Is there a simple interpretation of a ring action like a group or monoid one?

Question

A group action corresponds to a group of automorphisms. A monoid action corresponds to a monoid of endomorphisms. Is there a similar way to think about ring actions?

Answer 1

A ring action on an abelian group is a module.

A linear group action on a vector space, $V$, automatically produces a ring action (by the group ring $\Bbb{Z}G$ if $G$ is the group) on the free abelian group generated by $V$. (Although this sounds very different from the first line, it is less different than it sounds.)

(Linear group actions appear in many places, under the label "representations".)

This can be generalized to $\mathrm{Ab}$-enriched categories with one object. In fact, any category, $C$, has a free $\mathrm{Ab}$-enriched category $\Bbb{Z}[C]$ with a universal functor from $C$ to an $\mathrm{Ab}$-enriched category. (If this sounds similar to the group ring action above, it is.)

Answer 2

A group action on an object $X$ can be thought of as a group homomorphism $G \longrightarrow Aut(X)$. This is a special case of a monoid action, which would be a monoid homomorphism $M \longrightarrow End(X)$, the set of endomorphisms of $X$ under composition. It may sometimes be the case that $End(X)$ carries an additive structure as well. For instance, if $X$ is an abelian group then $End(X)$ has the addition $(f+g)(x) := f(x) + g(x)$. This makes $End(X)$ into a ring under composition and this pointwise addition. By analogy to the above definitions of group and ring actions, we can say that a ring $R$ acts linearly on $X$ by giving a ring homomorphism $R \longrightarrow End(X)$. More generally, this can be done for $X$ in any abelian (or even additive) category, such as chain complexes or sheaves.

This is also referred to as making $X$ into an $R$-module, when $X$ is an abelian group. You may have seen $R$-module structures defined as a map $R \times X \longrightarrow X$ satisfying some axioms. These two notions are compatible, in the same way thay a group action on $X$ can be thought of as a map $G \longrightarrow Aut(X)$ or as a map $G \times X \longrightarrow X$ satisfying some axioms. That is, a map $f: R \times X \longrightarrow X$ defines a map $R \longrightarrow End(X)$ via $r \mapsto (x \mapsto f(r, x))$. On the other hand, a map $f: R \longrightarrow End(X)$ corresponds to a map $(r, x) \mapsto f(r)(x)$.

Answer 3

I really like paul blart math cop's answer, and I'll give a slight modification with a more categorical flavor; note firstly that monoids and rings can be thought of as purely categorical gadgets, in the following sense; given a monoidal category $(\mathsf{C},\otimes, 1)$ (such as $\mathsf{Sets}$ with the cartesian product, or $\mathsf{Ab}$ with the tensor product) you can define a unital algebra to be an object $A \in \mathsf{C}$ with a multiplication map $m: A\otimes A \rightarrow A$ and a unit map $\eta:1\to A$ (where 1 is the unit of the monoidal product) such that the following associativity diagram commutes:

$$ \require{AMScd} \begin{CD} A \otimes A \otimes A @>m \otimes id>> A \otimes A \\ @Vid\otimes mVV @VVmV\\ A\otimes A @>m>> A \end{CD} $$

And the following morphisms both equal the identity $$ \require{AMScd} \begin{CD} A \cong A\otimes1 @>id \otimes \eta>> A \otimes A @>m>> A \\ \\ A \cong 1 \otimes A @>id \otimes \eta>> A \otimes A @>m>> A \\ \end{CD} $$

You can readily check that in the case of $(\mathsf{Sets}, \times, \{*\})$ or $(\mathsf{Ab}, \otimes_{\mathbb{Z}}, \mathbb{Z})$ that this categorifies the axioms for a monoid and a ring respectively. You can also define a module $M$ over an algebra via an action map $\alpha: A \otimes M \rightarrow M$ satisfying analogous associativity and unitality diagrams, which you can also readily check give you the usual notions of monoid action on a set and ring action on an abelian group (i.e. an $R$-module). I will leave to you the categorification of the notion of a group.

Now it remains to show that we have defined maps from the algebras into some sort of endomorphism object; this doesn't work in general, but it works in $\mathsf{Sets}$ and $\mathsf{Ab}$ because these categories have internal Homs, i.e. are enriched over themselves with the internal Hom being right adjoint to the tensor product (Currying and the classical tensor-hom adjunction).

Working in $\mathsf{Sets}$, let $A$ be a monoid and $\alpha: A \times M \to M$ a set with $A$ action. Notice that by currying (as described by paul blart math cop), this multiplication map corresponds to a map $\widetilde\alpha: A \to \operatorname{Hom}(M,M)$. Note here that $\operatorname{Hom}(M,M)$ is itself a monoid via the composition map $$\circ: \operatorname{Hom}(M,M)\times\operatorname{Hom}(M,M) \to \operatorname{Hom}(M,M)$$ and using the associativity of the $A$ action on $M$ and iterated currying, one can check that $\widetilde \alpha$ is in fact a homomorphism of monoids (and rings in $\mathsf{Ab}) using essentially the same argument as the one paul blart math cop describes (though it is a good exercise to categorify and do this without referring to elements at all! Essentially this comes down to checking how the tensor-hom adjunction plays with composition). This phenomenon is really quite general and works in other categories with monoidal structures and internal homs, which are called "closed".