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Let $X\subset\mathbb{P}^n$ be a (globally) complete intersection, let $(X_t)_{t\in\mathbb{C}^1}$ be a flat family, with $X_1=X$. Which types of schemes can we get as $X_0$?

Or, conversely, which (embedded, projective) schemes deform to complete intersections?

e.g. some non-ACM schemes (not arithmetically Cohen-Macaulay) deform to C.I.'s. Does every pure-dimensional subscheme of $\mathbb{P}^n$ deform to a C.I.? Two immediate obstructions are the degree and the arithmetic genus. Any other necessary conditions?

(Just for housekeeping, a somewhat related question on deformations of C.I.:)

upd: I meant the family over the germ $(\mathbb{C}^1,0)$, so there are no complications with the geometry of the base.

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1 Answer 1

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Anything with the same Hilbert polynomial as a globally complete intersection. This is true because two fibers of the same flat projective family have the same Hilbert polynomial, and because the Hilbert scheme is connected. EDIT: Since the base of the flat family is required to be irreducible, the projective scheme must be in the same irreducible component of the Hilbert scheme as a complete intersection. I don't know any nice characterization of when this is the case.

The Hilbert polynomial of a complete intersection is easy to calculate using the Koszul complex. The coefficients will be some set of polynomials in the degrees of the hypersurfaces you completely intersect. You can set them equal to the actual coefficients of the Hilbert polynomial and check if the system of equations has a positive integer solution.

Computationally, you can just check all possible factorizations of the degree.

For curves, degree, arithmetic genus, and dimension of the embedding are the only invariants needed. The last one is important. For instance, a genus $5$, degree $8$ curve is the complete intersection of three degree $2$ hypersurfaces in $\mathbb P^4$, but not the complete intersection of any pair of hypersurfaces in $\mathbb P^3$.

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Thanks for the response! The connectedness of Hilbert scheme is of course very relevant. But is the locus corresponding to complete intersections dense in the Hilbert scheme of subschemes with the fixed Hilb.polynomial? –  Dmitry Kerner May 26 '12 at 20:45
I don't know if it's dense. –  Will Sawin May 27 '12 at 0:36
I'm not sure this is correct, because qui-vadis assumed an irreducible base. So if the relevant Hilbert scheme has irreducible components that don't contain (but are merely connected to) complete intersections, the points that are in those components (and not in others) won't be $X_1$s. (I.e. density is not the issue; rather, that the locus has to meet each irreducible component.) –  Allen Knutson Jun 9 '12 at 17:48
Technically he asked for a family over $\mathbb A^1$. So this creates a very complicated issue, say if the Hilbert scheme were an elliptic curve then nothing would be a deformation of anything else by this definition. I missed that part of the question when I wrote my answer. –  Will Sawin Jun 9 '12 at 18:14
Oh no, I meant just the germ of an irreducible one dim. family, i.e. (C1,0). So, no global complications –  Dmitry Kerner Jun 20 '12 at 12:27

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