the unique maximal ideal $\mathfrak{m}_Z \subset k[x, y, z,w]$ associated to it as described in the construction above.
How can this idea be converted into a 'honest' map $p:S \to \mathbb{P}^4$? As soon as it is possible to construct such map $p$ we can use a result (reference ?) that for a proper surjective map $f: X \to Y$ with $Y$ and all fibers irreducible of same dimension, the scheme $X$ is irreducible, too.
Therefore the question reduces to 'How to construct $p:S \to \mathbb{P}^4$ from set map $p(k): S(k) \to \mathbb{P}^4(k)$?'
In addition note that that's just my suggestion how roughly I wanna to tackle this exercise. Maybe there are more effective ways to do it. All suggestions for alternative approaches are of course welcome!
 A: Honestly, I have no idea what you are on about connections and such…
Let $(a_{i,j})$ be an arbitrary $7\times 3$ matrix of scalars, and let $f_1,\dots,,f_7$ be the linear combinations of monomials indicated in the rows of following table:
$$\begin{array}{*{12}{c}}
x^2 & y^2 & z^3 & w^2 & xy & xz & xw & yz & yw & zw \\ \hline
 1 & & & & & & & a_{1,1} & a_{1,2} & a_{1,3} \\
 & 1 & & & & & & a_{2,1} & a_{2,2} & a_{2,3} \\
 & & 1 & & & & & a_{3,1} & a_{3,2} & a_{3,3} \\
 & & & 1 & & & & a_{4,1} & a_{4,2} & a_{4,3} \\
 & & & & 1 & & & a_{5,1} & a_{5,2} & a_{5,3} \\
 & & & & & 1 & & a_{6,1} & a_{6,2} & a_{6,3} \\
 & & & & & & 1 & a_{7,1} & a_{7,2} & a_{7,3} \\
\end{array}$$
Now let $a$, $b$, $c$ $d$ be four scalars and consider the ideal generated by the $7$ polynomials
$$
  f_1(x-a,y-b,z-d,w-d), \dots, f_7(x-a,y-b,z-d,w-d)
  $$
and all the polynomails $(x-a)^i(y-b)^j(z-c)^k(w-d)^l$ with $i+j+k+l=3$.
This gives you a $25$-dimensional family of ideals of colength $8$, parametrized by a point in $k^4\times M_{7,2}(k)$.
Viewing the entries of the matrix and the coordinates of the point $(a,b,c,d)$ as varibles now, the ideal generated by those seven polynials in $k[x,y,z,w,a,b,c,d,a_{1,1},\dots,a_{7,3}]$ define subscheme $Z$ in $k^4\times M_{7,2}(k)\times k^4$. The restriction of the map $p:k^4\times M_{7,2}(k)\times k^4\to k^4\times M_{7,2}(k)$ projecting on the first two factors to $Z$ is a map $Z\to k^4\times M_{7,2}(k)$ which is a flat family of subschemes of $k^4$, the fiber of $p$. The universal property of the Hilbert scheme tells you then that to this flat family corresponds a regular map into the Hilbert scheme.
