most recent 30 from http://mathoverflow.net 2013-05-22T08:13:40Z http://mathoverflow.net/feeds/question/100371 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100371/irreducible-components-of-the-hilbert-scheme Naga Venkata 2012-06-22T16:04:45Z 2012-06-23T05:34:46Z

<p>Let $P_1, P_2, Q$ denote the Hilbert scheme of a plane conic in $\mathbb{P}^3$, a quartic and a degree $d$ surface in $\mathbb{P}^3$. Then there is a natural inclusion map $i$ from Hilbert flag scheme $\mathrm{Hilb}_{P_1,Q}$ to $\mathrm{Hilb}_{P_2,Q}$ under the map, $(C,X) \mapsto (2C,X)$. Then is the image under the composition of the maps, $i$ with the natural projection map $\mathrm{pr}_2$,</p> <p>$\mathrm{Hilb}_{P_1,Q} \to\mathrm{Hilb}_{P_2,Q}\to \mathrm{Hilb}_{Q}$</p> <p>an irreducible component of the image of $\mathrm{pr}_2$? If so can this result be generalized to the case when we can replace plane conic and quartic by curves $C_1, C_2$ such that $rC_1$ has the same Hilbert polynomial as $C_2$? </p> <p>Note:Hilbert flag scheme $\mathrm{Hilb}_{P_i,Q}$ parametrize pairs of the form $C \subset X$ where $P_i$ is the Hilbert polynomial of $C$ and $X$ is a degree $d$ surface in $\mathbb{P}^3$.</p>

http://mathoverflow.net/questions/100371/irreducible-components-of-the-hilbert-scheme/100428#100428 Dan Petersen 2012-06-23T05:34:46Z 2012-06-23T05:34:46Z

<p>This question has been answered in the comments by Jason Starr. The morphism $i$ is in general not defined and the composition will in general not map to a component of the target. I am reposting this as a CW answer; if it gets upvoted, this question will not reappear on the front page.</p>