1
$\begingroup$

Let $R$ be a commutative ring with $1_R$.

I found in this post the definition of the action of a group acting on a ring. And then, the following questions came in my mind.

Question 1. What is the formal and common definition of a ring acting on a group? (Without use of equivalent definition of action with endomorphisms, as wiki does)

So, from this to conclude that "A module is a ring action on an abelian group".

Question 2. Now, let's remind to ourselves the following definition.

Definition. An abelian group $(A,+)$ is an associative $R$-algebra if: (1) $(A,+,\cdot) $ is a ring (not necessarily commutative), (2) $(A,+)$ together with the scalar multiplication $*:R\times A \longrightarrow A,\ (r,a) \longmapsto r*a$ is a left $R-$module and (3) the operations $*$ and $\cdot$ are compatible, that is $$\forall a,b \in A, \forall r\in R: r*(a\cdot b)=(r*a)\cdot b=a\cdot (r*b).$$

In this answer, we have a good way to think about the last axiom, as a result of an action.

Does the ring $R$ acts on the abelian group $(A,+)$ through the map (scalar multiplication) $R\times A \longrightarrow A,\ (r, a)\longmapsto r*a$ ? And if the answer is yes, why? In other words, what are precisely the properties which should are satisfied, in order to have this kind of action?

I'm trying to connect $R$-algebras with Group Theory.

Thank you very much.

$\endgroup$
12
  • 1
    $\begingroup$ @QiaochuYuan Thank you for your comment. What is the unclear point? $\endgroup$
    – Chris
    Commented Nov 21, 2018 at 1:26
  • 1
    $\begingroup$ The slogan "a module is a ring action in an abelian group" is unhelpful. It should read "a module is a homomorphism from a ring to the endomorphism ring of an abelian group". $\endgroup$
    – Rob Arthan
    Commented Nov 21, 2018 at 1:29
  • 1
    $\begingroup$ @RobArthan Ok, but these two, are equivalent. Why is this unhelpful? $\endgroup$
    – Chris
    Commented Nov 21, 2018 at 1:37
  • 1
    $\begingroup$ "Ring action in an abelian group" doesn't tell me what "ring action" means. "Homomorphism from ring to endomorphism ring" tells me exactly what is meant in a straightforward and memorable way. $\endgroup$
    – Rob Arthan
    Commented Nov 21, 2018 at 1:53
  • 1
    $\begingroup$ OK. That should answer your Q1. Can you now answer your Q2 for yourself? $\endgroup$
    – Rob Arthan
    Commented Nov 21, 2018 at 1:59

1 Answer 1

1
$\begingroup$

The common way to think about the ring R acting on a group G is for there to be a formal multiplication $$R \times G \rightarrow G$$ $$(r, g) \mapsto r\cdot g$$,

that moves group elements around. This multiplication agrees with the group action in the sense that it's distributive.

In practice, the action is clear--matrices on vectors, multiplication of numbers, etc. In general, the ring action has to be told.

$\endgroup$

You must log in to answer this question.

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