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
\[ \xymatrix{ 0 \ar[r] & L_ n \ar[r] \ar[d] & A^{\oplus s_ n} \ar[r] \ar[d] & I^ nM \ar[r] \ar[d] & 0 \\ 0 \ar[r] & K \ar[r] & A^{\oplus t} \ar[r] & M \ar[r] & 0 } \]
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
\[ \xymatrix{ \mathop{\mathrm{Hom}}\nolimits _ A(A^{\oplus s_ n}, N) \ar[r] & \mathop{\mathrm{Hom}}\nolimits _ A(L_ n, N) \ar[r] & \mathop{\mathrm{Ext}}\nolimits _ A^1(I^ nM, N) \ar[r] & 0 \\ \mathop{\mathrm{Hom}}\nolimits _ A((I^ n)^{\oplus t}, I^{n - c}N) \ar[r] \ar[u] & \mathop{\mathrm{Hom}}\nolimits _ A(K \cap (I^ n)^{\oplus t}, I^{n - c}N) \ar[r] \ar[u] & \mathop{\mathrm{Ext}}\nolimits _ A^1(I^ nM, I^{n - c}N) \ar[u] \\ \mathop{\mathrm{Hom}}\nolimits _ A(A^{\oplus t}, N) \ar[r] \ar[u] & \mathop{\mathrm{Hom}}\nolimits _ A(K, N) \ar[r] \ar[u] & \mathop{\mathrm{Ext}}\nolimits _ A^1(M, N) \ar@{..>}[u] \ar[r] & 0 } \]
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
\[ \xymatrix{ \mathop{\mathrm{Ext}}\nolimits ^{p - 1}_ A(I^{n - c}K, N) \ar[r] & \mathop{\mathrm{Ext}}\nolimits ^{p - 1}_ A(L_ n, N) \ar[r] & \mathop{\mathrm{Ext}}\nolimits _ A^ p(I^ nM, N) \\ \mathop{\mathrm{Ext}}\nolimits ^{p - 1}_ A(K, N) \ar[rr] \ar[u] & & \mathop{\mathrm{Ext}}\nolimits _ A^ p(M, N) \ar[u] } \]
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 $c + c'$). $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)