# Algebra and geometry in several complex variables by Palamodov V. By Palamodov V.

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Additional resources for Algebra and geometry in several complex variables

Example text

It is dense in F in the sense that an arbitrary series a is equal to a polynomial up to an element of mk for arbitrary k. e. for any two different elements we have either i j or i ≺ j. Note the property: if i, j, k ∈ Nn are arbitrary, then i j is equivalent to i + k j + k. Now for an arbitrary i ∈ Nn the subspace F(i) ⊂F of series that contains only monomials z j for j i, is an ideal in F. Consider the quotient . Q (i) = I ∩ F (i) + F i+ /F i+ where i+ is the next to i vector. The dimension of Q (i) is equal to 0 or 1.

Fn ) . 2. , fn ) ⊂ Ow . Indeed, for g = afj , a ∈Ow we take εj → 0 and check that the integral tends to zero. It . follows that the residue is defined on the Artin algebra A = Ow /Iw Proposition 6 We have Res Jadz . fn Theorem 7 For an arbitrary non zero element g ∈ A there exists an element h ∈ A such that B (g, h) = 1. fn defined on the Artin algebra A is non degenerated. fn )−1 adz is a differential operator of order < m. The operator Q : A → CN , Q (a) = q (i) (a) , |i| < m is a N¨other operator for Iw .

2) where E means the identity operator in F, (ii) D vanishes in I and G vanishes in FB . 2) in O. Examples 3. Let a = 0, I= (f ) where f ∈ mk and f (z) = z1k + φ1 (z ) z1k−1 + ... , 0)} . Example 4. , fk ) . , 0) . Problem 8. Let I be a complete intersection ideal generated by homogeneous polynomials, such that dim A/I= 0. , q where Γp is the set of elements i ∈ Γ, |i| = p and Γk = 0 for k > q. Problem 9. Is it true that for an arbitrary complete intersection ideal I in On of dimension d the set B is generated by n − d vertices?