Lemma 3

Let A be a noetherian ring, p ⊂ p_1 ⊂ ... ⊂ p_{r−1} ⊂ q a saturated chain of length r of prime ideals of A, M a finitely generated Λ-module and n an integer. If the A-module Ext^n_{A_p}(κ(p), M_p) is nonzero, the same is true of Ext^{n+r}_{A_q}(κ(q), M_q).

It is obviously enough to treat the case r = 1; then, replacing A, M, p and q by A_q, M_q, pA_q and qA_q respectively, one is reduced to treating the case where A is local and q = m_A. Let x be an element of m_A − p. The A_p-module Ext^n_A(A/p, M) ⊗_A A_p is isomorphic to Ext^n_{A_p}(κ(p), M_p) (A, X, p. 111, Prop. 10 b)), and is therefore nonzero by assumption; a fortiori Ext^n_A(A/p, M) is nonzero. The exact sequence

yields an exact sequence of extension modules

where u is the homothety with ratio x. By Nakayama's lemma, this is not surjective, hence the A-module Ext^{n+1}_A(A/(p + xA), M) is nonzero. But the only prime ideal of A containing p + xA is m_A, so the A-module A/(p + xA) is of finite length (VIII, § 3, No. 2, Lemma 2). If Ext^{n+1}_A(κ_A, M) were zero, one would deduce from this, by induction on the length of N, that Ext^{n+1}_A(N, M) = 0 for every A-module N of finite length. This contradiction completes the proof.