# The Stacks Project

## Tag 08WW

#### 34.4.14. Descent for modules and their morphisms

Throughout this subsection, fix a ring map $f: R \to S$. As seen in Section 34.3 we can use the language of cosimplicial algebras to talk about descent data for modules, but in this subsection we prefer a more down to earth terminology.

For $i = 1, 2, 3$, let $S_i$ be the $i$-fold tensor product of $S$ over $R$. Define the ring homomorphisms $\delta_0^1, \delta_1^1: S_1 \to S_2$, $\delta_{01}^1, \delta_{02}^1, \delta_{12}^1: S_1 \to S_3$, and $\delta_0^2, \delta_1^2, \delta_2^2: S_2 \to S_3$ by the formulas \begin{align*} \delta^1_0 (a_0) & = 1 \otimes a_0 \\ \delta^1_1 (a_0) & = a_0 \otimes 1 \\ \delta^2_0 (a_0 \otimes a_1) & = 1 \otimes a_0 \otimes a_1 \\ \delta^2_1 (a_0 \otimes a_1) & = a_0 \otimes 1 \otimes a_1 \\ \delta^2_2 (a_0 \otimes a_1) & = a_0 \otimes a_1 \otimes 1 \\ \delta_{01}^1(a_0) & = 1 \otimes 1 \otimes a_0 \\ \delta_{02}^1(a_0) & = 1 \otimes a_0 \otimes 1 \\ \delta_{12}^1(a_0) & = a_0 \otimes 1 \otimes 1. \end{align*} In other words, the upper index indicates the source ring, while the lower index indicates where to insert factors of 1. (This notation is compatible with the notation introduced in Section 34.3.)

Recall1 from Definition 34.3.1 that for $M \in \text{Mod}_S$, a descent datum on $M$ relative to $f$ is an isomorphism $$\theta : M \otimes_{S,\delta^1_0} S_2 \longrightarrow M \otimes_{S,\delta^1_1} S_2$$ of $S_2$-modules satisfying the cocycle condition $$\tag{34.4.14.1} (\theta \otimes \delta_2^2) \circ (\theta \otimes \delta_2^0) = (\theta \otimes \delta_2^1): M \otimes_{S, \delta^1_{01}} S_3 \to M \otimes_{S,\delta^1_{12}} S_3.$$ Let $DD_{S/R}$ be the category of $S$-modules equipped with descent data relative to $f$.

For example, for $M_0 \in \text{Mod}_R$ and a choice of isomorphism $M \cong M_0 \otimes_R S$ gives rise to a descent datum by identifying $M \otimes_{S,\delta^1_0} S_2$ and $M \otimes_{S,\delta^1_1} S_2$ naturally with $M_0 \otimes_R S_2$. This construction in particular defines a functor $f^*: \text{Mod}_R \to DD_{S/R}$.

Definition 34.4.15. The functor $f^*: \text{Mod}_R \to DD_{S/R}$ is called base extension along $f$. We say that $f$ is a descent morphism for modules if $f^*$ is fully faithful. We say that $f$ is an effective descent morphism for modules if $f^*$ is an equivalence of categories.

Our goal is to show that for $f$ universally injective, we can use $\theta$ to locate $M_0$ within $M$. This process makes crucial use of some equalizer diagrams.

Lemma 34.4.16. For $(M,\theta) \in DD_{S/R}$, the diagram $$\tag{34.4.16.1} \xymatrix@C=8pc{ M \ar[r]^{\theta \circ (1_M \otimes \delta_0^1)} & M \otimes_{S, \delta_1^1} S_2 \ar@<1ex>[r]^{(\theta \otimes \delta_2^2) \circ (1_M \otimes \delta^2_0)} \ar@<-1ex>[r]_{1_{M \otimes S_2} \otimes \delta^2_1} & M \otimes_{S, \delta_{12}^1} S_3 }$$ is a split equalizer.

Proof. Define the ring homomorphisms $\sigma^0_0: S_2 \to S_1$ and $\sigma_0^1, \sigma_1^1: S_3 \to S_2$ by the formulas \begin{align*} \sigma^0_0 (a_0 \otimes a_1) & = a_0a_1 \\ \sigma^1_0 (a_0 \otimes a_1 \otimes a_2) & = a_0a_1 \otimes a_2 \\ \sigma^1_1 (a_0 \otimes a_1 \otimes a_2) & = a_0 \otimes a_1a_2. \end{align*} We then take the auxiliary morphisms to be $1_M \otimes \sigma_0^0: M \otimes_{S, \delta_1^1} S_2 \to M$ and $1_M \otimes \sigma_0^1: M \otimes_{S,\delta_{12}^1} S_3 \to M \otimes_{S, \delta_1^1} S_2$. Of the compatibilities required in (34.4.2.1), the first follows from tensoring the cocycle condition (34.4.14.1) with $\sigma_1^1$ and the others are immediate. $\square$

Lemma 34.4.17. For $(M, \theta) \in DD_{S/R}$, the diagram $$\tag{34.4.17.1} \xymatrix@C=8pc{ C(M \otimes_{S, \delta_{12}^1} S_3) \ar@<1ex>[r]^{C((\theta \otimes \delta_2^2) \circ (1_M \otimes \delta^2_0))} \ar@<-1ex>[r]_{C(1_{M \otimes S_2} \otimes \delta^2_1)} & C(M \otimes_{S, \delta_1^1} S_2 ) \ar[r]^{C(\theta \circ (1_M \otimes \delta_0^1))} & C(M). }$$ obtained by applying $C$ to (34.4.16.1) is a split coequalizer.

Proof. Omitted. $\square$

Lemma 34.4.18. The diagram $$\tag{34.4.18.1} \xymatrix@C=8pc{ S_1 \ar[r]^{\delta^1_1} & S_2 \ar@<1ex>[r]^{\delta^2_2} \ar@<-1ex>[r]_{\delta^2_1} & S_3 }$$ is a split equalizer.

Proof. In Lemma 34.4.16, take $(M, \theta) = f^*(S)$. $\square$

This suggests a definition of a potential quasi-inverse functor for $f^*$.

Definition 34.4.19. Define the functor $f_*$ $: DD_{S/R} \to \text{Mod}_R$ by taking $f_*(M, \theta)$ to be the $R$-submodule of $M$ for which the diagram $$\tag{34.4.19.1} \xymatrix@C=8pc{f_*(M,\theta) \ar[r] & M \ar@<1ex>^{\theta \circ (1_M \otimes \delta_0^1)}[r] \ar@<-1ex>_{1_M \otimes \delta_1^1}[r] & M \otimes_{S, \delta_1^1} S_2 }$$ is an equalizer.

Using Lemma 34.4.16 and the fact that the restriction functor $\text{Mod}_S \to \text{Mod}_R$ is right adjoint to the base extension functor $\bullet \otimes_R S: \text{Mod}_R \to \text{Mod}_S$, we deduce that $f_*$ is right adjoint to $f^*$.

We are ready for the key lemma. In the faithfully flat case this is a triviality (see Remark 34.4.21), but in the general case some argument is needed.

Lemma 34.4.20. If $f$ is universally injective, then the diagram $$\tag{34.4.20.1} \xymatrix@C=8pc{ f_*(M, \theta) \otimes_R S \ar[r]^{\theta \circ (1_M \otimes \delta_0^1)} & M \otimes_{S, \delta_1^1} S_2 \ar@<1ex>[r]^{(\theta \otimes \delta_2^2) \circ (1_M \otimes \delta^2_0)} \ar@<-1ex>[r]_{1_{M \otimes S_2} \otimes \delta^2_1} & M \otimes_{S, \delta_{12}^1} S_3 }$$ obtained by tensoring (34.4.19.1) over $R$ with $S$ is an equalizer.

Proof. By Lemma 34.4.12 and Remark 34.4.13, the map $C(1_N \otimes f): C(N \otimes_R S) \to C(N)$ can be split functorially in $N$. This gives the upper vertical arrows in the commutative diagram $$\xymatrix@C=8pc{ C(M \otimes_{S, \delta_1^1} S_2) \ar@<1ex>^{C(\theta \circ (1_M \otimes \delta_0^1))}[r] \ar@<-1ex>_{C(1_M \otimes \delta_1^1)}[r] \ar[d] & C(M) \ar[r]\ar[d] & C(f_*(M,\theta)) \ar@{-->}[d] \\ C(M \otimes_{S,\delta_{12}^1} S_3) \ar@<1ex>^{C((\theta \otimes \delta_2^2) \circ (1_M \otimes \delta^2_0))}[r] \ar@<-1ex>_{C(1_{M \otimes S_2} \otimes \delta^2_1)}[r] \ar[d] & C(M \otimes_{S, \delta_1^1} S_2 ) \ar[r]^{C(\theta \circ (1_M \otimes \delta_0^1))} \ar[d]^{C(1_M \otimes \delta_1^1)} & C(M) \ar[d] \ar@{=}[dl] \\ C(M \otimes_{S, \delta_1^1} S_2) \ar@<1ex>[r]^{C(\theta \circ (1_M \otimes \delta_0^1))} \ar@<-1ex>[r]_{C(1_M \otimes \delta_1^1)} & C(M) \ar[r] & C(f_*(M,\theta)) }$$ in which the compositions along the columns are identity morphisms. The second row is the coequalizer diagram (34.4.17.1); this produces the dashed arrow. From the top right square, we obtain auxiliary morphisms $C(f_*(M,\theta)) \to C(M)$ and $C(M) \to C(M\otimes_{S,\delta_1^1} S_2)$ which imply that the first row is a split coequalizer diagram. By Remark 34.4.11, we may tensor with $S$ inside $C$ to obtain the split coequalizer diagram $$\xymatrix@C=8pc{ C(M \otimes_{S,\delta_2^2 \circ \delta_1^1} S_3) \ar@<1ex>^{C((\theta \otimes \delta_2^2) \circ (1_M \otimes \delta^2_0))}[r] \ar@<-1ex>_{C(1_{M \otimes S_2} \otimes \delta^2_1)}[r] & C(M \otimes_{S, \delta_1^1} S_2 ) \ar[r]^{C(\theta \circ (1_M \otimes \delta_0^1))} & C(f_*(M,\theta) \otimes_R S). }$$ By Lemma 34.4.10, we conclude (34.4.20.1) must also be an equalizer. $\square$

Remark 34.4.21. If $f$ is a split injection in $\text{Mod}_R$, one can simplify the argument by splitting $f$ directly, without using $C$. Things are even simpler if $f$ is faithfully flat; in this case, the conclusion of Lemma 34.4.20 is immediate because tensoring over $R$ with $S$ preserves all equalizers.

Theorem 34.4.22. The following conditions are equivalent.

1. (a)    The morphism $f$ is a descent morphism for modules.
2. (b)    The morphism $f$ is an effective descent morphism for modules.
3. (c)    The morphism $f$ is universally injective.

Proof. It is clear that (b) implies (a). We now check that (a) implies (c). If $f$ is not universally injective, we can find $M \in \text{Mod}_R$ such that the map $1_M \otimes f: M \to M \otimes_R S$ has nontrivial kernel $N$. The natural projection $M \to M/N$ is not an isomorphism, but its image in $DD_{S/R}$ is an isomorphism. Hence $f^*$ is not fully faithful.

We finally check that (c) implies (b). By Lemma 34.4.20, for $(M, \theta) \in DD_{S/R}$, the natural map $f^* f_*(M,\theta) \to M$ is an isomorphism of $S$-modules. On the other hand, for $M_0 \in \text{Mod}_R$, we may tensor (34.4.18.1) with $M_0$ over $R$ to obtain an equalizer sequence, so $M_0 \to f_* f^* M$ is an isomorphism. Consequently, $f_*$ and $f^*$ are quasi-inverse functors, proving the claim. $\square$

1. To be precise, our $\theta$ here is the inverse of $\varphi$ from Definition 34.3.1.

The code snippet corresponding to this tag is a part of the file descent.tex and is located in lines 1035–1312 (see updates for more information).

\subsection{Descent for modules and their morphisms}
\label{subsection-descent-modules-morphisms}

\noindent
Throughout this subsection, fix a ring map $f: R \to S$. As seen in
Section \ref{section-descent-modules} we can use the language of cosimplicial
algebras to talk about descent data for modules, but in this
subsection we prefer a more down to earth terminology.

\medskip\noindent
For $i = 1, 2, 3$, let $S_i$ be the $i$-fold tensor product of $S$ over $R$.
Define the ring homomorphisms $\delta_0^1, \delta_1^1: S_1 \to S_2$,
$\delta_{01}^1, \delta_{02}^1, \delta_{12}^1: S_1 \to S_3$, and
$\delta_0^2, \delta_1^2, \delta_2^2: S_2 \to S_3$ by the formulas
\begin{align*}
\delta^1_0  (a_0) & =  1 \otimes a_0 \\
\delta^1_1  (a_0) & = a_0 \otimes 1 \\
\delta^2_0  (a_0 \otimes a_1) & =  1 \otimes a_0 \otimes a_1 \\
\delta^2_1  (a_0 \otimes a_1) & =  a_0 \otimes 1 \otimes a_1 \\
\delta^2_2  (a_0 \otimes a_1) & =  a_0 \otimes a_1 \otimes 1 \\
\delta_{01}^1(a_0) & = 1 \otimes 1 \otimes a_0 \\
\delta_{02}^1(a_0) & = 1 \otimes a_0 \otimes 1 \\
\delta_{12}^1(a_0) & = a_0 \otimes 1 \otimes 1.
\end{align*}
In other words, the upper index indicates the source ring, while the lower
index indicates where to insert factors of 1. (This notation is compatible
with the notation introduced in Section \ref{section-descent-modules}.)

\medskip\noindent
Recall\footnote{To be precise, our $\theta$ here is the inverse of
$\varphi$ from Definition \ref{definition-descent-datum-modules}.}
from Definition \ref{definition-descent-datum-modules} that for
$M \in \text{Mod}_S$, a {\it descent datum} on $M$ relative to $f$ is
an isomorphism
$$\theta : M \otimes_{S,\delta^1_0} S_2 \longrightarrow M \otimes_{S,\delta^1_1} S_2$$
of $S_2$-modules satisfying the {\it cocycle condition}

\label{equation-cocycle-condition}
(\theta \otimes \delta_2^2) \circ (\theta \otimes \delta_2^0) = (\theta \otimes
\delta_2^1):
M \otimes_{S, \delta^1_{01}} S_3 \to M \otimes_{S,\delta^1_{12}} S_3.

Let $DD_{S/R}$ be the category of $S$-modules equipped with descent data
relative to $f$.

\medskip\noindent
For example, for $M_0 \in \text{Mod}_R$ and a choice of isomorphism
$M \cong M_0 \otimes_R S$ gives rise to a descent datum by identifying
$M \otimes_{S,\delta^1_0} S_2$ and $M \otimes_{S,\delta^1_1} S_2$
naturally with $M_0 \otimes_R S_2$. This construction in particular
defines a functor $f^*: \text{Mod}_R \to DD_{S/R}$.

\begin{definition}
\label{definition-effective-descent}
The functor $f^*: \text{Mod}_R \to DD_{S/R}$
is called {\it base extension along $f$}. We say that $f$ is a
{\it descent morphism for modules} if $f^*$ is fully
faithful. We say that $f$ is an {\it effective descent morphism for modules}
if $f^*$ is an equivalence of categories.
\end{definition}

\noindent
Our goal is to show that for $f$ universally injective, we can use $\theta$ to
locate $M_0$ within $M$. This process makes crucial use of some equalizer
diagrams.

\begin{lemma}
\label{lemma-equalizer-M}
For $(M,\theta) \in DD_{S/R}$, the diagram

\label{equation-equalizer-M}
\xymatrix@C=8pc{
M \ar[r]^{\theta \circ (1_M \otimes \delta_0^1)} &
M \otimes_{S, \delta_1^1} S_2
\ar@<1ex>[r]^{(\theta \otimes \delta_2^2) \circ (1_M \otimes \delta^2_0)}
\ar@<-1ex>[r]_{1_{M \otimes S_2} \otimes \delta^2_1} &
M \otimes_{S, \delta_{12}^1} S_3
}

is a split equalizer.
\end{lemma}

\begin{proof}
Define the ring homomorphisms $\sigma^0_0: S_2 \to S_1$ and $\sigma_0^1, \sigma_1^1: S_3 \to S_2$ by the formulas
\begin{align*}
\sigma^0_0 (a_0 \otimes a_1) & = a_0a_1 \\
\sigma^1_0 (a_0 \otimes a_1 \otimes a_2) & = a_0a_1 \otimes a_2 \\
\sigma^1_1 (a_0 \otimes a_1 \otimes a_2) & = a_0 \otimes a_1a_2.
\end{align*}
We then take the auxiliary morphisms to be
$1_M \otimes \sigma_0^0: M \otimes_{S, \delta_1^1} S_2 \to M$
and $1_M \otimes \sigma_0^1: M \otimes_{S,\delta_{12}^1} S_3 \to M \otimes_{S, \delta_1^1} S_2$.
Of the compatibilities required in (\ref{equation-split-equalizer-conditions}),
the first follows from tensoring the cocycle condition
(\ref{equation-cocycle-condition}) with $\sigma_1^1$
and the others are immediate.
\end{proof}

\begin{lemma}
\label{lemma-equalizer-CM}
For $(M, \theta) \in DD_{S/R}$, the diagram

\label{equation-coequalizer-CM}
\xymatrix@C=8pc{
C(M \otimes_{S, \delta_{12}^1} S_3)
\ar@<1ex>[r]^{C((\theta \otimes \delta_2^2) \circ (1_M \otimes \delta^2_0))}
\ar@<-1ex>[r]_{C(1_{M \otimes S_2} \otimes \delta^2_1)} &
C(M \otimes_{S, \delta_1^1} S_2 )
\ar[r]^{C(\theta \circ (1_M \otimes \delta_0^1))} & C(M).
}

obtained by applying $C$ to (\ref{equation-equalizer-M}) is a split
coequalizer.
\end{lemma}

\begin{proof}
Omitted.
\end{proof}

\begin{lemma}
\label{lemma-equalizer-S}
The diagram

\label{equation-equalizer-S}
\xymatrix@C=8pc{
S_1 \ar[r]^{\delta^1_1} &
S_2 \ar@<1ex>[r]^{\delta^2_2} \ar@<-1ex>[r]_{\delta^2_1} &
S_3
}

is a split equalizer.
\end{lemma}

\begin{proof}
In Lemma \ref{lemma-equalizer-M}, take $(M, \theta) = f^*(S)$.
\end{proof}

\noindent
This suggests a definition of a potential quasi-inverse functor for $f^*$.

\begin{definition}
\label{definition-pushforward}
Define the functor {\it $f_*$} $: DD_{S/R} \to \text{Mod}_R$ by taking
$f_*(M, \theta)$ to be the $R$-submodule of $M$ for which the diagram

\label{equation-equalizer-f}
\xymatrix@C=8pc{f_*(M,\theta) \ar[r] & M \ar@<1ex>^{\theta \circ (1_M \otimes
\delta_0^1)}[r] \ar@<-1ex>_{1_M \otimes \delta_1^1}[r] &
M \otimes_{S, \delta_1^1} S_2
}

is an equalizer.
\end{definition}

\noindent
Using Lemma \ref{lemma-equalizer-M} and the fact that the restriction functor
$\text{Mod}_S \to \text{Mod}_R$ is right adjoint to the base extension
functor $\bullet \otimes_R S: \text{Mod}_R \to \text{Mod}_S$,
we deduce that $f_*$ is right adjoint to $f^*$.

\medskip\noindent
We are ready for the key lemma. In the faithfully flat case this is a
triviality (see Remark \ref{remark-descent-lemma}),
but in the general case some argument is needed.

\begin{lemma}
\label{lemma-descent-lemma}
If $f$ is universally injective, then the diagram

\label{equation-equalizer-f2}
\xymatrix@C=8pc{
f_*(M, \theta) \otimes_R S
\ar[r]^{\theta \circ (1_M \otimes \delta_0^1)} &
M \otimes_{S, \delta_1^1} S_2
\ar@<1ex>[r]^{(\theta \otimes \delta_2^2) \circ (1_M \otimes \delta^2_0)}
\ar@<-1ex>[r]_{1_{M \otimes S_2} \otimes \delta^2_1} &
M \otimes_{S, \delta_{12}^1} S_3
}

obtained by tensoring (\ref{equation-equalizer-f}) over $R$ with $S$ is an
equalizer.
\end{lemma}

\begin{proof}
By
Lemma \ref{lemma-split-surjection} and
Remark \ref{remark-functorial-splitting},
the map $C(1_N \otimes f): C(N \otimes_R S) \to C(N)$ can be split functorially
in $N$. This gives the upper vertical arrows in the commutative diagram
$$\xymatrix@C=8pc{ C(M \otimes_{S, \delta_1^1} S_2) \ar@<1ex>^{C(\theta \circ (1_M \otimes \delta_0^1))}[r] \ar@<-1ex>_{C(1_M \otimes \delta_1^1)}[r] \ar[d] & C(M) \ar[r]\ar[d] & C(f_*(M,\theta)) \ar@{-->}[d] \\ C(M \otimes_{S,\delta_{12}^1} S_3) \ar@<1ex>^{C((\theta \otimes \delta_2^2) \circ (1_M \otimes \delta^2_0))}[r] \ar@<-1ex>_{C(1_{M \otimes S_2} \otimes \delta^2_1)}[r] \ar[d] & C(M \otimes_{S, \delta_1^1} S_2 ) \ar[r]^{C(\theta \circ (1_M \otimes \delta_0^1))} \ar[d]^{C(1_M \otimes \delta_1^1)} & C(M) \ar[d] \ar@{=}[dl] \\ C(M \otimes_{S, \delta_1^1} S_2) \ar@<1ex>[r]^{C(\theta \circ (1_M \otimes \delta_0^1))} \ar@<-1ex>[r]_{C(1_M \otimes \delta_1^1)} & C(M) \ar[r] & C(f_*(M,\theta)) }$$
in which the compositions along the columns are identity morphisms.
The second row is the coequalizer diagram
(\ref{equation-coequalizer-CM}); this produces the dashed arrow.
From the top right square, we obtain auxiliary morphisms $C(f_*(M,\theta)) \to C(M)$
and $C(M) \to C(M\otimes_{S,\delta_1^1} S_2)$ which imply that the first row is
a split coequalizer diagram.
By Remark \ref{remark-adjunction}, we may tensor with $S$ inside $C$ to obtain
the split coequalizer diagram
$$\xymatrix@C=8pc{ C(M \otimes_{S,\delta_2^2 \circ \delta_1^1} S_3) \ar@<1ex>^{C((\theta \otimes \delta_2^2) \circ (1_M \otimes \delta^2_0))}[r] \ar@<-1ex>_{C(1_{M \otimes S_2} \otimes \delta^2_1)}[r] & C(M \otimes_{S, \delta_1^1} S_2 ) \ar[r]^{C(\theta \circ (1_M \otimes \delta_0^1))} & C(f_*(M,\theta) \otimes_R S). }$$
By Lemma \ref{lemma-C-is-faithful}, we conclude
(\ref{equation-equalizer-f2}) must also be an equalizer.
\end{proof}

\begin{remark}
\label{remark-descent-lemma}
If $f$ is a split injection in $\text{Mod}_R$, one can simplify the argument by
splitting $f$ directly,
without using $C$. Things are even simpler if $f$ is faithfully flat; in this
case,
the conclusion of Lemma \ref{lemma-descent-lemma}
is immediate because tensoring over $R$ with $S$ preserves all equalizers.
\end{remark}

\begin{theorem}
\label{theorem-descent}
The following conditions are equivalent.
\begin{enumerate}
\item[(a)] The morphism $f$ is a descent morphism for modules.
\item[(b)] The morphism $f$ is an effective descent morphism for modules.
\item[(c)] The morphism $f$ is universally injective.
\end{enumerate}
\end{theorem}

\begin{proof}
It is clear that (b) implies (a). We now check that (a) implies (c). If $f$ is
not universally injective, we can find $M \in \text{Mod}_R$ such that the map
$1_M \otimes f: M \to M \otimes_R S$ has nontrivial kernel $N$.
The natural projection $M \to M/N$ is not an isomorphism, but its image in
$DD_{S/R}$ is an isomorphism.
Hence $f^*$ is not fully faithful.

\medskip\noindent
We finally check that (c) implies (b). By Lemma \ref{lemma-descent-lemma}, for
$(M, \theta) \in DD_{S/R}$,
the natural map $f^* f_*(M,\theta) \to M$ is an isomorphism of $S$-modules. On
the other hand, for $M_0 \in \text{Mod}_R$,
we may tensor (\ref{equation-equalizer-S}) with $M_0$ over $R$ to obtain an
equalizer sequence,
so $M_0 \to f_* f^* M$ is an isomorphism. Consequently, $f_*$ and $f^*$ are
quasi-inverse functors, proving the claim.
\end{proof}

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