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Algebraic Geometry

By Dan Laksov

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It follows from the assumptions of the Lemma that the A /IA = A ⊗A A/Imodule M /IM = A ⊗A M/IM is flat. Consequently we have that the A /IA module MQ /IMQ is flat. ) that it suffices to prove that KQ = 0. ) that the map A ⊗A K → K is surjective. Since the composite map K → K → KQ is zero by assumption it follows that the map K → KQ is zero. Since B is noetherian, and M is a finitely generated B module by assumption we have that K is a finitely generated B -module. Hence, since the map K → KQ is zero, we have that KQ = 0.

Pt,s ) for P . We can choose the pi such that the pi,j have the same degree dj for i = 1, . . , s. Let m be the maximum of d1 , . . , dt . s Given an element l ∈ I n M ∩ N . We can write l = i=1 fi (a1 , . . , ar )mi , with (f1 , . . , fs ) ∈ Jn . consequently we get t (f1 , . . , fs ) = gj (x1 , . . , xr )(pj,1 , . . , pj,s ) j=1 with gj ∈ A[x1 , . . , xr ]. On the left hand side we have homogeneious polynomials of degree n. Consequently, we may, after possibly removing terms on the right hand side, assume that deg gj + dj = n for j = 1, .

Consequently the formula holds with equality in both cases. Next assume that B is generated by one element x over A, but that it is not a polynomial ring. We can then write B = A[x]/I, where I is a non zero prime ideal in A[x]. We have that td. A B = 0. Since A ⊆ B we have that I ∩ A = 0. Consequently, if we denote by K the quotient field of A we have that ht I = ht IK[t] = 1. Let Q be the inverse image of Q by the canonical surjection A[x] → B. Then we have that Q = Q /I and κ(Q) = κ(Q ). We obtain, using the case when B is a polynomial ring over A, that ht Q ≤ ht Q − ht I = ht Q − 1 = ht P + 1 + td.

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