## 15.100 Systems of modules

Let $I$ be an ideal of a Noetherian ring $A$. In this section we add to our knowledge of the relationship between finite modules over $A$ and systems of finite $A/I^ n$-modules.

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

$K/I^ nK \xrightarrow {\alpha _ n} L/I^ nL \xrightarrow {\beta _ n} M/I^ nM$

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

$\xymatrix{ & & & H \ar[lld] \ar[ld] \ar[d] \\ \ldots \ar[r] & H_3 \ar[r] & H_2 \ar[r] & H_1 }$

Moreover, there exists a $c > 0$ and canonical $A$-module maps $H_ n \to H/I^{n - c}H$ for $n \geq c$ such that the compositions

$H/I^ n H \to H_ n \to H/I^{n - c}H \quad \text{and}\quad H_ n \to H/I^{n - c}H \to H_{n - c}$

are the canonical ones. Moreover, we have

1. $(H_ n)$ and $(H/I^ nH)$ are isomorphic as pro-objects of $\text{Mod}_ A$,

2. $\mathop{\mathrm{lim}}\nolimits H_ n = \mathop{\mathrm{lim}}\nolimits H/I^ n H$,

3. the inverse system $(H_ n)$ is Mittag-Leffler,

4. the image of $H_{n + c} \to H_ n$ is equal to the image of $H \to H_ n$,

5. the composition $I^ cH_ n \to H_ n \to H/I^{n - c}H \to H_ n/I^{n - c}H_ n$ is the inclusion $I^ cH_ n \to H_ n$ followed by the quotient map $H_ n \to H_ n/I^{n - c}H_ n$, and

6. the kernel and cokernel of $H/I^ nH \to H_ n$ is annihilated by $I^ c$.

Proof. Observe that $H_ n = \beta ^{-1}(I^ nM)/\mathop{\mathrm{Im}}(\alpha ) + I^ nL$. For $n \geq 2$ we have $\beta ^{-1}(I^ nM) \subset \beta ^{-1}(I^{n - 1}M)$ and $\mathop{\mathrm{Im}}(\alpha ) + I^ nL \subset \mathop{\mathrm{Im}}(\alpha ) + I^{n - 1}L$. Thus we obtain our canonical map $H_ n \to H_{n - 1}$. Similarly, we have $\mathop{\mathrm{Ker}}(\beta ) \subset \beta ^{-1}(I^ nM)$ and $\mathop{\mathrm{Im}}(\alpha ) \subset \mathop{\mathrm{Im}}(\alpha ) + I^ nL$ which produces the canonical map $H \to H_ n$. We omit the verification that the diagram commutes.

By Artin-Rees we may choose $c_1, c_2 \geq 0$ such that $\beta ^{-1}(I^ nM) \subset \mathop{\mathrm{Ker}}(\beta ) + I^{n - c_1}L$ 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 $c = c_1 + c_2$.

Let $n \geq c$. We define $\psi _ n : H_ n \to H/I^{n - c}H$ as follows. Say $x \in H_ n$. Choose $y \in \beta ^{-1}(I^ nM)$ representing $x$. Write $y = z + w$ with $z \in \mathop{\mathrm{Ker}}(\beta )$ and $w \in I^{n - c_1}L$ (this is possible by our choice of $c_1$). We set $\psi _ n(x)$ equal to the class of $z$ in $H/I^{n - c}H$. To see this is well defined, suppose we have a second set of choices $y', z', w'$ as above for $x$ with obvious notation. Then $y' - y \in \mathop{\mathrm{Im}}(\alpha ) + I^ nL$, say $y' - y = \alpha (v) + u$ with $v \in K$ and $u \in I^ nL$. Thus

$y' = z' + w' = \alpha (v) + u + z + w \Rightarrow z' = z + \alpha (v) + u + w - w'$

Since $\beta (z' - z - \alpha (v)) = 0$ we find that $u + w - w' \in \mathop{\mathrm{Ker}}(\beta ) \cap I^{n - c_1}L$ which is contained in $I^{n - c_1 - c_2}\mathop{\mathrm{Ker}}(\beta ) = I^{n - c}\mathop{\mathrm{Ker}}(\beta )$ by our choice of $c_2$. Thus $z'$ and $z$ have the same image in $H/I^{n - c}H$ as desired.

The composition $H/I^ n H \to H_ n \to H/I^{n - c}H$ is the canonical map because if $z \in \mathop{\mathrm{Ker}}(\beta )$ represents an element $x$ in $H/I^ nH = \mathop{\mathrm{Ker}}(\beta )/\mathop{\mathrm{Im}}(\alpha ) + I^ n\mathop{\mathrm{Ker}}(\beta )$ then it is clear from the above that $x$ maps to the class of $z$ in $H/I^{n - c}H$ under the maps constructed above.

Let us consider the composition $H_ n \to H/I^{n - c}H \to H_{n - c}$. Given $x, y, z, w$ as in the construction of $\psi _ n$ above, we see that $x$ is mapped to the cass of $z$ in $H_{n - c}$. On the other hand, the canonical map $H_ n \to H_{n - c}$ from the first paragraph of the proof sends $x$ to the class of $y$. Thus we have to show that $y - z \in \mathop{\mathrm{Im}}(\alpha ) + I^{n - c}L$ which is the case because $y - z = w \in I^{n - c_1}L \subset I^{n - c}L$.

Statements (1) – (4) are formal consequences of what we just proved. Namely, (1) follows from the existence of the maps and the definition of morphisms of pro-objects in Categories, Remark 4.22.5. Part (2) holds because isomorphic pro-objects have isomorphic limits. Part (3) is immediate from part (4). Part (4) follows from the factorization $H_{n + c} \to H/I^ nH \to H_ n$ of the canonical map $H_{n + c} \to H_ n$.

Proof of part (5). Let $x \in I^ cH_ n$. Write $x = \sum f_ i x_ i$ with $x_ i \in H_ n$ and $f_ i \in I^ c$. Choose $y_ i, z_ i, w_ i$ as in the construction of $\psi _ n$ for $x_ i$. Then for the computation of $\psi _ n$ of $x$ we may choose $y = \sum f_ iy_ i$, $z = \sum f_ i z_ i$ and $w = \sum f_ i w_ i$ and we see that $\psi _ n(x)$ is given by the class of $z$. The image of this in $H_ n/I^{n - c}H_ n$ is equal to the class of $y$ as $w = \sum f_ i w_ i$ is in $I^ nL$. This proves (5).

Proof of part (6). Let $y \in \mathop{\mathrm{Ker}}(\beta )$ whose class is $x$ in $H$. If $x$ maps to zero in $H_ n$, then $y \in I^ nL + \mathop{\mathrm{Im}}(\alpha )$. Hence $y - \alpha (v) \in \mathop{\mathrm{Ker}}(\beta ) \cap I^ nL$ for some $v \in K$. Then $y - \alpha (v) \in I^{n - c_2}\mathop{\mathrm{Ker}}(\beta )$ and hence the class of $y$ in $H/I^ nH$ is annihilated by $I^{c_2}$. Finally, let $x \in H_ n$ be the class of $y \in \beta ^{-1}(I^ nM)$. Then we write $y = z + w$ with $z \in \mathop{\mathrm{Ker}}(\beta )$ and $w \in I^{n - c_1}L$ as above. Clearly, if $f \in I^{c_1}$ then $fx$ is the class of $fy + fw \equiv fy$ modulo $\mathop{\mathrm{Im}}(\alpha ) + I^ nL$ and hence $fx$ is the image of the class of $fy$ in $H$ as desired. $\square$

Lemma 15.100.2. Let $I$ be an ideal of a Noetherian ring $A$. Let $K \in D(A)$ be pseudo-coherent. Set $K_ n = K \otimes _ A^\mathbf {L} A/I^ n$. Then for all $i \in \mathbf{Z}$ the system $H^ i(K_ n)$ satisfies Mittag-Leffler and $\mathop{\mathrm{lim}}\nolimits H^ i(K)/I^ nH^ i(K)$ is equal to $\mathop{\mathrm{lim}}\nolimits H^ i(K_ n)$.

Proof. We may represent $K$ by a bounded above complex $P^\bullet$ of finite free $A$-modules. Then $K_ n$ is represented by $P^\bullet /I^ nP^\bullet$. Hence the Mittag-Leffler property by Lemma 15.100.1. The final statement follows then from Lemma 15.97.6. $\square$

Lemma 15.100.3. 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

$M^\bullet \otimes _ A^\mathbf {L} A/I^ n \longrightarrow M^\bullet /I^ nM^\bullet$

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

Proof. Say $I = (f_1, \ldots , f_ r)$. Let $K_ n \in D(A)$ be the object represented by the Koszul complex on $f_1^ n, \ldots , f_ r^ n$. Recall that we have maps $K_ n \to A/I^ n$ which induce a pro-isomorphism of inverse systems, see Lemma 15.94.1. Hence it suffices to show that

$M^\bullet \otimes _ A^\mathbf {L} K_ n \longrightarrow M^\bullet /I^ nM^\bullet$

defines an isomorphism of pro-objects of $D(A)$. Since $K_ n$ is represented by a complex of finite free $A$-modules sitting in degrees $-r, \ldots , 0$ there exist $a, b \in \mathbf{Z}$ such that the source and target of the displayed arrow have vanishing cohomology in degrees outside $[a, b]$ for all $n$. Thus we may apply Derived Categories, Lemma 13.42.5 and we find that it suffices to show that the maps

$H^ i(M^\bullet \otimes _ A^\mathbf {L} A/I^ n) \to H^ i(M^\bullet /I^ nM^\bullet )$

define isomorphisms of pro-systems of $A$-modules for any $i \in \mathbf{Z}$. To see this choose a quasi-isomorphism $P^\bullet \to M^\bullet$ where $P^\bullet$ is a bounded above complex of finite free $A$-modules. The arrows above are given by the maps

$H^ i(P^\bullet /I^ nP^\bullet ) \to H^ i(M^\bullet /I^ nM^\bullet )$

These define an isomorphism of pro-systems by Lemma 15.100.1. Namely, the lemma shows both are isomorphic to the pro-system $H^ i/I^ nH^ i$ with $H^ i = H^ i(M^\bullet ) = H^ i(P^\bullet )$. $\square$

Lemma 15.100.4. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $M$, $N$ be finite $A$-modules. Set $M_ n = M/I^ nM$ and $N_ n = N/I^ nN$. Then

1. the systems $(\mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n))$ and $(\text{Isom}_ A(M_ n, N_ n))$ are Mittag-Leffler,

2. there exists a $c \geq 0$ such that the kernels and cokernels of

$\mathop{\mathrm{Hom}}\nolimits _ A(M, N)/I^ n\mathop{\mathrm{Hom}}\nolimits _ A(M, N) \to \mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n)$

are killed by $I^ c$ for all $n$,

3. we have $\mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n) =\mathop{\mathrm{Hom}}\nolimits _ A(M, N)^\wedge = \mathop{\mathrm{Hom}}\nolimits _{A^\wedge }(M^\wedge , N^\wedge )$

4. $\mathop{\mathrm{lim}}\nolimits \text{Isom}_ A(M_ n, N_ n) = \text{Isom}_{A^\wedge }(M^\wedge , N^\wedge )$.

Here ${}^\wedge$ denotes usual $I$-adic completion.

Proof. Note that $\mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n) = \mathop{\mathrm{Hom}}\nolimits _ A(M, N_ n)$. Choose a presentation

$A^{\oplus t} \to A^{\oplus s} \to M \to 0$

Applying the right exact functor $\mathop{\mathrm{Hom}}\nolimits _ A(-, N)$ we obtain a complex

$0 \xrightarrow {\alpha } N^{\oplus s} \xrightarrow {\beta } N^{\oplus t}$

whose cohomology in the middle is $\mathop{\mathrm{Hom}}\nolimits _ A(M, N)$ and such that for $n \geq 0$ the cohomology of

$0 \xrightarrow {\alpha _ n} N_ n^{\oplus s} \xrightarrow {\beta _ n} N_ n^{\oplus t}$

is $\mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n)$. Let $c \geq 0$ be as in Lemma 15.100.1 for this $A$, $I$, $\alpha$, and $\beta$. By part (3) of the lemma we deduce the Mittag-Leffler property for $(\mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n))$. The kernel and cokernel of the maps $\mathop{\mathrm{Hom}}\nolimits _ A(M, N)/I^ n\mathop{\mathrm{Hom}}\nolimits _ A(M, N) \to \mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n)$ are killed by $I^ c$ by [art part (6) of the lemma. We find that $\mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n) = \mathop{\mathrm{Hom}}\nolimits _ A(M, N)^\wedge$ by part (2) of the lemma. The equality

$\mathop{\mathrm{Hom}}\nolimits _{A^\wedge }(M^\wedge , N^\wedge ) = \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n)$

follows formally from the fact that $M^\wedge = \mathop{\mathrm{lim}}\nolimits M_ n$ and $M_ n = M^\wedge /I^ nM^\wedge$ and the corresponding facts for $N$, see Algebra, Lemma 10.97.4.

The result for isomorphisms follows from the case of homomorphisms applied to both $(\mathop{\mathrm{Hom}}\nolimits (M_ n, N_ n))$ and $(\mathop{\mathrm{Hom}}\nolimits (N_ n, M_ n))$ and the following fact: for $n > m > 0$, if we have maps $\alpha : M_ n \to N_ n$ and $\beta : N_ n \to M_ n$ which induce an isomorphisms $M_ m \to N_ m$ and $N_ m \to M_ m$, then $\alpha$ and $\beta$ are isomorphisms. Namely, then $\alpha \circ \beta$ is surjective by Nakayama's lemma (Algebra, Lemma 10.20.1) hence $\alpha \circ \beta$ is an isomorphism by Algebra, Lemma 10.16.4. $\square$

Lemma 15.100.5. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $M$, $N$ be finite $A$-modules. Set $M_ n = M/I^ nM$ and $N_ n = N/I^ nN$. If $M_ n \cong N_ n$ for all $n$, then $M^\wedge \cong N^\wedge$ as $A^\wedge$-modules.

Proof. By Lemma 15.100.4 the system $(\text{Isom}_ A(M_ n, N_ n))$ is Mittag-Leffler. By assumption each of the sets $\text{Isom}_ A(M_ n, N_ n)$ is nonempty. Hence $\mathop{\mathrm{lim}}\nolimits \text{Isom}_ A(M_ n, N_ n)$ is nonempty. Since $\mathop{\mathrm{lim}}\nolimits \text{Isom}_ A(M_ n, N_ n) = \text{Isom}_{A^\wedge }(M^\wedge , N^\wedge )$ we obtain an isomorphism. $\square$

Remark 15.100.6. Let $I$ be an ideal of a Noetherian ring $A$. Set $A_ n = A/I^ n$ for $n \geq 1$. Consider the following category:

1. An object is a sequence $\{ E_ n\} _{n \geq 1}$ where $E_ n$ is a finite $A_ n$-module.

2. A morphism $\{ E_ n\} \to \{ E'_ n\}$ is given by maps

$\varphi _ n : I^ cE_ n \longrightarrow E'_ n/E'_ n[I^ c] \quad \text{for }n \geq c$

where $E'_ n[I^ c]$ is the torsion submodule (Section 15.88) up to equivalence: we say $(c, \varphi _ n)$ is the same as $(c + 1, \overline{\varphi }_ n)$ where $\overline{\varphi }_ n : I^{c + 1}E_ n \longrightarrow E'_ n/E'_ n[I^{c + 1}]$ is the induced map.

Composition of $(c, \varphi _ n) : \{ E_ n\} \to \{ E'_ n\}$ and $(c', \varphi '_ n) : \{ E'_ n\} \to \{ E''_ n\}$ is defined by the obvious compositions

$I^{c + c'}E_ n \to I^{c'}E'_ n/E'_ n[I^{c}] \to E''_ n/E''_ n[I^{c + c'}]$

for $n \geq c + c'$. We omit the verification that this is a category.

Lemma 15.100.7. A morphism $(c, \varphi _ n)$ of the category of Remark 15.100.6 is an isomorphism if and only if there exists a $c' \geq 0$ such that $\mathop{\mathrm{Ker}}(\varphi _ n)$ and $\mathop{\mathrm{Coker}}(\varphi _ n)$ are $I^{c'}$-torsion for all $n \gg 0$.

Proof. We may and do assume $c' \geq c$ and that the $\mathop{\mathrm{Ker}}(\varphi _ n)$ and $\mathop{\mathrm{Coker}}(\varphi _ n)$ are $I^{c'}$-torsion for all $n$. For $n \geq c'$ and $x \in I^{c'}E'_ n$ we can choose $y \in I^ cE_ n$ with $x = \varphi _ n(y) \bmod E'_ n[I^ c]$ as $\mathop{\mathrm{Coker}}(\varphi _ n)$ is annihilated by $I^{c'}$. Set $\psi _ n(x)$ equal to the class of $y$ in $E_ n/E_ n[I^{c'}]$. For a different choice $y' \in I^ cE_ n$ with $x = \varphi _ n(y') \bmod E'_ n[I^ c]$ the difference $y - y'$ maps to zero in $E'_ n/E'_ n[I^ c]$ and hence is annihilated by $I^{c'}$ in $I^ cE_ n$. Thus the maps $\psi _ n : I^{c'}E'_ n \to E_ n/E_ n[I^{c'}]$ are well defined. We omit the verification that $(c', \psi _ n)$ is the inverse of $(c, \varphi _ n)$ in the category. $\square$

Lemma 15.100.8. Let $I$ be an ideal of the Noetherian ring $A$. Let $M$ and $N$ be finite $A$-modules. Write $A_ n = A/I^ n$, $M_ n = M/I^ nM$, and $N_ n = N/I^ nN$. For every $i \geq 0$ the objects

$\{ \mathop{\mathrm{Ext}}\nolimits ^ i_ A(M, N)/I^ n\mathop{\mathrm{Ext}}\nolimits ^ i_ A(M, N)\} _{n \geq 1} \quad \text{and}\quad \{ \mathop{\mathrm{Ext}}\nolimits ^ i_{A_ n}(M_ n, N_ n)\} _{n \geq 1}$

are isomorphic in the category $\mathcal{C}$ of Remark 15.100.6.

Proof. Choose a short exact sequence

$0 \to K \to A^{\oplus r} \to M \to 0$

and set $K_ n = K/I^ nK$. For $n \geq 1$ define $K(n) = \mathop{\mathrm{Ker}}(A_ n^{\oplus r} \to M_ n)$ so that we have exact sequences

$0 \to K(n) \to A_ n^{\oplus r} \to M_ n \to 0$

and surjections $K_ n \to K(n)$. In fact, by Lemma 15.100.1 there is a $c \geq 0$ and maps $K(n) \to K_ n/I^{n - c}K_ n$ which are “almost inverse”. Since $I^{n - c}K_ n \subset K_ n[I^ c]$ these maps which witness the fact that the systems $\{ K(n)\} _{n \geq 1}$ and $\{ K_ n\} _{n \geq 1}$ are isomorphic in $\mathcal{C}$.

We claim the systems

$\{ \mathop{\mathrm{Ext}}\nolimits ^ i_{A_ n}(K(n), N_ n)\} _{n \geq 1} \quad \text{and}\quad \{ \mathop{\mathrm{Ext}}\nolimits ^ i_{A_ n}(K_ n, N_ n)\} _{n \geq 1}$

are isomorphic in the category $\mathcal{C}$. Namely, the surjective maps $K_ n \to K(n)$ have kernels annihilated by $I^ c$ and therefore determine maps

$\mathop{\mathrm{Ext}}\nolimits ^ i_{A_ n}(K(n), N_ n) \to \mathop{\mathrm{Ext}}\nolimits ^ i_{A_ n}(K_ n, N_ n)$

whose kernel and cokernel are annihilated by $I^ c$. Hence the claim by Lemma 15.100.7.

For $i \geq 2$ we have isomorphisms

$\mathop{\mathrm{Ext}}\nolimits ^{i - 1}_ A(K, N) = \mathop{\mathrm{Ext}}\nolimits ^ i_ A(M, N) \quad \text{and}\quad \mathop{\mathrm{Ext}}\nolimits ^{i - 1}_{A_ n}(K(n), N_ n) = \mathop{\mathrm{Ext}}\nolimits ^ i_{A_ n}(M_ n, N_ n)$

In this way we see that it suffices to prove the lemma for $i = 0, 1$.

For $i = 0, 1$ we consider the commutative diagram

$\xymatrix{ 0 \ar[r] & \mathop{\mathrm{Hom}}\nolimits (M, N) \ar[r] \ar[dd] & N^{\oplus r} \ar[r]_-\varphi \ar[dd] & \mathop{\mathrm{Hom}}\nolimits (K, N) \ar[r] \ar[d] & \mathop{\mathrm{Ext}}\nolimits ^1(M, N) \ar[r] & 0 \\ & & & \mathop{\mathrm{Hom}}\nolimits (K_ n, N_ n) \\ 0 \ar[r] & \mathop{\mathrm{Hom}}\nolimits (M_ n, N_ n) \ar[r] & N_ n^{\oplus r} \ar[r] & \mathop{\mathrm{Hom}}\nolimits (K(n), N_ n) \ar[r] \ar[u] & \mathop{\mathrm{Ext}}\nolimits ^1(M_ n, N_ n) \ar[r] & 0 }$

By Lemma 15.100.4 we see that the kernel and cokernel of $\mathop{\mathrm{Hom}}\nolimits (M, N)/I^ n \mathop{\mathrm{Hom}}\nolimits (M, N) \to \mathop{\mathrm{Hom}}\nolimits (M_ n, N_ n)$ and $\mathop{\mathrm{Hom}}\nolimits (K, N)/I^ n \mathop{\mathrm{Hom}}\nolimits (K, N) \to \mathop{\mathrm{Hom}}\nolimits (K_ n, N_ n)$ and are $I^ c$-torsion for some $c \geq 0$ independent of $n$. Above we have seen the cokernel of the injective maps $\mathop{\mathrm{Hom}}\nolimits (K(n), N_ n) \to \mathop{\mathrm{Hom}}\nolimits (K_ n, N_ n)$ are annihilated by $I^ c$ after possibly increasing $c$. For such a $c$ we obtain maps $\delta _ n : I^ c\mathop{\mathrm{Hom}}\nolimits (K, N)/I^ n\mathop{\mathrm{Hom}}\nolimits (K, N) \to \mathop{\mathrm{Hom}}\nolimits (K(n), N_ n)$ fitting into the diagram (precise formulation omitted). The kernel and cokernel of $\delta _ n$ are annihilated by $I^ c$ after possibly increasing $c$ since we know that the same thing is true for $\mathop{\mathrm{Hom}}\nolimits (K, N)/I^ n \mathop{\mathrm{Hom}}\nolimits (K, N) \to \mathop{\mathrm{Hom}}\nolimits (K_ n, N_ n)$ and $\mathop{\mathrm{Hom}}\nolimits (K(n), N_ n) \to \mathop{\mathrm{Hom}}\nolimits (K_ n, N_ n)$. Then we can use commutativity of the solid diagram

$\xymatrix{ \varphi ^{-1}(I^ c\mathop{\mathrm{Hom}}\nolimits (K, N)) \ar[r]_-\varphi \ar[d] & I^ c\mathop{\mathrm{Hom}}\nolimits (K, N)/I^ n\mathop{\mathrm{Hom}}\nolimits (K, N) \ar[r] \ar[d]^{\delta _ n} & I^ c\mathop{\mathrm{Ext}}\nolimits ^1(M, N)/I^ n\mathop{\mathrm{Ext}}\nolimits ^1(M, N) \ar[r] \ar@{..>}[d] & 0 \\ N_ n^{\oplus r} \ar[r] & \mathop{\mathrm{Hom}}\nolimits (K(n), N_ n) \ar[r] & \mathop{\mathrm{Ext}}\nolimits ^1(M_ n, N_ n) \ar[r] & 0 }$

to define the dotted arrow. A straightforward diagram chase (omitted) shows that the kernel and cokernel of the dotted arrow are annihilated buy $I^ c$ after possibly increasing $c$ one final time. $\square$

Remark 15.100.9. The awkwardness in the statement of Lemma 15.100.8 is partly due to the fact that there are no obvious maps between the modules $\mathop{\mathrm{Ext}}\nolimits ^ i_{A_ n}(M_ n, N_ n)$ for varying $n$. What we may conclude from the lemma is that there exists a $c \geq 0$ such that for $m \gg n \gg 0$ there are (canonical) maps

$I^ c\mathop{\mathrm{Ext}}\nolimits ^ i_{A_ n}(M_ m, N_ m)/I^ n\mathop{\mathrm{Ext}}\nolimits ^ i_{A_ n}(M_ m, N_ m) \to \mathop{\mathrm{Ext}}\nolimits ^ i_{A_ n}(M_ n, N_ n)/\mathop{\mathrm{Ext}}\nolimits ^ i_{A_ n}(M_ n, N_ n)[I^ c]$

whose kernel and cokernel are annihilated by $I^ c$. This is the (weak) sense in which we get a system of modules.

Example 15.100.10. Let $k$ be a field. Let $A = k[[x, y]]/(xy)$. By abuse of notation we denote $x$ and $y$ the images of $x$ and $y$ in $A$. Let $I = (x)$. Let $M = A/(y)$. There is a free resolution

$\ldots \to A \xrightarrow {y} A \xrightarrow {x} A \xrightarrow {y} A \to M \to 0$

We conclude that

$\mathop{\mathrm{Ext}}\nolimits ^2_ A(M, N) = N[y]/xN$

where $N[y] = \mathop{\mathrm{Ker}}(y : N \to N)$. We denote $A_ n = A/I^ n$, $M_ n = M/I^ nM$, and $N_ n = N/I^ nN$. For each $n$ we have a free resolution

$\ldots \to A_ n^{\oplus 2} \xrightarrow {y, x^{n - 1}} A_ n \xrightarrow {x} A_ n \xrightarrow {y} A_ n \to M_ n \to 0$

We conclude that

$\mathop{\mathrm{Ext}}\nolimits ^2_{A_ n}(M_ n, N_ n) = (N_ n[y] \cap N_ n[x^{n - 1}])/xN_ n$

where $N_ n[y] = \mathop{\mathrm{Ker}}(y : N_ n \to N_ n)$ and $N[x^{n - 1}] = \mathop{\mathrm{Ker}}(x^{n - 1} : N_ n \to N_ n)$. Take $N = A/(y)$. Then we see that

$\mathop{\mathrm{Ext}}\nolimits ^2_ A(M, N) = N[y]/xN = N/xN \cong k$

but

$\mathop{\mathrm{Ext}}\nolimits ^2_{A_ n}(M_ n, N_ n) = (N_ n[y] \cap N_ n[x^{n - 1}])/xN_ n = N_ n[x^{n - 1}]/xN_ n = 0$

for all $r$ because $N_ n = k[x]/(x^ n)$ and the sequence

$N_ n \xrightarrow {x} N_ n \xrightarrow {x^{n - 1}} N_ n$

is exact. Thus ignoring some kind of $I$-power torsion is necessary to get a result as in Lemma 15.100.8.

Lemma 15.100.11. Let $A \to B$ be a flat homomorphism of Noetherian rings. Let $I \subset A$ be an ideal. Let $M, N$ be $A$-modules. Set $B_ n = B/I^ nB$, $M_ n = M/I^ nM$, $N_ n = N/I^ nN$. If $M$ is flat over $A$, then we have

$\mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^ i_ B(M, N)/I^ n \mathop{\mathrm{Ext}}\nolimits ^ i_ B(M, N) = \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^ i_{B_ n}(M_ n, N_ n)$

for all $i \in \mathbf{Z}$.

Proof. Choose a resolution

$\ldots \to P_2 \to P_1 \to P_0 \to M \to 0$

by finite free $B$-modues $P_ i$. Set $P_{i, n} = P_ i/I^ nP_ i$. Since $M$ and $B$ are flat over $A$, the sequence

$\ldots \to P_{2, n} \to P_{1, n} \to P_{0, n} \to M_ n \to 0$

is exact. We see that on the one hand the complex

$\mathop{\mathrm{Hom}}\nolimits _ B(P_0, N) \to \mathop{\mathrm{Hom}}\nolimits _ B(P_1, N) \to \mathop{\mathrm{Hom}}\nolimits _ B(P_2, N) \to \ldots$

computes the modules $\mathop{\mathrm{Ext}}\nolimits ^ i_ B(M, N)$ and on the other hand the complex

$\mathop{\mathrm{Hom}}\nolimits _{B_ n}(P_{0, n}, N_ n) \to \mathop{\mathrm{Hom}}\nolimits _{B_ n}(P_{1, n}, N_ n) \to \mathop{\mathrm{Hom}}\nolimits _{B_ n}(P_{2, n}, N_ n) \to \ldots$

computes the modules $\mathop{\mathrm{Ext}}\nolimits ^ i_{B_ n}(M_ n, N_ n)$. Since

$\mathop{\mathrm{Hom}}\nolimits _{B_ n}(P_{i, n}, N_ n) = \mathop{\mathrm{Hom}}\nolimits _ B(P_ i, N)/I^ n \mathop{\mathrm{Hom}}\nolimits _ B(P_ i, N)$

we obtain the result from Lemma 15.100.1 part (2). $\square$

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