By Mumford.

ISBN-10: 0387112901

ISBN-13: 9780387112909

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Sample text

Assume that A is universally catenary. Then A/P is universally catenary so we can assume that A is an integral domain and that B contains A. We have that B = A[x1 , . . , xn ]/I where A[x1 , . . , xn ] is a polynomial ring over A and I a prime ideal in A[x1 , . . , xn ]. Let Q be a prime ideal in B. Then we have that Q = R/I, where R is a prime ideal in A[x1 , . . , xn ]. However the ring A[x1 , . . ) that we have ht Q = ht R − ht I. 4) for polynomial rings we have that ht R + td. κ(P ) κ(R) = ht P + td.

To prove the Lemma it suffices to show that the map A /IA ⊗A/I T → T obtained from the left vertical map of the latter diagram is surjective. 12 Reduction to noetherian rings 3 of A/I-modules. Tensorize the last two sequences with A /IA over A/I. We obtain two exact sequences 0 → A /IA ⊗A/I Q → A /IA ⊗A L0 → A /IA ⊗A M → 0 and A /IA ⊗A/I T → A /IA ⊗A L1 → A /IA ⊗A Q → 0 of A /IA -modules, where ethe first is exact since M/IM is a flat A/I-module by assumption. Consequently we obtain a commutative diagram B ⊗A/I T −−−→ B ⊗A L1 −−−→ B ⊗A L0 −−−→ B ⊗A M −−−→ 0         0 −−−→ T −−−→ B ⊗A L1 −−−→ B ⊗A Lo −−−→ B ⊗A M −−−→ 0 of B = A /IA -modules, where the three right vertical maps are isomorphisms.

Assume that F is flat over A. Since BQ is flat over B the functor that sends an AP -module N to BQ ⊗B (N ⊗A F ) is exact. However BQ ⊗B (F ⊗A N ) = FQ ⊗A N = FQ ⊗AP N . Consequently the functor that sends N to FQ ⊗AP N is exact, that is, the AP -module FQ is flat. 6 Flatness 6 with P = ϕ−1 (Q). The functor that sends an A-module N ot the AP -module NP is exact. Consequently the functor that sends N to FQ ⊗AP NP is exact. However, we have that FQ ⊗AP NP = FQ ⊗AP (AP ⊗A N ) = FQ ⊗A N . Hence the functor that sends an A–module N to FQ ⊗A N is exact.