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I do not work in algebra, so i apologize in advance if there are some unclear/wrong sentences. Let us consider the ring $\mathbb{C}[X_1,\ldots,X_q]$ of polynomials in $q$ variables. For an ideal $I$ of this ring we associate its variety $$V(I) = \left\lbrace(x_1,\ldots,x_q)\in\mathbb{C}^q \quad\middle | \quad P(x_1,\ldots,x_q) = 0\quad\forall P\in I\right\rbrace.$$ I'm interested in the set $$E = \left\lbrace I\text{ ideals of } \mathbb{C}[X_1,\ldots,X_q]\quad\middle |\quad V(I) = \lbrace(0,\ldots,0)\rbrace\right\rbrace$$ That is, the sets of ideals whose corresponding algebraic variety consists only in the origin point. One can define the multiplicity of an element ideal $I$ in $E$ as $$mult(I) = \dim \mathbb{C}[X_1,\ldots,X_q]/I,$$ and for $p\geq 1$ the space $$E_p = \left\lbrace I\in E\quad\middle |\quad mult(I) = p\right\rbrace.$$ Is there any known natural differentiable structure on the space $E_p$ ?

Following this article, an ideal in $E_p$ can be described as a collection of $p$ differential conditions at the origin, who must satisfies some closeness condition (in order to be an ideal). Thus a natural topology would be the topology induced by the Grassmannian $Gr(p,V)$, where $V$ is the set of differential operators whose order do not exceed $p-1$.

Second question, can we prove that topologically, $$\dim(E_p) = (p-1)(q-1)\quad ?$$ For instance if $q=1$ then $E_p$ consists in the only ideal $\langle X^p\rangle$ and thus $\dim(E_p) = 0$. For $q>1$, there is a one-to-one correspondence between $E_2$ and $\mathbb{S}^{q-1}$ given by the mapping $$\theta\mapsto \left\lbrace\quad P \quad\middle|\quad P(0) = 0\quad\text{and}\quad \partial_\theta P(0) = 0 \right\rbrace,$$ it yields $\dim(E_2) = q-1$. On can also prove in a similar fashion that $\dim(E_3) = 2(q-1)$ but for $p>3$ it gets a bit messy.

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    $\begingroup$ Are you sure you wish to ask about real ideals $I$ whose only real point is the origin rather than those whose only complex (=geometric) point is the origin? For example, with the definition you wrote, the ideal $(X_1^2+X_2^2)$ in $\mathbb{R}[X_1,X_2]$ is in $E$, although it is not zero-dimensional in any way, and in particular $\operatorname{mult}(I)$ is infinite, which seems to go against the fact that you seem to want to know geometric things about zero-dimensional ideals. So I think your question needs fixing. $\endgroup$
    – Gro-Tsen
    Commented Mar 31, 2023 at 13:25
  • $\begingroup$ (contd.) So: either your questions are really geometrical in nature, in which case you should fix them (change $\mathbb{R}$ to $\mathbb{C}$ in the definition of $V(I)$ and maybe elsewhere as well; so then $E$ is the set of ideals having $(X_1,\ldots,X_q)$ as radical); or alternatively, if you really mean $\mathbb{R}$ then you should explain $\operatorname{mult}(I)$ better, and start by stating what you already know about ideals such that $V(I)$ has no real point. $\endgroup$
    – Gro-Tsen
    Commented Mar 31, 2023 at 13:29
  • $\begingroup$ Thank you for your answer, indeed it makes sense to replace $\mathbb{R}$ by $\mathbb{C}$, i edited my question. My question is indeed really geometrical in nature. I want to understand "how" $p$ points can be 'infinitely' close to each other. If a generic sequence of polynomial cancels on $p$ distinct points that collapse to a single point, what can I say about the limiting polynomial if it exists ? Well it has to lie in some $E_p$, so now the goal is to understand who is $E_p$. I figure it's been done already but I cannot find any reference. $\endgroup$
    – Pii_jhi
    Commented Mar 31, 2023 at 15:43
  • $\begingroup$ In this case, a keyword which you should try searching for if you do not already know about it is “Hilbert scheme of zero-dimensional subschemes” (there is a book by Lothar Göttsche with this title) or “punctual Hilbert schemes” (there is a book by Anthony Iarrobino with this title). What you call $E_p$ is the “Hilbert scheme of $0$-dimensional subschemes of length $p$ supported at the origin” (or some analogous description). And unfortunately, it tends to be not at all smooth, nor even irreducible. $\endgroup$
    – Gro-Tsen
    Commented Mar 31, 2023 at 22:27
  • $\begingroup$ (In fact, if I remember correctly, of the the surprises about these objects is that, for $p$ large enough, there exist entire components of this Hilbert scheme which are not specializations of $p$ distinct points. I think this result is by Iarrobino, but I don't remember where I learned it.) $\endgroup$
    – Gro-Tsen
    Commented Mar 31, 2023 at 22:30

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