Lemma 15.101.2. Let A be a Noetherian ring. Let I \subset A be an ideal. Let M, N be A-modules with M finite. For each p > 0 there exists a c \geq 0 such that for n \geq c the map \mathop{\mathrm{Ext}}\nolimits _ A^ p(M, N) \to \mathop{\mathrm{Ext}}\nolimits _ A^ p(I^ nM, N) factors through \mathop{\mathrm{Ext}}\nolimits ^ p_ A(I^ nM, I^{n - c}N) \to \mathop{\mathrm{Ext}}\nolimits _ A^ p(I^ nM, N).
Proof. For p = 0, if \varphi : M \to N is an A-linear map, then \varphi (\sum f_ i m_ i) = \sum f_ i \varphi (m_ i) for f_ i \in A and m_ i \in M. Hence \varphi induces a map I^ nM \to I^ nN for all n and the result is true with c = 0.
Choose a short exact sequence 0 \to K \to A^{\oplus t} \to M \to 0. For each n we pick a short exact sequence 0 \to L_ n \to A^{\oplus s_ n} \to I^ nM \to 0. It is clear that we can construct a map of short exact sequences
such that A^{\oplus s_ n} \to A^{\oplus t} has image in (I^ n)^{\oplus t}. By Artin-Rees (Algebra, Lemma 10.51.2) there exists a c \geq 0 such that L_ n \to K factors through I^{n - c}K if n \geq c.
For p = 1 our choices above induce a solid commutative diagram
whose horizontal arrows are exact. The lower middle vertical arrow arises because K \cap (I^ n)^{\oplus t} \subset I^{n - c}K and hence any A-linear map K \to N induces an A-linear map (I^ n)^{\oplus t} \to I^{n - c}N by the argument of the first paragraph. Thus we obtain the dotted arrow as desired.
For p > 1 we obtain a commutative diagram
whose bottom horizontal arrow is an isomorphism. By induction on p the left vertical map factors through \mathop{\mathrm{Ext}}\nolimits ^{p - 1}_ A(I^{n - c}K, I^{n - c - c'}N) for some c' \geq 0 and all n \geq c + c'. Using the composition \mathop{\mathrm{Ext}}\nolimits ^{p - 1}_ A(I^{n - c}K, I^{n - c - c'}N) \to \mathop{\mathrm{Ext}}\nolimits ^{p - 1}_ A(L_ n, I^{n - c - c'}N) \to \mathop{\mathrm{Ext}}\nolimits ^ p_ A(I^ nM, I^{n - c - c'}N) we obtain the desired factorization (for n \geq c + c' and with c replaced by c + c'). \square
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