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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!

user267839
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  • If you know S(k) for any ring k (not only a field), and the map p(k) you define is functorial in k then you get that p is coming from a morphism of schemes from the Yoneda Lemma. – Danny Ofek Sep 09 '22 at 22:21
  • @DannyOfek: That's true. But from the assumpions I was only able to construct the map $p(k)$ for $k=$ base field. I'm not sure if it is possible to prolonge from this piece of information it somehow to an arbitrary ring. What we assume is that $k$ is algebraically closed ... maybe that might be enough but I don't know... – user267839 Sep 09 '22 at 22:31
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    Show trhat that set has dense subset you can parametrize by an irreducible variety using a regular map. Describe one of those ideals involves picking a point in an $A^4$ and the a subspace of dimension $7$ in a $10$ dimensional vector space, so a point in a grassmanian. – Mariano Suárez-Álvarez Sep 10 '22 at 02:01
  • @MarianoSuárez-Álvarez: I'm a bit confused with the usage of word 'set' in this context. When you say ' Show that set has dense subset... ' then you consider S as ' set ' literally as 'bare' underlying set forgetting the scheme structure which comprises of elements corresponding to prime ideals of the scheme? And where the closed points are exactly the k-valued points, because k is algebraically closed. It's known that the later is dense. Is then the set of $k$-valued points exactly the 'dense subset' you suggested in your last comment to consider? – user267839 Sep 10 '22 at 19:47
  • If yes, then I'm not sure how to carry out the next step, namely to construct a regular map to an ireducible variety, presumingly to the product $\mathbb{A}^4 \times \mathbb{G}_{10,7}$. Which polynomials I should take to establish this regular map? – user267839 Sep 10 '22 at 19:47
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    You described a construction that from a point $p$ in $A^4$ and a $7$-dimensional subspace of $(x,y,z,w)^2/(x,y,z,w)^3$ produces a subscheme of $A^4$. Show that that map is regular. – Mariano Suárez-Álvarez Sep 10 '22 at 21:23
  • @MarianoSuárez-Álvarez: I'm not sure if you rephrased the map in the construction correctly: set theoretically it takes pair $(p, V)$ \to $Z:= V(I = V + \mathfrak{m}^3) \subset \mathbb{P}^4$, where $p \in A^4$ and $V$ is a $7$-dimensional subspace of $ \mathfrak{m}_p^2/\mathfrak{m}_p^3 $ ! where the maximal ideal $\mathfrak{m}_p \subset k[x,y,z,w]$ corresponds to point $p$! Note that $\mathfrak{m}_p$ varies with $p$; therefore the $V$ is not neccessary always contained in $(x,y,z,w)^2/(x,y,z,w)^3$. $(x,y,z,w)$ is only one possible maximal ideal... – user267839 Sep 10 '22 at 23:07
  • Therefore I not know how to show that this map is regular. recall, if we go back to old fashioned variety world, then by definition for two closed subvarieties $X \subset A^n, Y \subset A^m$ a map $f: X \to Y$ is called regular if is the restriction of a polynomial map $A^n \to A^m$ given explicitely in the form $x \mapsto (f_1(x),..., f_m(x))$ where the $f_i$ are living in coordinate ring of $X$: $k[x_1,..., x_n]/I(X)$. – user267839 Sep 10 '22 at 23:12
  • Let come back to our case: we have as above $(p, V) \mapsto Z_{p,V}:= V(I = V + \mathfrak{m}^3) \subset \mathbb{P}^4$ and we want to show it's regular, that is of the form above. First problem: In which variety $X$ are the pairs $(p,V)$ living. $p \in A^4$, but note that $V$ varies with $p$ as I explained above, so $V$ is not an element of Grassmanian $\mathbb{G}_7((x,y,z,w)^2/(x,y,z,w)^3)$ !... – user267839 Sep 10 '22 at 23:18
  • @MarianoSuárez-Álvarez: ok, I think I see now what you mean. You suggest to predecompose to this map from (a) - let call it $f$ - which as before maps $(p, V) \mapsto Z_{p,V}:= V(I = V + \mathfrak{m}^3) \subset \mathbb{P}^4$ - an operation which 'shifts' fiberwise $ V $ living in $ \mathfrak{m}_p^2/\mathfrak{m}_p^3$ to $ (x,y,z,w)^2/(x,y,z,w)^3 $, right? And then check the regularity of the composition of this 'shift map' with $ f $. Is this the idea you suggest? – user267839 Sep 14 '22 at 22:43
  • If yes, then the problem that I see is that this 'fiberwise shift' map of the $V$'s seems not to be a regular map in sense above, ie it is not given by polynomials. So I think the point is not if the is always an isomorphism between $ \mathfrak{m}p^2/\mathfrak{m}_p^3$ to $(x,y,z,w)^2/(x,y,z,w)^3 $ (which abstractly of course always exists), but how it evolve when we vary the point $p$. I think the problem with this approach is similar to the reason why the need connections in differential geometry: https://en.m.wikipedia.org/wiki/Connection(vector_bundle)#Motivation – user267839 Sep 14 '22 at 22:59
  • Abstractly tangent spaces of a manifold are at every point isomorphic but in order to impose a reasonable concept of differentiation we have to identify the tangent spaces, and this requires noncanonical choices from point to point – user267839 Sep 14 '22 at 23:01

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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.

Mariano Suárez-Álvarez
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  • alright I see the idea now, once having constructed this flat family the existence of the desired regular map to thr Hilbert scheme follows automatically since it represents this functor. The association with connections from diffgeo in the comments above are nonsense as I see now ; previously I misunderstood your comments in that way that I thought that your intention was to construct explicitly / directly a regular map to the Hilbert scheme 'shifting fiberwise each subspace $ V \subset \mathfrak{m}_p^2/\mathfrak{m}_p ^3$ to $(x,y,z,w)^2/(x,y,z,w)^3$ and this strategy seemed – user267839 Sep 19 '22 at 20:38
  • quite 'non algebraic' to me since there would be noncanonical choices be involved. And this reminded me a bit on this business with connections. That's it, but as I see was thinking in wrong direction... – user267839 Sep 19 '22 at 20:40
  • sorry for digging out this old issue, but is it clear that the family you constructed is flat over $k^4\times M_{7,2}(k)$? Previously I thought that it follows from an abstract result that as all fibres over closed points are equidimensional (...precisely zero dimensional), that this already suffice to imply flatness, but in general that's not the case. Could you maybe loose few words on how you actually see that this family is flat? – user267839 Dec 10 '24 at 16:16