## 15.100 Systems of modules, bis

Let $I$ be an ideal of a Noetherian ring $A$. In Section 15.99 we considered what happens when considering systems of the form $M/I^ nM$ for finite $A$-modules $M$. In this section we consider the systems $I^ nM$ instead.

Lemma 15.100.1. Let $I$ be an ideal of a Noetherian ring $A$. Let $ K \xrightarrow {\alpha } L \xrightarrow {\beta } M $ be a complex of finite $A$-modules. Set $H = \mathop{\mathrm{Ker}}(\beta )/\mathop{\mathrm{Im}}(\alpha )$. For $n \geq 0$ let

\[ I^ nK \xrightarrow {\alpha _ n} I^ nL \xrightarrow {\beta _ n} I^ nM \]

be the induced complex. Set $H_ n = \mathop{\mathrm{Ker}}(\beta _ n)/\mathop{\mathrm{Im}}(\alpha _ n)$. Then there are canonical $A$-module maps

\[ \ldots \to H_3 \to H_2 \to H_1 \to H \]

There exists a $c > 0$ such that for $n \geq c$ the image of $H_ n \to H$ is contained in $I^{n - c}H$ and there is a canonical $A$-module map $I^ nH \to H_{n - c}$ such that the compositions

\[ I^ n H \to H_{n - c} \to I^{n - 2c}H \quad \text{and}\quad H_ n \to I^{n - c}H \to H_{n - 2c} \]

are the canonical ones. In particular, the inverse systems $(H_ n)$ and $(I^ nH)$ are isomorphic as pro-objects of $\text{Mod}_ A$.

**Proof.**
We have $H_ n = \mathop{\mathrm{Ker}}(\beta ) \cap I^ nL/\alpha (I^ nK)$. Since $\mathop{\mathrm{Ker}}(\beta ) \cap I^ nL \subset \mathop{\mathrm{Ker}}(\beta ) \cap I^{n - 1}L$ and $\alpha (I^ nK) \subset \alpha (I^{n - 1}K)$ we get the maps $H_ n \to H_{n - 1}$. Similarly for the map $H_1 \to H$.

By Artin-Rees we may choose $c_1, c_2 \geq 0$ such that $\mathop{\mathrm{Im}}(\alpha ) \cap I^ nL \subset \alpha (I^{n - c_1}K)$ for $n \geq c_1$ and $\mathop{\mathrm{Ker}}(\beta ) \cap I^ nL \subset I^{n - c_2}\mathop{\mathrm{Ker}}(\beta )$ for $n \geq c_2$, see Algebra, Lemmas 10.51.3 and 10.51.2. Set $c = c_1 + c_2$.

It follows immediately from our choice of $c \geq c_2$ that for $n \geq c$ the image of $H_ n \to H$ is contained in $I^{n - c}H$.

Let $n \geq c$. We define $\psi _ n : I^ nH \to H_{n - c}$ as follows. Say $x \in I^ nH$. Choose $y \in I^ n\mathop{\mathrm{Ker}}(\beta )$ representing $x$. We set $\psi _ n(x)$ equal to the class of $y$ in $H_{n - c}$. To see this is well defined, suppose we have a second choice $y'$ as above for $x$. Then $y' - y \in \mathop{\mathrm{Im}}(\alpha )$. By our choice of $c \geq c_1$ we conclude that $y' - y \in \alpha (I^{n - c}K)$ which implies that $y$ and $y'$ represent the same element of $H_{n - c}$. Thus $\psi _ n$ is well defined.

The statements on the compositions $I^ n H \to H_{n - c} \to I^{n - 2c}H$ and $H_ n \to I^{n - c}H \to H_{n - 2c}$ follow immediately from our definitions.
$\square$

Lemma 15.100.2. 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 $c + c'$).
$\square$

Lemma 15.100.3. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $M$, $N$ be $A$-modules with $M$ finite and $N$ annihilated by a power of $I$. For each $p > 0$ there exists an $n$ such that the map $\mathop{\mathrm{Ext}}\nolimits _ A^ p(M, N) \to \mathop{\mathrm{Ext}}\nolimits _ A^ p(I^ nM, N)$ is zero.

**Proof.**
Immediate consequence of Lemma 15.100.2 and the fact that $I^ mN = 0$ for some $m > 0$.
$\square$

Lemma 15.100.4. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $K \in D(A)$ be pseudo-coherent and let $M$ be a finite $A$-module. For each $p \in \mathbf{Z}$ there exists an $c$ such that the image of $\mathop{\mathrm{Ext}}\nolimits _ A^ p(K, I^ nM) \to \mathop{\mathrm{Ext}}\nolimits _ A^ p(K, M)$ is contained in $I^{n - c}\mathop{\mathrm{Ext}}\nolimits _ A^ p(K, M)$ for $n \geq c$.

**Proof.**
Choose a bounded above complex $P^\bullet $ of finite free $A$-modules representing $K$. Then $\mathop{\mathrm{Ext}}\nolimits _ A^ p(K, M)$ is the cohomology of

\[ \mathop{\mathrm{Hom}}\nolimits _ A(F^{-p + 1}, M) \xrightarrow {a} \mathop{\mathrm{Hom}}\nolimits _ A(F^{-p}, M) \xrightarrow {b} \mathop{\mathrm{Hom}}\nolimits _ A(F^{-p - 1}, M) \]

and $\mathop{\mathrm{Ext}}\nolimits _ A^ p(K, I^ nM)$ is computed by replacing these finite $A$-modules by $I^ n$ times themselves. Thus the result by Lemma 15.100.1 (and much more is true).
$\square$

In Situation 15.90.15 we define complexes $I_ n^\bullet $ such that we have distinguished triangles

\[ I_ n^\bullet \to A \to K_ n^\bullet \to I_ n^\bullet [1] \]

in the triangulated category $K(A)$ of complexes of $A$-modules up to homotopy. Namely, we set $I_ n^\bullet = \sigma _{\leq -1}K_ n^\bullet [-1]$. We have termwise split short exact sequences of complexes

\[ 0 \to A \to K_ n^\bullet \to I_ n^\bullet [1] \to 0 \]

defining distinguished triangles by definition of the triangulated structure on $K(A)$. Their rotations determine the desired distinguished triangles above. Note that $I_ n^0 = A^{\oplus r} \to A$ is given by multiplication by $f_ i^ n$ on the $i$th factor. Hence $I_ n^\bullet \to A$ factors as

\[ I_ n^\bullet \to (f_1^ n, \ldots , f_ r^ n) \to A \]

In fact, there is a short exact sequence

\[ 0 \to H^{-1}(K_ n^\bullet ) \to H^0(I_ n^\bullet ) \to (f_1^ n, \ldots , f_ r^ n) \to 0 \]

and for every $i < 0$ we have $H^ i(I_ n^\bullet ) = H^{i - 1}(K_ n^\bullet $. The maps $K_{n + 1}^\bullet \to K_ n^\bullet $ induce maps $I_{n + 1}^\bullet \to I_ n^\bullet $ and we obtain a commutative diagram

\[ \xymatrix{ \ldots \ar[r] & I_3^\bullet \ar[d] \ar[r] & I_2^\bullet \ar[d] \ar[r] & I_1^\bullet \ar[d] \\ \ldots \ar[r] & (f_1^3, \ldots , f_ r^3) \ar[r] & (f_1^2, \ldots , f_ r^2) \ar[r] & (f_1, \ldots , f_ r) } \]

in $K(A)$.

Lemma 15.100.5. In Situation 15.90.15 assume $A$ is Noetherian. With notation as above, the inverse system $(I^ n)$ is pro-isomorphic in $D(A)$ to the inverse system $(I_ n^\bullet )$.

**Proof.**
It is elementary to show that the inverse system $I^ n$ is pro-isomorphic to the inverse system $(f_1^ n, \ldots , f_ r^ n)$ in the category of $A$-modules. Consider the inverse system of distinguished triangles

\[ I_ n^\bullet \to (f_1^ n, \ldots , f_ r^ n) \to C_ n^\bullet \to I_ n^\bullet [1] \]

where $C_ n^\bullet $ is the cone of the first arrow. By Derived Categories, Lemma 13.41.4 it suffices to show that the inverse system $C_ n^\bullet $ is pro-zero. The complex $I_ n^\bullet $ has nonzero terms only in degrees $i$ with $-r + 1 \leq i \leq 0$ hence $C_ n^\bullet $ is bounded similarly. Thus by Derived Categories, Lemma 13.41.3 it suffices to show that $H^ p(C_ n^\bullet )$ is pro-zero. By the discussion above we have $H^ p(C_ n^\bullet ) = H^ p(K_ n^\bullet )$ for $p \leq -1$ and $H^ p(C_ n^\bullet ) = 0$ for $p \geq 0$. The fact that the inverse systems $H^ p(K_ n^\bullet )$ are pro-zero was shown in the proof of Lemma 15.93.1 (and this is where the assumption that $A$ is Noetherian is used).
$\square$

Lemma 15.100.6. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $M^\bullet $ be a bounded complex of finite $A$-modules. The inverse system of maps

\[ I^ n \otimes _ A^\mathbf {L} M^\bullet \longrightarrow I^ nM^\bullet \]

defines an isomorphism of pro-objects of $D(A)$.

**Proof.**
Choose generators $f_1, \ldots , f_ r \in I$ of $I$. The inverse system $I^ n$ is pro-isomorphic to the inverse system $(f_1^ n, \ldots , f_ r^ n)$ in the category of $A$-modules. With notation as in Lemma 15.100.5 we find that it suffices to prove the inverse system of maps

\[ I_ n^\bullet \otimes _ A^\mathbf {L} M^\bullet \longrightarrow (f_1^ n, \ldots , f_ r^ n)M^\bullet \]

defines an isomorphism of pro-objects of $D(A)$. Say we have $a \leq b$ such that $M^ i = 0$ if $i \not\in [a, b]$. Then source and target of the arrows above have cohomology only in degrees $[-r + a, b]$. Thus it suffices to show that for any $p \in \mathbf{Z}$ the inverse system of maps

\[ H^ p(I_ n^\bullet \otimes _ A^\mathbf {L} M^\bullet ) \longrightarrow H^ p((f_1^ n, \ldots , f_ r^ n)M^\bullet ) \]

defines an isomorphism of pro-objects of $A$-modules, see Derived Categories, Lemma 13.41.5. Using the pro-isomorphism between $I_ n^\bullet \otimes _ A^\mathbf {L} M^\bullet $ and $I^ n \otimes _ A^\mathbf {L} M^\bullet $ and the pro-isomorphism between $(f_1^ n, \ldots , f_ r^ n)M^\bullet $ and $I^ nM^\bullet $ this is equivalent to showing that the inverse system of maps

\[ H^ p(I^ n \otimes _ A^\mathbf {L} M^\bullet ) \longrightarrow H^ p(I^ nM^\bullet ) \]

defines an isomorphism of pro-objects of $A$-modules Choose a bounded above complex of finite free $A$-modules $P^\bullet $ and a quasi-isomorphism $P^\bullet \to M^\bullet $. Then it suffices to show that the inverse system of maps

\[ H^ p(I^ nP^\bullet ) \longrightarrow H^ p(I^ nM^\bullet ) \]

is a pro-isomorphism. This follows from Lemma 15.100.1 as $H^ p(P^\bullet ) = H^ p(M^\bullet )$.
$\square$

Lemma 15.100.7. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $M$ be a finite $A$-module. There exists an integer $n > 0$ such that $I^ nM \to M$ factors through the map $I \otimes _ A^\mathbf {L} M \to M$ in $D(A)$.

**Proof.**
This follows from Lemma 15.100.6. It can also been seen directly as follows. Consider the distinguished triangle

\[ I \otimes _ A^\mathbf {L} M \to M \to A/I \otimes _ A^\mathbf {L} M \to I \otimes _ A^\mathbf {L} M[1] \]

By the axioms of a triangulated category it suffices to prove that $I^ nM \to A/I \otimes _ A^\mathbf {L} M$ is zero in $D(A)$ for some $n$. Choose generators $f_1, \ldots , f_ r$ of $I$ and let $K = K_\bullet (A, f_1, \ldots , f_ r)$ be the Koszul complex and consider the factorization $A \to K \to A/I$ of the quotient map. Then we see that it suffices to show that $I^ nM \to K \otimes _ A M$ is zero in $D(A)$ for some $n > 0$. Suppose that we have found an $n > 0$ such that $I^ nM \to K \otimes _ A M$ factors through $\tau _{\geq t}(K \otimes _ A M)$ in $D(A)$. Then the obstruction to factoring through $\tau _{\geq t + 1}(K \otimes _ A M)$ is an element in $\mathop{\mathrm{Ext}}\nolimits ^ t(I^ nM, H_ t(K \otimes _ A M))$. The finite $A$-module $H_ t(K \otimes _ A M)$ is annihilated by $I$. Then by Lemma 15.100.3 we can after increasing $n$ assume this obstruction element is zero. Repeating this a finite number of times we find $n$ such that $I^ nM \to K \otimes _ A M$ factors through $0 = \tau _{\geq r + 1}(K \otimes _ A M)$ in $D(A)$ and we win.
$\square$

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