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By Dan Laksov

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Hence χF ,P is independent of P . Conversely, assume that A is integral and that χF ,P is independent of the prime ideal P of A. Denote by s the point corresponding to the prime ideal P . Let K be the fraction field of A. 1) respectively K ⊗AP H 0 (XSpec AP , FSpec AP (m)) → H 0 (XSpec K , FSpec K (m)). 2) have the same dimension over κ(P ) respectively over K. 9) that H 0 (XSpec AP , FSpec AP (m)) is a free AP – module. 4) that we, for each m, have an isomorphism AP ⊗A H 0 (X, F (m)) → H 0 (XSpec AP , FSpec AP (m)).

Hence L = 0, and F is isomorphic to Ad . 10) Theorem. Assume that Spec A is connected. (1) If F is flat over Spec A then the polynomial χF ,s is independent of s ∈ Spec A. (2) If A is integral and χF ,s is independent of s ∈ Spec A, then F is flat over Spec A. → → → Proof. Assume that F is flat over Spec A. 2) that H i (X, F (m)) = 0 for i > 0 and for big m. 7) that f∗ F (m) is coherent for all m. 16)(1) that f∗ F (m) is locally free. Since Spec A is connected we have that f∗ F (m) has constant rank r(m) on Spec A.

We choose a basis e0 , . . , er of E. Denote by R = SymA (E) the symmetric algebra of E over A and write P(E) = Proj(R). Let X be a closed subscheme of P(E) with inclusion ι: X → P(E), and F a coherent OX –module. 2) Definition. Denote by Q[t] the polynomial ring in the variable t over the rational numbers. For each positive integer d we define a polynomial dt in Q[t] by t t(t − 1)(t − 2) · · · (t − d + 1) = td /d! + cd−1 td−1 + · · · + c0 = d d! and we let t 0 = 1. 3) Note. For each positive integer e we define an operator ∆e on all functions f : Z → Z by ∆e f (m) = f (m + e) − f (m).

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