By Karl-Heinz Fieseler and Ludger Kaup

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Since both ideals are homogeneous, we may assume, that g is homogeneous (of degree q) and not divisible by T0 - a factor T0 does not contribute to the zeros on k ∗ × k n . , Tn ) vanishes on Y , so (g1 ) ∈ a for some > 0. But then g = g1 resp. g ∈ I(N (a)). 9. If a = (f ) is a principal ideal, then a = (f ), because of the multiplicativity gf = f · g of the homogeneization. , fr for a. ) of Y is an algebraic variety X together with an isomorphism Y ∼ = U ⊂ X, where U ⊂ X is a dense open subset.

As their common set of zeros in k ∗ × k n , and the same holds for the fˆ, f ∈ a. , Tn ], it follows k ∗ · ({1} × Y ) ⊂ N (a). , Tn ] vanishing on k ∗ · ({1} × Y ). Since both ideals are homogeneous, we may assume, that g is homogeneous (of degree q) and not divisible by T0 - a factor T0 does not contribute to the zeros on k ∗ × k n . , Tn ) vanishes on Y , so (g1 ) ∈ a for some > 0. But then g = g1 resp. g ∈ I(N (a)). 9. If a = (f ) is a principal ideal, then a = (f ), because of the multiplicativity gf = f · g of the homogeneization.

Proof. 2 it suffices to show that Pn is complete. g. affine variety and Y → Pn × Z a closed set. With B := prZ (Y ) we have the commutative diagram Y ↓ → Pn × Z ↓ ? B → Z 61 and want to see, why we are allowed to remove the question mark. For z ∈ Z we consider the sectional variety Yz := {y ∈ Pn ; (y, z) ∈ Y } , such that B = {z ∈ Z; Yz = ∅}. Denote C(Y ) → k n+1 × Z the closure of (π × idZ )−1 (Y ) → (k n+1 \ {0}) × Z, where π × idZ : (k n+1 \ {0}) × Z −→ Pn × Z. Indeed (π × idZ )−1 (Y ) = (C(Yz )∗ × {z}) z∈B and C(Yz ) × {z} C(Y ) = ∪ ({0} × B).