
## 15.88 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 15.88.1. 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 $H = \mathop{\mathrm{Ker}}(\alpha )/\mathop{\mathrm{Im}}(\beta )$. 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 $H_ n = \mathop{\mathrm{Ker}}(\alpha _ n)/\mathop{\mathrm{Im}}(\beta _ n)$. Then $H_ n$ is an inverse system of $A$-modules, there are canonical maps $H \to H_ n$ compatible for varying $n$, the system $(H_ n)$ is Mittag-Leffler, there is a $c \geq 0$ such that the image of $H_{n + c} \to H_ n$ is equal to the image of $H \to H_ n$, and $\mathop{\mathrm{lim}}\nolimits H_ n = \mathop{\mathrm{lim}}\nolimits H/I^ n H$. Moreover, there is a $c \geq 0$ and canonical maps

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

such that the composition $H \to H_ n \to H/I^{n - c}H$ is the quotient map and 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 $H_ n \to H_ n/I^{n - c}H_ n$.

Proof. The construction of the canonical maps $H_{n + 1} \to H_ n$ and $H \to H_ n$ is left to the reader. Choose $c_1 \geq 0$ such that $\beta ^{-1}(I^ nM) \subset \mathop{\mathrm{Ker}}(\beta ) + I^{n - c_1}L$ for $n \geq c_1$ (Algebra, Lemma 10.50.3). Then $\mathop{\mathrm{Im}}(H_{n + c_1} \to H_ n) = \mathop{\mathrm{Im}}(H \to H_ n)$ and the Mittag-Leffler property holds. It follows that $\mathop{\mathrm{lim}}\nolimits H/I^ nH \to \mathop{\mathrm{lim}}\nolimits H_ n$ is surjective (small detail omitted).

Choose $c_2 \geq 0$ such that $\mathop{\mathrm{Ker}}(\beta ) \cap I^ nL \subset I^{n - c_2}L$ for $n \geq c_2$ (Algebra, Lemma 10.50.2). Take $n \geq c_1 + c_2 = c$. Then we can define $H_ n \to H/I^{n - c}H$ for $n \geq c$ as follows. Given $x \in H_ n$ choose $y \in \beta ^{-1}(I^ nM)$ representing $x$. Write $y = z + z'$ with $z \in \mathop{\mathrm{Ker}}(\beta )$ and $z' \in I^{n - c_1}L$ (possible by choice of $c_1$). Then we map $x$ to the class of $z$ in $H/I^{n - c}H$. This is well defined by choice of $c_2$ and the fact that $c = c_1 + c_2$. The statement on the composition $H \to H_ n \to H/I^{n - c}H$ is immediate. The composition $H_ n \to H/I^{n - c}H \to H_ n/I^{n - c}H_ n$ may not be the identity as $y$ and $z$ may not have the same class in $H_ n/I^{n - c}H_ n$. However, if $x \in I^{c_1}H_ n$, then writing $x = \sum f_ i x_ i$ with $x_ i \in H_ n$ and $f_ i \in I^{c_1}$ we see that we can choose $y = \sum f_ iy_ i$, $z = \sum f_ i z_ i$ and $z' = \sum f_ i z'_ i$ where $y_ i, z_ i, z'_ i$ are suitable choices for $x_ i$. Then $z' \in I^ nL$ and we conclude $y$ and $z$ do determine the same class in $H_ n$ which proves the final statement.

It follows that $\mathop{\mathrm{lim}}\nolimits H/I^ n H \to \mathop{\mathrm{lim}}\nolimits H_ n$ is injective (as well as surjective, see above) because this map has a left inverse using the maps we just constructed and the obvious equality $\mathop{\mathrm{lim}}\nolimits H/I^{n - c}H = \mathop{\mathrm{lim}}\nolimits H/I^ nH$. This finishes the proof. $\square$

Lemma 15.88.2. 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 15.88.1. The final statement follows then from Lemma 15.85.6. $\square$

Lemma 15.88.3. 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 $\mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n)$. From Lemma 15.88.1 we deduce the Mittag-Leffler property for $(\mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n))$, we deduce part (2), and we find that $\mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ A(M_ n, N_ n) =\mathop{\mathrm{Hom}}\nolimits _ A(M, N)^\wedge$. 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 10.96.4.

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 10.19.1) hence $\alpha \circ \beta$ is an isomorphism by Algebra, Lemma 10.15.4. $\square$

Lemma 15.88.4. 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 $N_ n = N/I^ nN$. If $M_ n \cong N_ n$ for all $n$, then $M^\wedge \cong N^\wedge$ as $A^\wedge$-modules.

Proof. By Lemma 15.88.3 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 15.88.5. 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 and 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 15.79) 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 15.88.6. A morphism $(c, \varphi _ n)$ of the category of Remark 15.88.5 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 15.88.7. 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 15.88.5.

Proof. Choose a short exact sequence

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

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 15.88.1 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 15.88.6.

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 15.88.3 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 15.88.8. The awkwardness in the statement of Lemma 15.88.7 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 15.88.9. 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 15.88.7.

Lemma 15.88.10. 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 conclude that the base change of the complex

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

which computes the modules $\mathop{\mathrm{Ext}}\nolimits ^ i_ B(M, N)$ by the map $B \to B_ n$ is the complex

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

which computes the modules $\mathop{\mathrm{Ext}}\nolimits ^ i_{B_ n}(M_ n, N_ n)$. Thus the result by Lemma 15.88.1. $\square$

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