Hilbert polynomial

In IG2) Grothendieck introduced the Hilbert scheme of lPr I prefer to associate one Hilbert scheme to each Hilbert polynomial p(t), so that in this volume the Hilbert scheme of JPr in the sense of [G2] is the disjoint union of all the Hilbert schemes of Pr relative to all different polynomials p(t)... 

By taking the pullbacks XxgK we may assume that S is irreducible. Let p(t) be the Hilbert polynomial of the fibres of f. Suppose that X X1 U Xz is reducible and let U c S be a non-empty open set such that... 

In fact from corollary (5.3) it follows that at most as many Hilbert polynomials occur as the number of connected components of the sets... 

Going back to the example given at the beginning and noting that every closed subscheme X c Pr with Hilbert polynomial equal to (t+n) is a linear subspace of dimension n < >, we see that the incidence relation has the above defined universal property with respect to p(t) (t+n) , hence... 

From theorem (1.10) it follows that there is an integer m0 such that for every closed subscheme X c Pr with Hilbert polynomial p(t) the sheaf of ideals is m0-regular it suffices to take... 

Since deg(X)= 1 and dim(X) n, X contains a projective subspace L of dimension n as a closed subscheme the equality of their Hilbert polynomials implies that the inclusion L c X is an isomorphism. 

By the choice of H, tt is a flat family of closed subschemes of Pr with Hilbert polynomial equal to p[Pg.82]

Notation. From the universal property it follows immediately that if XcPr has Hilbert polynomial equal to p(t), there is a unique closed point or Hllbrp(t) which parametrizes X (i.e. whose fibre in the universal family is X ). We will denote this point by 1X1... 

Conversely, with an argument similar to the case of projective subspaces (see the footnote on page 7-3 ) one can show that if a closed subscheme of Pr has Hilbert polynomial h(t) then it is a hypersurface of degree d. 

Every X c Pr is obviously of maximal rank in every degree i 0, moreover it follows from theorem (1.10) that there is a k0, depending only on the Hilbert polynomial p(t) of X, such that X is of maximal rank in every degree k>k0 take k0 such that h1(Pr(tx(IO) 0 for all kik0. 

From the semicontinutty theorem it follows that condition (c) is open in HmrP(t)l in other words the locus of points that parametrize schemes of. maximal rank in degree k is an open ( and possibly empty ) subset Uk of HilbW Since any X with Hilbert polynomial p(t) is possibly not of maximal rank only in degrees 1 i k i k0, it follows that the locus of points of H1lbrpft) which parametrize schemes of maximal rank is open, being equal to u, n... n ... 

Applying theorem (1.10) twice we can find an integer p such that for every closed subscheme X,cPr ( resp. X c Pr) with Hilbert polynomial Pi[Pg.121]

Hence X, c Xz c PrxS are flat over S with Hilbert polynomials p,(t) and P2(t) respectively. Let f. PrxS - S be the projection. We have induced classifying morphisms ... 

From the definition it follows that the closed points of FH ) are in natural 1-1 correspondence with the m-tuples (X1f.of closed subschemes of Pr such that X, has Hilbert polynomial p,(t), and... 

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