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I admit freely that the following question is a bit of a fishing expedition inspired by this lovely "definition" of a module as found on Wikipedia:

A module is a ring action on an abelian group.

It takes a while for the novice to unzip this definition into the usual list of identities, but for someone with a basic understanding of (group) actions who needs a quick shorthand to remember what being a module entails, it seems hard to beat this definition.

Now, volumes have been written about group actions. This makes sense, because set theory underlies modern mathematics and for any set $X$, the automorphisms $\operatorname{Aut}(X)$ naturally have a group structure. And of course, a group $G$ acting on $X$ is just a group homomorphism $G \to \operatorname{Aut}X$.

On the other hand — aside from this definition of a module — it is hard to come across a general theory of ring actions. In order to find interesting actions of a ring $R$, one analogously needs the set of endomorphisms $\operatorname{End}X$ to have the structure of a ring so that one may search for ring morphisms $R \to \operatorname{End}X$.

What are the most general objects $O$ for which $\operatorname{End}O$ canonically has the structure of a ring?

Certainly, $O$ at least contains abelian groups, but is there anything else?

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    $\begingroup$ Regarding the title question: Group actions on vector spaces (or abelian groups) automatically give you ring actions - in fact a linear action of a group $G$ gives you an action by the group ring $\mathbb{Z}G$. If you have an action of $G$ on a set $X$, you automatically get an action of $\mathbb{Z}G$ on the free abelian group with basis $X$. Linear actions (known as representations) are quite pervasive in mathematics, perhaps because linear algebra is a powerful set of tools for studying symmetry, so ring actions are in practice not much harder to find than group actions. $\endgroup$
    – S. Carnahan
    Jun 25, 2012 at 7:00

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First, I am not really sure what you mean by "it is hard to come across a general theory of ring actions." This is precisely module theory! If $f : R \to S$ is any ring homomorphism whatsoever, then composition with the natural map $S \to \text{End}(S)$ (where $S$ is regarded as an abelian group and $S$ acts on this abelian group by left multiplication; this proves "Cayley's theorem for rings") gives $S$ the structure of a left $R$-module.

Anyway, the objects you're looking for are the objects in categories such that Hom-sets canonically have the structure of an abelian group. These are the categories enriched over $(\text{Ab}, \otimes)$, or $\text{Ab}$-enriched categories. As Will Sawin says, a more traditional name is preadditive category, but I don't like this name because I don't think I should have to remember the distinction between preadditive, additive, and abelian categories to refer to something as fundamental as $\text{Ab}$-enrichment.

$\text{Ab}$-enriched categories are abundant. In fact, any ordinary category $C$ has a free $\text{Ab}$-enriched category $\mathbb{Z}[C]$ equipped with the universal functor from $C$ to an $\text{Ab}$-enriched category. This functor $\mathbb{Z}[-]$ is left adjoint to the forgetful functor from $\text{Ab}$-enriched categories to ordinary categories and can be explicitly described by taking free abelian groups on hom-sets. This is a slight generalization of the point Scott Carnahan makes in the comments, as this functor is monoidal, so sends monoids to their monoid algebras and sends monoid actions to ring actions.

More generally, let $V$ be any monoidal category. Then one can define $V$-enriched categories, and a $V$-enriched category with one object (more precisely the monoid of endomorphisms of such a category) is precisely a monoid object in $V$, and one gets various generalizations of monoids and rings this way. For example:

  • A $\text{Set}$-enriched category with one object is a monoid.
  • A $\text{Top}$-enriched category with one object is a topological monoid.
  • An $\text{Ab}$-enriched category with one object is a ring.
  • A $\text{Vect}$-enriched category with one object is a (unital, associative) algebra.
  • A $\text{Ban}$-enriched category with one object is a (unital) Banach algebra.
  • A $\text{Ch}$-enriched category with one object is a dg-algebra.

And so forth. (Note that $V$ itself need not be $V$-enriched; I believe this condition is equivalent to $V$ being closed monoidal). It is also possible to give a uniform definition of what it means for a $V$-monoid $A$ to act on an element $M$ of $V$ in terms of a map $A \otimes M \to M$ satisfying the usual axioms (this circumvents the need for $V$ to be $V$-enriched) which reproduces the notion of action of a monoid, module of a ring, representation of an algebra, etc.

Just as it is fruitful to generalize groups to groupoids by allowing multiple objects, one can generalize monoids to monoidoids (that is, categories!) in any of the above settings by allowing multiple objects. From this perspective, an $\text{Ab}$-enriched category is just a ringoid: a ring with many objects!

Thinking in this way makes certain aspects of ring theory look more natural, I think. For example, for certain nice $V$ there is a notion of Cauchy completion of a $V$-enriched category $C$ generalizing both the notion of Karoubi envelope and the notion of Cauchy completion of a metric space. The Cauchy completion of an $\text{Ab}$-enriched category is obtained by formally adjoining finite biproducts and then splitting idempotents. Now:

The Cauchy completion of a ring $R$ (as an $\text{Ab}$-enriched category) is the category of finitely-generated projective right $R$-modules.

This result gives a very conceptual approach to Morita equivalence: as it turns out, two rings are Morita equivalent if and only if their Cauchy completions are equivalent!

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  • $\begingroup$ Fair enough; I'll edit. $\endgroup$ Jun 25, 2012 at 6:53
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    $\begingroup$ Quite exhaustive answer :). The question of the OP is already answered in the 2nd paragraph. $\endgroup$ Jun 25, 2012 at 8:35
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$\newcommand{\Hom}{\operatorname{Hom}}$An example for a class of rings where the ring actions are very close to group actions are Hopf algebras. As a consequence many theorems from representation theory of groups generalize to the representation theory of Hopf algebras.

In order to give a feeling on how definitions from group actions generalize to Hopf algebras consider a group $G$ and $kG$-Modules $M,N$ ($k$ a commutative ring). Then we have the invariants $M^G := \lbrace m \in M\mid \forall g \in G:\;gm=m\rbrace$ and $G$ also acts on $\Hom_k(M,N)$ via
$(g\cdot f)(m) := gf(g^{-1}m)$. Then the $G$-linear maps can be recovered as invariants by $\Hom_{kG}(M,N) = \Hom_k(M,N)^G$.

Now let $H$ be a Hopf algebra over $k$ (corresponds to $kG$) with antipode $S: H \to H$ (corresponds to $g \mapsto g^{-1}$), augmentation $\epsilon: H \to k$ (corresponds to $g \mapsto 1$) and coproduct $\Delta: H \to H \otimes_k H$ (corresponds to $g \mapsto g \otimes g$). Then one defines for $H$-modules $M,N$: $$M^H := \lbrace m \in M\mid \forall h \in H: \;hm=\epsilon(h)m\rbrace$$ and if $\Delta(h)=\sum_i h_i^{(1)} \otimes h_i^{(2)}$ then $H$ acts on $\Hom_k(M,N)$ by $$(h\cdot f)(m):= \sum_i h_i^{(1)}f\big(S(h_i^{(2)})m\big).$$ Now, as expected, the identity $\Hom_H(M,N)=\Hom_k(M,N)^H$ holds again.

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Preadditive categories are probably the most general context. They have endomorphism rings, since one can both add and multiply morphisms. Most examples of preadditive categories are abelian categories, such as abelian groups, modules over a ring, representations of a group, sheaves of abelian groups on a site, etc. Some are not, such as the category of abelian topological groups.

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    $\begingroup$ Or if one does not feeling like checking the wikipedia article, preadditive categories are just categories enriched over the category $Ab$ of abelian groups. There are lots of these! $\endgroup$
    – David Roberts
    Jun 25, 2012 at 5:54
  • $\begingroup$ Did you mean "abelian groups" when you said "sets"? $\endgroup$
    – S. Carnahan
    Jun 25, 2012 at 6:49
  • $\begingroup$ I would imagine so :) $\endgroup$
    – David Roberts
    Jun 25, 2012 at 6:59

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