The Stacks project

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. $\square$

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

Proof. Immediate consequence of Lemma 15.101.2 and the fact that $I^ mN = 0$ for some $m > 0$. $\square$

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

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 15.101.1 (and much more is true). $\square$

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

where $C_ n^\bullet$ is the cone of the first arrow. By Derived Categories, Lemma 13.42.4 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 13.42.3 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 15.94.1 (and this is where the assumption that $A$ is Noetherian is used). $\square$

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 15.101.5 we find that it suffices to prove the inverse system of maps

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

defines an isomorphism of pro-objects of $A$-modules, see Derived Categories, Lemma 13.42.5. 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

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

is a pro-isomorphism. This follows from Lemma 15.101.1 as $H^ p(P^\bullet ) = H^ p(M^\bullet )$. $\square$

Proof. This follows from Lemma 15.101.6. It can also been seen directly as follows. Consider the distinguished triangle

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 15.101.3 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. $\square$