In a paper by Fogarty titled "Algebraic Family On An Algebraic Surface," he conjectured that $\bf Hilb^n(\mathbb P^N)$ is always variety-- reduced and irreducible.
Is this still a conjecture; any special cases have been treated?
Thanks.
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1$\begingroup$ Assuming that you mean the Hilbert scheme of length $n$ subschemes of projective space, this is false. See e.g. arxiv.org/abs/0803.0341. Actually, it is easy to disprove for fixed $N\geq 3$ and large enough $n$: There is a unique component $Z$ that is the closure of the locus of reduced subschemes, and it can easily be checked that the dimension of the locus of subschemes supported at a point is greater than the dimension of $Z.$ $\endgroup$– dhyMay 17, 2016 at 0:29
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$\begingroup$ @dhy Thank you so much. Yes. I mean lenght n, 0-dim subschemes. So I see its irreducible, but is it also reduced? even in the case $Hilb^n(S)$ where S is a surface? not necessarily smooth. $\endgroup$– AlgeometryMay 17, 2016 at 1:26
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$\begingroup$ Not precisely the same, but closely related; and see references in answers: mathoverflow.net/questions/20288/… $\endgroup$– Zach TeitlerMay 17, 2016 at 3:40
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1 Answer
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The conjecture is located at the bottom of page 520 in:
Fogarty, John Algebraic families on an algebraic surface. Amer. J. Math 90 1968 511–521.
In this paper it is shown that $\mathrm{Hilb}^n(\mathbb{P}^N)$ is connected (Proposition 2.3), and that the $N=1,2$ cases are non-singular rational varieties (Corollary 2.10); although irreducibility was shown by Hartshorne in 1966.
On the other hand, in:
Iarrobino, A. Reducibility of the families of 0-dimensional schemes on a variety. Invent. Math. 15 (1972), 72–77.
it is shown for $N\geq 3$ and $n>\!\!>0$ that $\mathrm{Hilb}^n(\mathbb{P}^N)$ is reducible.
So the conjecture is resolved negatively, as stated.
answered May 17, 2016 at 16:15
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1$\begingroup$ For online access to Iarrobino's article: gdz.sub.uni-goettingen.de/dms/load/img/?PID=GDZPPN002089513 $\endgroup$ May 17, 2016 at 22:26