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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.

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  • $\begingroup$ Isn't it isomorphic to an appropriate Grassmannian? $\endgroup$
    – Yosemite Stan
    Commented Mar 22, 2023 at 19:45
  • $\begingroup$ @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? $\endgroup$
    – JackYo
    Commented Mar 22, 2023 at 20:03

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – JackYo
    Commented Mar 22, 2023 at 21:22
  • $\begingroup$ 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 $X \to Hilb^8(\mathbb{P}^4)$? $\endgroup$
    – JackYo
    Commented Mar 22, 2023 at 21:31
  • $\begingroup$ 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. $\endgroup$
    – Cranium Clamp
    Commented Mar 23, 2023 at 0:12
  • $\begingroup$ 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? $\endgroup$
    – JackYo
    Commented Mar 23, 2023 at 1:35
  • $\begingroup$ Very important result: math.stackexchange.com/questions/579527/… $\endgroup$
    – Cranium Clamp
    Commented Mar 23, 2023 at 1:41

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