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Tag 08XA

Chapter 34: Descent > Section 34.4: Descent for universally injective morphisms

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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