# The Stacks Project

## Tag 08XA

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. $\square$

The code snippet corresponding to this tag is a part of the file descent.tex and is located in lines 1284–1292 (see updates for more information).

\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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