Let $I$ be an ideal of a Noetherian ring $A$. Let $M$ be a finite $A$-module. Let $i > 0$. By More on Algebra, Lemma 15.27.3 there exists a $c = c(A, I, M, i)$ such that $\text{Tor}^ A_ i(M, A/I^ n) \to \text{Tor}^ A_ i(M, A/I^{n - c})$ is zero for all $n \geq c$. In this section, we discuss some results which show that one sometimes can choose a constant $c$ which works for all $A$-modules $M$ simultaneously (and for a range of indices $i$). This material is related to uniform Artin-Rees as discussed in [Huneke-uniform] and [AHS].
In Remark 51.21.9 we will apply this to show that various pro-systems related to derived completion are (or are not) strictly pro-isomorphic.
The following lemma can be significantly strengthened.
Lemma 51.21.1. Let $I$ be an ideal of a Noetherian ring $A$. For every $m \geq 0$ and $i > 0$ there exist a $c = c(A, I, m, i) \geq 0$ such that for every $A$-module $M$ annihilated by $I^ m$ the map is zero for all $n \geq c$.
\[ \text{Tor}^ A_ i(M, A/I^ n) \to \text{Tor}^ A_ i(M, A/I^{n - c}) \]
Proof. By induction on $i$. Base case $i = 1$. The short exact sequence $0 \to I^ n \to A \to A/I^ n \to 0$ determines an injection $\text{Tor}_1^ A(M, A/I^ n) \subset I^ n \otimes _ A M$, see Algebra, Remark 10.75.9. As $M$ is annihilated by $I^ m$ we see that the map $I^ n \otimes _ A M \to I^{n - m} \otimes _ A M$ is zero for $n \geq m$. Hence the result holds with $c = m$.
Induction step. Let $i > 1$ and assume $c$ works for $i - 1$. By More on Algebra, Lemma 15.27.3 applied to $M = A/I^ m$ we can choose $c' \geq 0$ such that $\text{Tor}_ i(A/I^ m, A/I^ n) \to \text{Tor}_ i(A/I^ m, A/I^{n - c'})$ is zero for $n \geq c'$. Let $M$ be annihilated by $I^ m$. Choose a short exact sequence
The corresponding long exact sequence of tors gives an exact sequence
for all integers $n \geq 0$. If $n \geq c + c'$, then the map $\text{Tor}_{i - 1}^ A(S, A/I^ n) \to \text{Tor}_{i - 1}^ A(S, A/I^{n - c})$ is zero and the map $\text{Tor}_ i^ A(A/I^ m, A/I^{n - c}) \to \text{Tor}_ i^ A(A/I^ m, A/I^{n - c - c'})$ is zero. Combined with the short exact sequences this implies the result holds for $i$ with constant $c + c'$. $\square$
Lemma 51.21.2. Let $I = (a_1, \ldots , a_ t)$ be an ideal of a Noetherian ring $A$. Set $a = a_1$ and denote $B = A[\frac{I}{a}]$ the affine blowup algebra. There exists a $c > 0$ such that $\text{Tor}_ i^ A(B, M)$ is annihilated by $I^ c$ for all $A$-modules $M$ and $i \geq t$.
Proof. Recall that $B$ is the quotient of $A[x_2, \ldots , x_ t]/(a_1x_2 - a_2, \ldots , a_1x_ t - a_ t)$ by its $a_1$-torsion, see Algebra, Lemma 10.70.6. Let
viewed as a chain complex sitting in degrees $(t - 1), \ldots , 0$. The complex $B_\bullet [1/a_1]$ is isomorphic to the Koszul complex on $x_2 - a_2/a_1, \ldots , x_ t - a_ t/a_1$ which is a regular sequence in $A[1/a_1][x_2, \ldots , x_ t]$. Since regular sequences are Koszul regular, we conclude that the augmentation
is a quasi-isomorphism after inverting $a_1$. Since the homology modules of the cone $C_\bullet $ on $\epsilon $ are finite $A[x_2, \ldots , x_ n]$-modules and since $C_\bullet $ is bounded, we conclude that there exists a $c \geq 0$ such that $a_1^ c$ annihilates all of these. By Derived Categories, Lemma 13.12.5 this implies that, after possibly replacing $c$ by a larger integer, that $a_1^ c$ is zero on $C_\bullet $ in $D(A)$. The proof is finished once the reader contemplates the distinguished triangle
Namely, the first term is represented by $B_\bullet \otimes _ A M$ which is sitting in homological degrees $(t - 1), \ldots , 0$ in view of the fact that the terms in the Koszul complex $B_\bullet $ are free (and hence flat) $A$-modules. Whence $\text{Tor}_ i^ A(B, M) = H_ i(C_\bullet \otimes _ A^\mathbf {L} M)$ for $i > t - 1$ and this is annihilated by $a_1^ c$. Since $a_1^ cB = I^ cB$ and since the tor module is a module over $B$ we conclude. $\square$
For the rest of the discussion in this section we fix a Noetherian ring $A$ and an ideal $I \subset A$. We denote
the blowing up of $\mathop{\mathrm{Spec}}(A)$ in the ideal $I$. In other words, $X$ is the $\text{Proj}$ of the Rees algebra $\bigoplus _{n \geq 0} I^ n$. By Cohomology of Schemes, Lemmas 30.14.2 and 30.14.3 we can choose an integer $q(A, I) \geq 0$ such that for all $q \geq q(A, I)$ we have $H^ i(X, \mathcal{O}_ X(q)) = 0$ for $i > 0$ and $H^0(X, \mathcal{O}_ X(q)) = I^ q$.
Lemma 51.21.3. In the situation above, for $q \geq q(A, I)$ and any $A$-module $M$ we have in $D(A)$.
\[ R\Gamma (X, Lp^*\widetilde{M}(q)) \cong M \otimes _ A^\mathbf {L} I^ q \]
Proof. Choose a free resolution $F_\bullet \to M$. Then $\widetilde{F}_\bullet $ is a flat resolution of $\widetilde{M}$. Hence $Lp^*\widetilde{M}$ is given by the complex $p^*\widetilde{F}_\bullet $. Thus $Lp^*\widetilde{M}(q)$ is given by the complex $p^*\widetilde{F}_\bullet (q)$. Since $p^*\widetilde{F}_ i(q)$ are right acyclic for $\Gamma (X, -)$ by our choice of $q \geq q(A, I)$ and since we have $\Gamma (X, p^*\widetilde{F}_ i(q)) = I^ qF_ i$ by our choice of $q \geq q(A, I)$, we get that $R\Gamma (X, Lp^*\widetilde{M}(q))$ is given by the complex with terms $I^ qF_ i$ by Derived Categories of Schemes, Lemma 36.4.3. The result follows as the complex $I^ qF_\bullet $ computes $M \otimes _ A^\mathbf {L} I^ q$ by definition. $\square$
Lemma 51.21.4. In the situation above, let $t$ be an upper bound on the number of generators for $I$. There exists an integer $c = c(A, I) \geq 0$ such that for any $A$-module $M$ the cohomology sheaves $H^ j(Lp^*\widetilde{M})$ are annihilated by $I^ c$ for $j \leq -t$.
Proof. Say $I = (a_1, \ldots , a_ t)$. The question is affine local on $X$. For $1 \leq i \leq t$ let $B_ i = A[\frac{I}{a_ i}]$ be the affine blowup algebra. Then $X$ has an affine open covering by the spectra of the rings $B_ i$, see Divisors, Lemma 31.32.2. By the description of derived pullback given in Derived Categories of Schemes, Lemma 36.3.8 we conclude it suffices to prove that for each $i$ there exists a $c \geq 0$ such that
is annihilated by $I^ c$ for $j \geq t$. This is Lemma 51.21.2. $\square$
Lemma 51.21.5. In the situation above, let $t$ be an upper bound on the number of generators for $I$. There exists an integer $c = c(A, I) \geq 0$ such that for any $A$-module $M$ the tor modules $\text{Tor}_ i^ A(M, A/I^ q)$ are annihilated by $I^ c$ for $i > t$ and all $q \geq 0$.
Proof. Let $q(A, I)$ be as above. For $q \geq q(A, I)$ we have
by Lemma 51.21.3. We have a bounded and convergent spectral sequence
by Derived Categories of Schemes, Lemma 36.4.4. Let $d$ be an integer as in Cohomology of Schemes, Lemma 30.4.4 (actually we can take $d = t$, see Cohomology of Schemes, Lemma 30.4.2). Then we see that $H^{-i}(X, Lp^*\widetilde{M}(q)) = \text{Tor}_ i^ A(M, I^ q)$ has a finite filtration with at most $d$ steps whose graded are subquotients of the modules
If $i \geq t$ then all of these modules are annihilated by $I^ c$ where $c = c(A, I)$ is as in Lemma 51.21.4 because the cohomology sheaves $H^{- i - a}(Lp^*\widetilde{M})$ are all annihilated by $I^ c$ by the lemma. Hence we see that $\text{Tor}_ i^ A(M, I^ q)$ is annihilated by $I^{dc}$ for $q \geq q(A, I)$ and $i \geq t$. Using the short exact sequence $0 \to I^ q \to A \to A/I^ q \to 0$ we find that $\text{Tor}_ i(M, A/I^ q)$ is annihilated by $I^{dc}$ for $q \geq q(A, I)$ and $i > t$. We conclude that $I^ m$ with $m = \max (dc, q(A, I) - 1)$ annihilates $\text{Tor}_ i^ A(M, A/I^ q)$ for all $q \geq 0$ and $i > t$ as desired. $\square$
Lemma 51.21.6. Let $I$ be an ideal of a Noetherian ring $A$. Let $t \geq 0$ be an upper bound on the number of generators of $I$. There exist $N, c \geq 0$ such that the maps are zero for any $A$-module $M$ and all $n \geq N$.
\[ \text{Tor}_{t + 1}^ A(M, A/I^ n) \to \text{Tor}_{t + 1}^ A(M, A/I^{n - c}) \]
Proof. Let $c_1$ be the constant found in Lemma 51.21.5. Please keep in mind that this constant $c_1$ works for $\text{Tor}_ i$ for all $i > t$ simultaneously.
Say $I = (a_1, \ldots , a_ t)$. For an $A$-module $M$ we set
This is an element of $\{ 0, 1, \ldots , t\} $. We will prove by descending induction on $0 \leq s \leq t$ the following statement $H_ s$: there exist $N, c \geq 0$ such that for every module $M$ with $\ell (M) \geq s$ the maps
are zero for $i = 0, \ldots , s$ for all $n \geq N$.
Base case: $s = t$. If $\ell (M) = t$, then $M$ is annihilated by $(a_1^{c_1}, \ldots , a_ t^{c_1}\} $ and hence by $I^{t(c_1 - 1) + 1}$. We conclude from Lemma 51.21.1 that $H_ t$ holds by taking $c = N$ to be the maximum of the integers $c(A, I, t(c_1 - 1) + 1, t + 1), \ldots , c(A, I, t(c_1 - 1) + 1, 2t + 1)$ found in the lemma.
Induction step. Say $0 \leq s < t$ we have $N, c$ as in $H_{s + 1}$. Consider a module $M$ with $\ell (M) = s$. Then we can choose an $i$ such that $a_ i^{c_1}$ is nonzero on $M$. It follows that $\ell (M[a_ i^ c]) \geq s + 1$ and $\ell (M/a_ i^{c_1}M) \geq s + 1$ and the induction hypothesis applies to them. Consider the exact sequence
Denote $E \subset M$ the image of the middle arrow. Consider the corresponding diagram of Tor modules
with exact rows and columns (for every $q$). The south-east arrow is zero by our choice of $c_1$. We conclude that the module $\text{Tor}_ i(M, A/I^ q)$ is sandwiched between a quotient module of $\text{Tor}_ i(M[a_ i^{c_1}], A/I^ q)$ and a submodule of $\text{Tor}_{i + 1}(M/a_ i^{c_1}M, A/I^ q)$. Hence we conclude $H_ s$ holds with $N$ replaced by $N + c$ and $c$ replaced by $2c$. Some details omitted. $\square$
Proposition 51.21.7. Let $I$ be an ideal of a Noetherian ring $A$. Let $t \geq 0$ be an upper bound on the number of generators of $I$. There exist $N, c \geq 0$ such that for $n \geq N$ the maps satisfy the equivalent conditions of Lemma 51.20.2 with $e = t$.
\[ A/I^ n \to A/I^{n - c} \]
Proof. Immediate consequence of Lemmas 51.21.6 and 51.20.2. $\square$
Remark 51.21.8. The paper [AHS] shows, besides many other things, that if $A$ is local, then Proposition 51.21.7 also holds with $e = t$ replaced by $e = \dim (A)$. Looking at Lemma 51.20.3 it is natural to ask whether Proposition 51.21.7 holds with $e = t$ replaced with $e = \text{cd}(A, I)$. We don't know.
Remark 51.21.9. Let $I$ be an ideal of a Noetherian ring $A$. Say $I = (f_1, \ldots , f_ r)$. Denote $K_ n^\bullet $ the Koszul complex on $f_1^ n, \ldots , f_ r^ n$ as in More on Algebra, Situation 15.91.15 and denote $K_ n \in D(A)$ the corresponding object. Let $M^\bullet $ be a bounded complex of finite $A$-modules and denote $M \in D(A)$ the corresponding object. Consider the following inverse systems in $D(A)$:
$M^\bullet /I^ nM^\bullet $, i.e., the complex whose terms are $M^ i/I^ nM^ i$,
$M \otimes _ A^\mathbf {L} A/I^ n$,
$M \otimes _ A^\mathbf {L} K_ n$, and
$M \otimes _ P^\mathbf {L} P/J^ n$ (see below).
All of these inverse systems are isomorphic as pro-objects: the isomorphism between (2) and (3) follows from More on Algebra, Lemma 15.94.1. The isomorphism between (1) and (2) is given in More on Algebra, Lemma 15.100.3. For the last one, see below.
However, we can ask if these isomorphisms of pro-systems are “strict”; this terminology and question is related to the discussion in [pages 61, 62, quillenhomology]. Namely, given a category $\mathcal{C}$ we can define a “strict pro-category” whose objects are inverse systems $(X_ n)$ and whose morphisms $(X_ n) \to (Y_ n)$ are given by tuples $(c, \varphi _ n)$ consisting of a $c \geq 0$ and morphisms $\varphi _ n : X_ n \to Y_{n - c}$ for all $n \geq c$ satisfying an obvious compatibility condition and up to a certain equivalence (given essentially by increasing $c$). Then we ask whether the above inverse systems are isomorphic in this strict pro-category.
This clearly cannot be the case for (1) and (3) even when $M = A[0]$. Namely, the system $H^0(K_ n) = A/(f_1^ n, \ldots , f_ r^ n)$ is not strictly pro-isomorphic in the category of modules to the system $A/I^ n$ in general. For example, if we take $A = \mathbf{Z}[x_1, \ldots , x_ r]$ and $f_ i = x_ i$, then $H^0(K_ n)$ is not annihilated by $I^{r(n - 1)}$.1
It turns out that the results above show that the natural map from (2) to (1) discussed in More on Algebra, Lemma 15.100.3 is a strict pro-isomorphism. We will sketch the proof. Using standard arguments involving stupid truncations, we first reduce to the case where $M^\bullet $ is given by a single finite $A$-module $M$ placed in degree $0$. Pick $N, c \geq 0$ as in Proposition 51.21.7. The proposition implies that for $n \geq N$ we get factorizations
of the transition maps in the system (2). On the other hand, by More on Algebra, Lemma 15.27.3, we can find another constant $c' = c'(M) \geq 0$ such that the maps $\text{Tor}_ i^ A(M, A/I^{n'}) \to \text{Tor}_ i(M, A/I^{n' - c'})$ are zero for $i = 1, 2, \ldots , t$ and $n' \geq c'$. Then it follows from Derived Categories, Lemma 13.12.5 that the map
factors through $M \otimes _ A^\mathbf {L}A/I^{n + tc'} \to M/I^{n + tc'}M$. Combined with the previous result we obtain a factorization
which gives us what we want. If we ever need this result, we will carefully state it and provide a detailed proof.
For number (4) suppose we have a Noetherian ring $P$, a ring homomorphism $P \to A$, and an ideal $J \subset P$ such that $I = JA$. By More on Algebra, Section 15.60 we get a functor $M \otimes _ P^\mathbf {L} - : D(P) \to D(A)$ and we get an inverse system $M \otimes _ P^\mathbf {L} P/J^ n$ in $D(A)$ as in (4). If $P$ is Noetherian, then the system in (4) is pro-isomorphic to the system in (1) because we can compare with Koszul complexes. If $P \to A$ is finite, then the system (4) is strictly pro-isomorphic to the system (2) because the inverse system $A \otimes _ P^\mathbf {L} P/J^ n$ is strictly pro-isomorphic to the inverse system $A/I^ n$ (by the discussion above) and because we have
by More on Algebra, Lemma 15.60.1.
A standard example in (4) is to take $P = \mathbf{Z}[x_1, \ldots , x_ r]$, the map $P \to A$ sending $x_ i$ to $f_ i$, and $J = (x_1, \ldots , x_ r)$. In this case one shows that
and we reduce to one of the cases discussed above (although this case is strictly easier as $A[x_1, \ldots , x_ r]/(x_1, \ldots , x_ r)^ n$ has tor dimension at most $r$ for all $n$ and hence the step using Proposition 51.21.7 can be avoided). This case is discussed in the proof of [Proposition 3.5.1, BS].
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