# Exercise 1.5.8 from Robin Hartshorne's Deformation Theory

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.

• Isn't it isomorphic to an appropriate Grassmannian? Mar 22 at 19:45
• @YosemiteStan: yes, we can surely construct 'by hand' a map of sets $\text{Gr}(7, \mathfrak{m}^2/\mathfrak{m}^3) \to \operatorname{Hilb}^8(\mathbb{P}^4)(k)$ sending subvector space $V \subset \mathfrak{m}^2/\mathfrak{m}^3$ to $Z \subset \mathbb{P}^4$, a point of the Hilbert scheme. But can we prolonge it naturally to an closed embedding $\mathcal{Gras}(7, \dim_k \mathfrak{m}^2/\mathfrak{m}^3) \subset \operatorname{Hilb}^8(\mathbb{P}^4)$ of schemes? Mar 22 at 20:03

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.
• I'm getting a bit confused with your second statement. Do you claim that every morphism $X \to Hilb^8(\mathbb{P}^4)$ from arbitrary scheme $X$ can be defined by specifying it on $\mathbb{C}$-valued points, or is it here a special intrinsic feature of the space $Gr_3(\operatorname{Sym}^2 \Omega_{\mathbb{P}^4} )$ that in order to establish a map $Gr_3(\operatorname{Sym}^2 \Omega_{\mathbb{P}^4} ) \rightarrow Hilb^8(\mathbb{P}^4)$ it suffice to specify it on $\mathbb{C}$-valued points? Mar 22 at 21:22
• Or do you mean above that if $X$ is any scheme with the property that any $\mathbb{C}$-point of it can be intrinsically identified with a point of $Hilb^8(\mathbb{P}^4)$, then this already suffice to obtain a map $X \to Hilb^8(\mathbb{P}^4)$? Mar 22 at 21:31
• Any morphism from a variety $X$ to $Hilb$ can be understood by understanding it on its geometric points. This is in Hartshorne (his original book) as the t-functor in chapter 2. Mar 23 at 0:12
• Next, the dimension count is easy since every fiber of $Gr_3(\operatorname{Sym}^2 \Omega_{\mathbb{P}^4} ) \to \mathbb{P}^4$ is $Gr_3(m_p^2/m_p^3)$ and has therefore dimension $(10-3)3=21$ plus the dimnsion of the base, so $25$. But why is $Gr_3(\operatorname{Sym}^2 \Omega_{\mathbb{P}^4} )$ irreducible? Over affine charts $\mathbb{A}^4 \subset \mathbb{P}^4$ it becomes the Grassmanian over affine space $\operatorname{Sym}^2 \Omega_{\mathbb{A}^4}= \mathbb{A}^N$, which is irreducible. Does it follow from this that $Gr_3(\operatorname{Sym}^2 \Omega_{\mathbb{P}^4} ) \to \mathbb{P}^4$ is irreducible? Mar 23 at 1:35