3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Isn't it isomorphic to an appropriate Grassmannian? $\endgroup$ 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
    Mar 22, 2023 at 20:03

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ 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
    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
    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. $\endgroup$ 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
    Mar 23, 2023 at 1:35
  • $\begingroup$ Very important result: math.stackexchange.com/questions/579527/… $\endgroup$ Mar 23, 2023 at 1:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.