By Palamodov V.

**Read Online or Download Algebra and geometry in several complex variables PDF**

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**Additional info for Algebra and geometry in several complex variables**

**Sample text**

Proof. Fix j < d and consider the function −1 hj (zd , z ) = zj πd−1 (zd , z ) defined on the root set Z (p) = {p (zd , z ) = 0, D (z ) = 0} . 8) and is analytic. , ζm (z ) in Z (p). Take the interpolating pp for Dhj qj (zd , z ) = D (z ) hj (ζk (z )) (zd − ζl (z )) l=k (ζk (z ) − ζl (z )) l=k The sum is a pp and is equal hj (ζk ) as zd = ζk , consequently the pp q coincides with Dzj in the set Z (p) . It is a pp with coefficients that are analytic in Bd \∆. The coefficients are bounded because of the formula D = const Π (ζk − ζl )2 .

0 0 ... 0 a0 a1 ... am . S (p, q) = b0 ... bn−1 bn 0 ... 0 0 b0 ... b b ... 0 n−1 n ... ... ... ... 0 0 ... 0 b0 ... bn−1 bn 3 is called Sylvester resultant of the polynomials: R (p, q) = det S (p, q) . , bn with integer coefficients. Proposition 7 If A is a field, then R (p, q) = 0 if and only if there exist polynomials u, v ∈ A [t] such that uv = 0 and up + vq = 0 deg u < n, deg v < m. Proposition 8 If the field A is algebraically closed, then R (p, q) = 0 if and only if there exists, at least, one common root of p and q in A or at infinity.

2 For a proof we take a primary decomposition of I and apply the above Theorem to each component. This theory is generalized for arbitrary modules of finite type. Theorem 4 Let M be an arbitrary O n -module of finite type, N is a submodule N = N1 ∩ ... ,ps are associated prime ideals. For each j there exists an integer lj and an O n -differential operator qj : M → [O n /p]lj of N¨other type such that ∩ Ker qj = N . See [2] for a proof. Problem 4. Let F : [O n ]s → [On ]t be a morphism of O n -modules.