I initially asked this question on Mathoverflow as I thought it was to right place to do so. But it might not be so I will copy it here instead. I apologize for double posting and I will gladly erase the inapropriate one. It's the first time that I use these forums so I'm not sure if it is the right amount of details.

I tried to give a proof that fppf (faithfully flat) descent implies Galois descent purely at the level of modules and I stumble to obtain the Galois cocycle condition. I'm interested to consider some questions of twisted sheaves with a Galois cohomological description and understanding how to obtain the former would be useful to me.

I obtained the following the following conditions: Given a finite Galois extension $L/K$ of Galois group $G$ and $M$ an $L$-vector space $M$, we have for each $\sigma \in G$ an isomorphism of $L$-vector spaces $\psi_\sigma : M \to M^\sigma$ satisfying $\psi_\sigma(am) = \sigma(a) \psi_\sigma(m)$, where $a \in L$ and $m \in M$ and such that for every pair $(\sigma, \tau) \in G \times G$ we have

$$ \psi_{ ( \sigma, \tau), (\sigma, \tau), \sigma } \circ \psi_{ ( \sigma, \tau), (\sigma, \tau), \tau } = \psi_{ ( \sigma, \tau), (\sigma, \tau), \sigma \tau } $$

as isomorphisms of $L$-modules. The $L$-module structure of $M^\sigma$ is twisted by $\sigma$, i.e given by $a \cdot m:= \sigma(a)m$.

My issue is that because of what one obtains for sheaves of modules (https://stacks.math.columbia.edu/tag/0CDQ) I would expect the cocycle condition for modules to also have a twisting in the formula.

Translating what is done in stacks project into modules is certainly possible but I wasn't able to do so.

Edit: If someone can give me the details of how it is done at the level of modules (or a least some clear sketch), I would accept this as a correct answer.

Instead I will present a (somehow long) sketch of what I did. I can provide more details upon request and I apologize if there are too many. My difficulty is on Step 5. You can skip directly to this step if you want, the rest explains how I got there.

The context: Let $L/K$ be a finite Galois extension and let $M$ be an $L$-module together we an isomorphism of $L \otimes_K L$-modules $\phi: M \otimes_K L \to L \otimes_K M$ satisfying the cocyle condition $p_{13}^* \phi = p_{23}^* \phi \circ p_{12}^* \phi$ as isomorphisms of $L \otimes_K L \otimes_K L$-modules.

What I did:

Step 1: Describing some isomorphisms We have an isomorphism of $K$-algebras $L \otimes_K L \to \prod_{\sigma \in G} L$ given by $a \otimes 1 \mapsto ( a )_{\sigma \in G}$ and $1 \otimes a \mapsto ( \sigma(a) )_{\sigma \in G}$

and another one $ L \otimes_K L \otimes_K L \to \prod_{\sigma \in G} \Big( \prod_{\tau \in G} L \Big)$ given by

$$ a \otimes 1 \otimes 1 \mapsto \Big( (a)_{\tau \in G} \Big)_{\sigma \in G}, $$

$$ 1 \otimes a \otimes 1 \mapsto \Big( (\tau(a)_{\tau \in G} \Big)_{\sigma \in G}, $$

$$ 1 \otimes 1 \otimes a \mapsto \Big( \tau\sigma(a)_{\tau \in G} \Big)_{\sigma \in G}. $$

We can then describe the above isomorphisms as

$$ (1_L \coprod \sigma) : L \otimes_K L \to \prod_{\sigma \in G} L $$


$$ (1_L \coprod \tau \coprod \tau \sigma): L \otimes_K L \otimes_K L \to \prod_{\sigma \in G} \Big( \prod_{\tau \in G} L \Big). $$

Step 2: Obtain some $\prod_{\sigma \in G} L$-module structures

We then have a commutative diagram of modules.

$$ \begin{array}{ccccc} M \otimes_K L & \xrightarrow{} & \prod_{\sigma \in G} \Big( M \otimes_K L \Big) & \xrightarrow{} & \prod_{\sigma \in G} M\\ \downarrow & & \downarrow & & \downarrow \\ L \otimes_K M & \xrightarrow{} & \prod_{\sigma \in G} \Big( L\otimes_K M \Big) & \xrightarrow{} & \prod_{\sigma \in G} M^\sigma \end{array} $$

where the left most vertical arrow is $\phi$ and we denote by $\psi$ the induced the right most vertical arrow.

Using Step 1 we have ring morphisms $L \xrightarrow{1_L} L$ and $L \xrightarrow{\sigma} L$ for each $\sigma \in G$. Tensoring $M$ with these morphisms give $L$-module structures for $M \otimes_K L$ by $a \cdot ( m \otimes c ) = m \otimes ac$ and for $L \otimes_{L,\sigma} M$ by $a \cdot (c \otimes m) = \sigma(a)c \otimes m$. Now the isomorphism of $L$-modules $\mu: M \otimes_L L \to M: m \otimes c \mapsto cm$ then gives to $M$ the $L$-module structure $a \otimes m = am$. We also have the composite diagram

$$ L \otimes_{L,\sigma} M \xrightarrow{ \sigma^{-1} \otimes 1_M } L \otimes_L M \xrightarrow{\mu'} M : c \otimes m \mapsto \sigma^{-1}(c) \otimes m \mapsto \sigma^{-1}(c)m. $$

Then this $M$ has $L$-module structure given by $a \cdot m := \sigma(a)m$ and we denote it by $M^\sigma$ (We can relabel to get $M^\sigma$ instead of $M^{\sigma^{-1}}$.).

Now a structure of $\prod_{\sigma \in G} L$-module on $\prod_{\sigma \in G} M$ (resp. on $\prod_{\sigma \in G} M^\sigma$) is determined by an $L$-module structure on $M$ (resp. on $M^\sigma$) for each $\sigma \in G$. Therefore, if $(a_\sigma)_{\sigma \in G} \in \prod_{\sigma \in G}$ and $(m_\sigma)_{\sigma \in G} \in \prod_{\sigma \in G} M$ (resp. in $\prod_{\sigma \in G} M^\sigma$), then

$$ (a_\sigma)_{\sigma \in G} \bullet (m_\sigma)_{\sigma \in G} = (~a_\sigma m_\sigma)_{\sigma \in G} ( \text{ resp. } (a_\sigma)_{\sigma \in G} \circ (m_\sigma)_{\sigma \in G} = ( \sigma(a)_\sigma m_\sigma)_{\sigma \in G}~). $$

Step 3: Determine for each $\sigma \in G$ the isomorphisms of $L$-modules $\psi_\sigma$.

The isomorphism $\psi$ induced by $\phi$ must then satisfy

$$ \psi \big( a_\sigma m_\sigma)_{\sigma \in G} ) = (a_\sigma)_{\sigma \in G} \circ \psi( ( m_\sigma)_{\sigma \in G} ). $$

For each $\sigma \in G$ we have an isomorphism of $L$-modules

$$ M \xrightarrow{\iota_\sigma} \prod_{\sigma \in G} M \xrightarrow{\psi} \prod_{\sigma \in G} M^\sigma \xrightarrow{\pi_\sigma} M^\sigma $$

given by

$$ am \mapsto (0, \cdots, 0, am, 0, \cdots, 0) \mapsto \big( \sigma(a) \pi_\sigma \Big( \psi( \iota_\sigma(m) \Big) \big)_{\sigma \in G} \mapsto \sigma(a)\pi_\sigma \Big( \psi(m) \Big). $$

So for each $\sigma \in G$ we have an isomorphism $\psi_\sigma : M \to M^\sigma$ defined by $\psi_\sigma(m):=\pi_\sigma( \psi(m) )$ and such that $\psi_\sigma(am) = \sigma(a)\psi_\sigma(m)$.

Step 4: Determine some $\prod_{\sigma \in G} \prod_{\tau \in G} L$-module structures

I will skip some details, which I can provide upon request. I use the cocycle condition to determine three $\prod_{\sigma \in G} \prod_{\tau \in G} L$-module structures.

Consider $p_{12}^* p_1^* M$ (or equivalently $p_{13}^*p_1^*M$).

The $\prod_{(\sigma,\tau) \in G \times G } L$-module $\prod_{(\sigma,\tau) \in G \times G } M$ is

$$ (a_{g,h}) \cdot ( m_{g,h} ) = ( a_{g,h} m_{g,h} ). $$

Consider $p_{12}^* p_2^* M$ (or equivalently $p_{23}^*p_1^*M$).

The $\prod_{(\sigma,\tau) \in G \times G } L$-module $\prod_{(\sigma,\tau) \in G \times G } M^\tau$ is

$$ (a_{\sigma,\tau}) \cdot ( m_{\sigma,\tau} ) = ( \tau(a_{\sigma,\tau}) m_{\sigma,\tau} ). $$

Consider $p_{13}^* p_2^* M$ (or equivalently $p_{23}^*p_2^*M$).

The $\prod_{(\sigma,\tau) \in G \times G } L$-module $\prod_{(\sigma,\tau) \in G \times G } M^{\sigma \tau}$ is

$$ (a_{\sigma,\tau}) \cdot ( m_{\sigma,\tau} ) = ( (\sigma \circ \tau)(a_{\sigma,\tau}) m_{\sigma,\tau} ). $$

Step 5: Determine the cocycle condition

Finally, for each pair $(\sigma, \tau) \in G \times G$ we have three composite maps given as follows:

$$ M_{(\sigma, \tau)} \xrightarrow{ \iota_{(\sigma,\tau)}} \prod_{(\sigma,\tau) \in G \times G} M \xrightarrow{ p_{12}^* \psi } \prod_{(\sigma,\tau) \in G \times G} M^\tau \xrightarrow{ \pi_{\sigma,\tau}} M_{(\sigma, \tau)}^\tau $$

defining an $L$-module isomorphism $ \psi_{ ( \sigma, \tau), (\sigma, \tau), \tau } : M_{(\sigma,\tau)} \to M_{(\sigma,\tau)}^\tau$ satisfying

$$ \psi_{ ( \sigma, \tau), (\sigma, \tau), \tau }(am) = \tau(a)\psi_{ ( \sigma, \tau), (\sigma, \tau), \tau }(m) $$

$$ M_{(\sigma, \tau)} \xrightarrow{ \iota_{(\sigma,\tau)}} \prod_{(\sigma,\tau) \in G \times G} M \xrightarrow{ p_{13}^* \psi } \prod_{(\sigma,\tau) \in G \times G} M^{\sigma \tau} \xrightarrow{ \pi_{\sigma,\tau}} M_{(\sigma, \tau)}^{\sigma\tau} $$

defining an $L$-module isomorphism $ \psi_{ ( \sigma, \tau), (\sigma, \tau), \sigma \tau } : M_{(\sigma,\tau)} \to M_{(\sigma,\tau)}^{\sigma \tau}$ satisfying

$$ \psi_{ ( \sigma, \tau), (\sigma, \tau), \tau }(am) = (\sigma \circ \tau)(a)\psi_{ ( \sigma, \tau), (\sigma, \tau), \tau }(m) $$


$$ \psi_{ ( \sigma, \tau), (\sigma, \tau), \tau }(am) = \tau(a)\psi_{ ( \sigma, \tau), (\sigma, \tau), \tau }(m) $$

$$ M_{(\sigma, \tau)}^\tau \xrightarrow{ \iota_{(\sigma,\tau)}} \prod_{(\sigma,\tau) \in G \times G} M^\tau \xrightarrow{ p_{23}^* \psi } \prod_{(\sigma,\tau) \in G \times G} M^{\sigma \tau} \xrightarrow{ \pi_{\sigma,\tau}} M_{(\sigma, \tau)}^{\sigma \tau} $$

defining an $L$-module isomorphism $ \psi_{ ( \sigma, \tau), (\sigma, \tau), \sigma } : M_{(\sigma,\tau)}^\tau \to M_{(\sigma,\tau)}^{\sigma \tau}$ satisfying

$$ \psi_{ ( \sigma, \tau), (\sigma, \tau), \sigma }(am) = \sigma(a)\psi_{ ( \sigma, \tau), (\sigma, \tau), \tau }(m). $$

Indeed, $\psi_{ ( \sigma, \tau), (\sigma, \tau), \sigma }$ sends $\tau(a)m$ to $(\sigma \circ \tau)(a)m$ and so sends $am = \tau(\tau^{-1}(a))m$ to $\sigma( am )$.

Since $p_{23}^* \phi \circ p_{12}^*\phi = p_{13}^* \phi$, we have $p_{23}^* \psi \circ p_{12}^*\psi = p_{13}^* \psi$ and therefore for each pair $(\sigma, \tau) \in G \times G$ we have

$$ \psi_{ ( \sigma, \tau), (\sigma, \tau), \sigma } \circ \psi_{ ( \sigma, \tau), (\sigma, \tau), \tau } = \psi_{ ( \sigma, \tau), (\sigma, \tau), \sigma \tau } $$

as isomorphisms of $L$-modules.

Remarks: The map $\psi_{ ( \sigma, \tau), (\sigma, \tau), \sigma }$ is from $M^\tau$ to $M^{\sigma \tau}$ and it is not clear to me how to re-express it as starting from $M$ and twisting it by $\tau$. Another problem is that the maps I found on Step 3 are not in an obvious way related to those of Step 5 and there might be a need to twick something here as well.


I am not very comfortable with all of your notations (notably $\psi_{(\sigma,\tau),(\sigma,\tau),\sigma)}$ which I didn't understand the point of this notation). So let me propose another one (with a slight change of convention which I find more convenient, let me know if you prefer that I go back to yours).

So again, if $M$ is a $L$-module, let $M^\sigma$ be $M$ with the left $L$-module structure such that $l.m=\sigma(l)m$ (if $M$ has the structure of a $L\otimes_K R$-module, then $R$ acts as before on $M$, in particular $M^\sigma$ has a $L\otimes_K L$-module structure where $(l_1\otimes l_2)m=\sigma(l_1)l_2m$). We have for every $L$-module $M$ a morphism of $L\otimes_K L$-modules : $$L\otimes_K M\to \prod_{\sigma\in G} M^{\sigma}$$ such that $l\otimes m\mapsto (l.m)_{\sigma}=(\sigma(l)m)_\sigma$. This gives $$\psi_\tau :M\xrightarrow{.\otimes 1} M\otimes_K L\xrightarrow{\phi} L\otimes_K M\to \prod_{\sigma\in G}M^\sigma\xrightarrow{\pi_\tau} M^\tau$$ which is a morphism of $L$-modules (on the left) because each of the above arrows are $L$-linear on the left. This is actually just a $K$-linear map $M\to M$ such that $\psi_\tau(lm)=\tau(l)\psi_{\tau}(m)$. It actually deserve the abuse of notation $\psi_\tau=:\tau$ so that $\tau(lm)=\tau(l)\tau(m)$. The point is that the cocycle condition will tell us that this is indeed a $G$-action by semi-linear automorphisms.

So now, let us compute $\psi_{\tau\sigma}$. Note that any $L$-linear map $f:M\to N$ induces a $L$-linear map $f^\sigma:M^\sigma\to N^\sigma$ (which is simply $f$ on the underlying set) and we have an obvious equality $(M^{\sigma})^\tau=M^{\sigma\tau}$.

We also have the canonical isomorphism of $L\otimes_K L\otimes_K L$-modules $M^\sigma\otimes_K L\simeq (M\otimes_K L)^\sigma$ (which is the identity on the underlying set). Recall that the twisting only change the leftmost structure.

We have a commutative diagram where every map is $L\otimes_K L\otimes_K L$-linear and where equality mean canonical isomorphism : (To prove that this indeed commute, you will need the linearity of $\phi$, this is then straightforward) $$ \require{AMScd} \begin{CD} M\otimes_K L\otimes_K L@>p_{12}^*\phi>> L\otimes_K M\otimes_K L@>p_{23}^*\phi>> L\otimes_K L\otimes_K M\\ @.@VVV@VVV\\ @.\prod_{\sigma\in G}M^\sigma\otimes_K L@.\prod_{\sigma\in G}(L\otimes M)^\sigma\\ @.@|@|\\ @.\prod_{\sigma\in G}(M\otimes_K L)^\sigma@>\prod\phi^\sigma>>\prod_{\sigma\in G}(L\otimes M)^\sigma\\ @.@.@VVV\\ @.@.\prod_{\sigma\in G}\prod_{\tau\in G}M^{\tau\sigma} \end{CD} $$

Now just follow the path of $m\otimes 1\otimes 1$.

  • For $\rightarrow\downarrow\downarrow$, we get on the $\sigma$-component $\psi_\sigma(m)\otimes 1$. So if we continue along $\rightarrow\downarrow$, we get on the $\tau$-component $\psi_\tau(\psi_\sigma(m))$ (or rather $\psi_\tau^\sigma(\psi_\sigma(m))$).
  • For $\rightarrow\rightarrow$. Recall that this composition is actually $p_{13}^*\phi$. So, if we write $\phi(m\otimes 1)=\sum a_i\otimes m_i$, then $m\otimes 1\otimes 1\mapsto \sum a_i\otimes 1\otimes m_i$. So now if we go $\downarrow$, we get on the $\sigma$-component $\sigma(a_i)(1\otimes m_i)=\sigma(a_i)\otimes m_i$. And if we go $\downarrow$ again, we get on the $\tau$-component $\tau(\sigma(a_i))m_i$. But this is exactly $\psi_{\tau\sigma}(m)$.

Since this holds for every $m\in M$, $\psi_\tau\psi_\sigma=\psi_{\tau\sigma}$ (or rather $\psi_\tau^\sigma\psi_\sigma=\psi_{\tau\sigma}$). This exactly means that $G$ acts on $L$ by semi-linear automorphisms.

  • $\begingroup$ Thank you Roland for this clear and detailed solution. This approach to define the $\psi$'s is cleaner than to insert the module on the $(\sigma,\tau)$ th factor and then to project on the $(\sigma,\tau)$ th factor (the reason for my notation $\psi_{(\sigma,\tau),(\sigma,\tau),\sigma)}$). $\endgroup$ – FelixBB Jun 12 '19 at 11:15
  • $\begingroup$ You're welcome. I read your previous comment and now I understand your notation. Well, I haven't really thought about it and your cocycle condition is probably right, but I don't think it is obvious from the one I gave (which is the one used in practice : namely Galois descent follows from the datum of a semi-linear action of the Galois group). It is just that I don't have a good grasp on the different direct factors of $\prod M^\sigma$. $\endgroup$ – Roland Jun 12 '19 at 13:12
  • $\begingroup$ Just to make the obvious "obvious", given a map of L-modules $f: M \to N$, the induced map $f^\sigma: M^\sigma \to N^\sigma$ for each $\sigma \in G$ is $f$ as map of abelian group (as you said) but with the difference that it respects the new $L$-module structures, namely for $a \in L$ and $m \in M^\sigma$ we have $f^\sigma(a \cdot m) = f^\sigma( \sigma(a)m ) = \sigma(a) f^\sigma(m)$? $\endgroup$ – FelixBB Jun 12 '19 at 15:10
  • 1
    $\begingroup$ $f^\sigma=f$ as maps of sets. Now $f^\sigma(a.m)=f^\sigma(\sigma(a)m)=f(\sigma(a)m)=\sigma(a)f(m)=\sigma(a)f^\sigma(m)=a.f^\sigma(m)$ so $f^\sigma$ is $L$-linear with respect to the twisted structure. $\endgroup$ – Roland Jun 12 '19 at 15:35

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