The Stacks project

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

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. $\square$

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. $\square$

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

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

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

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 $H^ i = H^ i(M^\bullet ) = H^ i(P^\bullet )$. $\square$

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

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

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

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

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. $\square$

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. $\square$

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. $\square$

Proof. Choose a short exact sequence

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

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

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

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

For $i \geq 2$ we have isomorphisms

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

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

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. $\square$

Proof. Choose a resolution

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

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

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

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

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