The *punctual Hilbert scheme* in dimension $d$ parameterizes ideals $I$ of codimension $n$ in $k[x\_1,\dots, x\_d]$ which are contained in some power of the ideal $(x\_1,\dots, x\_d)$. In other words, it is the Hilbert scheme of $n$ points supported at the origin in $\mathbb A^d$.


Can anybody give me a reference for this object? In particular, I'd like to know if it is irreducible for arbitrary $d$.




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**A related question:** the *curvilinear Hilbert scheme* parameterizes ideals $I$ of codimension $n$ in $k[x\_1,\dots, x\_d]$ such that $I\not\subset(x\_1,\dots, x\_d)^{n-1}$. This object is irreducible; in fact, it's covered by dense open subschemes isomorphic to $\mathbb A^m$ for some $m$. Does anybody know of a reference that calculates what these spaces are? For $n=2$, it's $\mathbb P^{d-1}$. For $n=3$, I think it's the total space of $\mathcal O\_{\mathbb P^{d-1}}(1)$.



Reference request: is the punctual Hilbert scheme irreducible?
It is certainly not irreducible if *n=8* and *d>3*. This is analyzed nicely in the paper 


Hilbert schemes of 8 points, Dustin A. Cartwright, Daniel Erman, Mauricio Velasco, Bianca Viray available at <http://arxiv.org/abs/0803.0341> (and I think published in ANT).


From there you can look at the references, especially I think the first paper studying in detail this problem was 


Anthony Iarrobino. Reducibility of the family of 0-dimensional schemes on a variety. Inventiones Math., 15:72–77, 1972.


Note that, at least the arXiv reference above, deals with subschemes of $\mathbb{A}^d$ not necessarily supported at the origin. On the other hand, since for $n \leq 8$ the above is the only non-irreducible example, the "new" component must be supported at the origin. Indeed in the case *n=8* and *d=4*, the extra component is a product of a Grassmannian and an affine space.


EDIT: Let me expand on my initial answer. First a general remark: if a point on a scheme is non-singular, then it lies on a unique irreducible component of that scheme. Now, in the usual (not the punctual) Hilbert scheme of 8 points in A^4 there are two components, one being the closure of the component consisting of 8 distinct points, and one supported entirely at a single point (the point being allowed to vary; by translating the support, we can assume the point is the origin). In the arXiv paper they show that there are points supported at a single point that are non-singular and **smoothable** in the full Hilbert scheme (Section 4.4). It follows that such points cannot lie in the component whose points consist entirely of subschemes supported at the origin. We conclude that the punctual Hilbert scheme contains the "non-smoothable" component (necessarily supported at the origin, this one being a minimal example of a reducible Hilbert scheme), as well as more points, corresponding to the points described above (the ones appearing in Section 4.4). In particular, the punctual Hilbert scheme is not irreducible. Hope this clears up the doubts (or that it exposes my mistake!).


