Discrepancy in the definition of a module over a group

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 $$g \cdot (m + m') = g \cdot m + g \cdot m' , \forall m, m' \in M$$?

• 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. Commented Aug 23, 2023 at 8:29
• 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). Commented Aug 23, 2023 at 8:36

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.
• Thank you for your intriguing example. Could you tell me which open set in $M$ is not open under the pullback? Commented Aug 24, 2023 at 2:47
• 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. Commented Aug 24, 2023 at 9:48
• 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 ? Commented Aug 25, 2023 at 2:08
• "My" Galois group is profinite, as inverse limit of the Galois groups of $\Bbb Q(\sqrt{1},\ldots,\sqrt{n})/\Bbb Q$, i.e., $G=\varprojlim_n {\rm Gal}( \Bbb Q(\sqrt{1},\ldots,\sqrt{n})/\Bbb Q)$. Commented Aug 25, 2023 at 8:19