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Let $Z$ be a local complete intersection subscheme of dimension $m$ in $\mathbb{P}^{2m+1}$. Let $P$ be the Hilbert polynomial of $Z$. Denote by $\operatorname{Hilb}_P$ the Hilbert scheme of local complete intersection subschemes with Hilbert polynomial $P$. Is it true that for a generic subscheme $Z'$ in $\operatorname{Hilb}_P$, $I(Z')$ can be generated by polynomials that define smooth hypersurfaces in $\mathbb{P}^{2m+1}$ i.e., does there exists polynomials $P_1, ..., P_n$ such that $I(Z')=(P_1,...,P_n)$ and the zero locus of $P_i$ is smooth for all $i$?

A slightly weaker condition would be to ask if a generic hypersurface in $\operatorname{pr}_2 \operatorname{Hilb}_{P,Q}$ is smooth where $\operatorname{Hilb}_{P,Q}$ is the flag Hilbert scheme of pairs $(Z \subset X)$ where $Z$ is a local complete intersection subscheme of dimension $m$ with Hilbert polynomial $P$ contained is a hypersurface $X$ in $\mathbb{P}^{2m+1}$ with Hilbert polynomial $Q$ and $\operatorname{pr}_2$ denotes the natural projection map.

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    $\begingroup$ A necessary condition is that the tangent space at any point $P\in Z$ has dimension $\le 2m$. $\endgroup$
    – rita
    Commented Nov 15, 2012 at 15:10
  • $\begingroup$ @Rita: That is what I was thinking as well, but I cannot think of a local complete intersection with a singular point of embedding dimension $2m+1$ for which I can prove the same is true after a generic deformation (I have been trying to produce "rigid" nonreduced structures on a smooth scheme of dimension $m$, but the computations are beyond me). Since you are an expert on this, are there configurations of subschemes (e.g., configurations of degree 6 Del Pezzo surfaces in $\mathbb{P}^5$) that are LCI of embedding dimension $2m+1$ and also "rigid"? $\endgroup$
    – Jason Starr
    Commented Nov 15, 2012 at 15:21
  • $\begingroup$ @Jason: I had not seen your comment earlier. Unfortunately, I have no idea. $\endgroup$
    – rita
    Commented Dec 14, 2012 at 12:16

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I believe that the answer to the first question is no for the reason that a local complete intersection is pretty far from a complete intersection. Take $Z$ to be $3$ points in $\mathbb P^3$ which aren't on a line. Then the Hilbert scheme of local complete intersections is a dense open subset of the Hilbert scheme of $3$ points in $\mathbb P^3$. However, the complete intersections will have to be the intersection of a cubic hypersurface with two hyperplanes, for degree reasons. Thus, the complete intersections will all lie on a single line in $\mathbb P^3$, which is not true for 3 generic points.

If you assumed that $Z$ is a complete intersection, not just a local complete intersection, then what you want is probably true.

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    $\begingroup$ @Dustin: The OP does not ask whether $Z$ is a complete intersection. He asks whether $Z$ is an intersection of some (presumably huge) number $n$ of smooth hypersurfaces. The integer $n$ need not equal $m+1$, the codimension. The answer to the question is yes for $Z$ any finite collection of reduced points in $\mathbb{P}^3$. $\endgroup$
    – Jason Starr
    Commented Nov 15, 2012 at 15:15
  • $\begingroup$ @Starr: Is there some results when $Z$ is not zero dimensional? $\endgroup$
    – Naga Venkata
    Commented Nov 15, 2012 at 16:41
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    $\begingroup$ @Jason Starr: probably everybody thinks the question is over an algebraically closed fields, but this is not stated explicitely in the OP. Over an imperfect field, a reduced point can not be contained in a smooth hypersurface if its residue field has too high inseparability degree. $\endgroup$
    – Qing Liu
    Commented Nov 15, 2012 at 22:04
  • $\begingroup$ @Qing Liu: The OP explicitly states that he is taking a generic element of the corresponding Hilbert scheme. Thus the zero dimensional scheme will be smooth over the ground field. $\endgroup$
    – Jason Starr
    Commented Nov 16, 2012 at 3:28
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    $\begingroup$ yes we all assumed he meant complete intersection! $\endgroup$
    – roy smith
    Commented Feb 22, 2013 at 4:34

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