# Determine the cocycle condition in Galois descent induced by faithfully flat descent

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

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

and

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

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

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

and

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

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

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.

• 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 $\psi_{(\sigma,\tau),(\sigma,\tau),\sigma)}$). – FelixBB Jun 12 '19 at 11:15
• 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 $\prod M^\sigma$. – Roland Jun 12 '19 at 13:12
• 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 $f^\sigma(a \cdot m) = f^\sigma( \sigma(a)m ) = \sigma(a) f^\sigma(m)$? – FelixBB Jun 12 '19 at 15:10
• $f^\sigma=f$ 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. – Roland Jun 12 '19 at 15:35