1
$\begingroup$

Let $G$ be a group. A G-module M is defined as an abelian group on which $G$ acts through the map $ G \times M \to M$ where $ (g, m) \mapsto g \cdot m$ This action satisfies $g \cdot (m + m') = g \cdot m + g \cdot m', \forall m, m' \in M$.

When $G$ is a Galois group, a G-module M is described as an abelian group on which $G$ acts continuously, respecting the Krull topology on $G$ and the discrete topology on $M$.

Are these two definitions compatible? In particular, if $G$ acts continuously on $M$, does it necessarily imply that the action $G \times M \to M, (g, m) \mapsto g \cdot m$ satisfies $g \cdot (m + m') = g \cdot m + g \cdot m' , \forall m, m' \in M$?

$\endgroup$
2
  • 1
    $\begingroup$ The algebraic definitions are equal for both. But in the second, a topology is required in addition, also for the (infinite) Galois group, so that the action is continuous. See Milne's lecture notes on galois cohomology. $\endgroup$ Aug 23, 2023 at 8:29
  • 3
    $\begingroup$ In the second definition, "acts" implicitly means "by linear transformations," so it includes the additivity axiom. It should be "continuous $G$-module." In the first definition there aren't any topologies (or if you prefer the topologies are all discrete). $\endgroup$ Aug 23, 2023 at 8:36

1 Answer 1

4
$\begingroup$

The definition of a $G$-module for a profinite group is as follows.

Definition: Let $G$ be a profinite group. An abelian group $M$ is called a continuous (or discrete) $G$-module, if it is a $G$-module in the usual sense, and in addition the action $G\times M\rightarrow M$ is continuous when $M$ is endowed with the discrete topology, and $G\times M$ with the product topology.

So in particular, $g(m+m')=gm+gm'$ already holds.

Not every $G$-module $M$ is continuous. Consider the Galois extension $\Bbb Q(\sqrt{\Bbb N})/\Bbb Q$ with Galois group $G$. Then $$ M=\prod_p\Bbb Q(\sqrt{p}) $$ is a $G$-module, which is not continuous.

$\endgroup$
4
  • $\begingroup$ Thank you for your intriguing example. Could you tell me which open set in $M$ is not open under the pullback? $\endgroup$
    – Pont
    Aug 24, 2023 at 2:47
  • 1
    $\begingroup$ If $M$ is continuous, then $G\times \{m\}\rightarrow M$ is continuous and any $m\in M$ has only finitely many images under the action of $G$. This is not true, take for example $m=(\sqrt{2},\sqrt{3},\sqrt{5},\ldots)$, which has infinitely many images. $\endgroup$ Aug 24, 2023 at 9:48
  • $\begingroup$ I'm sorry, but I'm probably stuck on some fundamental aspects. Galois group can be regarded as inverse limit of finite groups, why is your Galois group $G$ is not pro finite ? $\endgroup$
    – Pont
    Aug 25, 2023 at 2:08
  • $\begingroup$ "My" Galois group is profinite, as inverse limit of the Galois groups of $\Bbb Q(\sqrt{1},\ldots,\sqrt{n})/\Bbb Q$, i.e., $G=\varprojlim_n {\rm Gal}( \Bbb Q(\sqrt{1},\ldots,\sqrt{n})/\Bbb Q)$. $\endgroup$ Aug 25, 2023 at 8:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .