By I.G. Macdonald

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Moreover, let N = B n1 + · · · B ns . We have that N is a finitely generated B module such that BN = N . It follows from the induction assumption used to the A-algebra B and the B -module N that we can find an element f ∈ A such that Nf is a free Af -module. It therefore remains to prove that we can find an element f ∈ A such that (N/N )f is a free Af -module. To this end we write Ni = N + uh N + · · · + uih N and Pi = {n ∈ N : ui+1 h n ∈ Ni }. Clearly Ni is a B -submodule of N and Pi a B -submodule of N .

We obtain the formula dim AP /KAP + dim AP /P AP + td. AP /K (BQ ) = dim BQ + dim BQ /QBQ + td. κ(P ) κ(Q ). Since A is catenary we have that AP /K is catenary and from the chain P AP ⊇ P AP /K we get the formula dim AP /P AP + dim AP /KAP = dim AP /K. Consequently we have that dim AP /K + td. AP /K (BQ ) = dim BQ + dim BQ /QBQ + td. κ(P ) κ(Q ). From the first formula we proved we thus obtain that dim BQ = dim BQ + dim BQ /QBQ , which is the formula we wanted to prove. 6) Remark. ) that fields are universally catenary.

Since B is flat over A we obtain an injection A/P → A. It follows that assB (B/P B) ⊆ ass(B). Conversely, let Q ∈ ass B and let P be the contraction of Q to A. We first show that P ∈ ass(A). ), a minimal decomposition 0 = N1 ∩· · ·∩Nr of zero in A, such that A/Ni has only one associated prime Pi . ) that the primes P1 , . . , Pr are the associated primes of A. We have an injective homomorphism A → ⊕ri=1 A/Ni and, since B is flat over A, we get an injective homomorphism B → ⊕ri=1 B/Ni B. Consequently we have that Q ∈ ass B/Ni B for some i.

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