Why is there no Cayley's Theorem for rings? - MathOverflow

Why is there no Cayley's Theorem for rings?

2010-06-12T20:41:20Z

Cayley's theorem makes groups nice: a closed set of bijections is a group and a group is a closed set of bijections- beautiful, natural and understandable canonically as symmetry. It is not so much a technical theorem as a glorious wellspring of intuition- something, at least from my perspective, that rings are missing; and I want to know why.

Certainly the axiom system is more complicated- so there is no way you're going to get as simple a characterisation as you do with groups- but surely there must be some sort of universal object for rings of a given cardinality, analogous to the symmetric group in group theory. I would be surprised if it was a ring- the multiplicative and additive properties of a ring could be changed (somewhat) independently of one another- but perhaps a fibration of automorphisms over a group? If so is there a natural(ish) way of interpreting it?

Perhaps it's possible for a certain subclass of rings, perhaps it's possible but useless, perhaps it's impossible for specific reasons, in which case: the more specific the better.

Edit: So Jack's answer seems to have covered it (and quickly!): endomorphisms of abelian groups is nice! But can we do better? Is there a chance that 'abelian' can be unwound to the extent we can make this about sets again- or is that too much to hope for?

Answer by Jack Schmidt for Why is there no Cayley's Theorem for rings?

2010-06-12T20:50:41Z

Every (associative, unital) ring is a subring of the endomorphism ring of its underlying additive group. Rings act on abelian groups; groups act on sets. The universal action on an abelian group is its endomorphism ring; the universal action on a set is the symmetric group. Modules are rings that remember their action on an abelian group; permutation groups are groups that remember their action on a set.

A set is determined by its cardinality, but for abelian groups cardinality is not a very useful invariant. Rather than "order" of a ring, consider the isomorphism class of its underlying additive group. This is even commonly done in the finite ring case, where the order still has some mild control, but not as much as the isomorphism type of the additive group.

Answer by Greg Kuperberg for Why is there no Cayley's Theorem for rings?

2010-06-12T21:52:07Z

One interpretation of Cayley's theorem is that it gives you a cofinal sequence of finite groups, with respect to injective group homomorphisms, or a cofinal family (or maybe a transfinite sequence) of not-necessarily-finite groups. It's important to note that it's not the only interesting cofinal sequence/family; you could also look at the automorphism groups of vector spaces (say over one favorite finite field).

In studying this question for rings, by Jack Schmidt's answer suggests (but does not rigorously imply) looking at cofinal families of abelian groups first. Even in the category of finite abelian groups, there aren't any cofinal sequences that look as nice as the sequence of symmetric groups $S_n$ in Cayley's theorem. For instance $(\mathbb{Z}/n!)^n$ is a cofinal sequence, but this is decidedly less nice.

Jack Schmidt's answer can be further defend by this analogy: Cayley's theorem for groups is analogous to Cayley's theorem for semigroups, using the semigroup of all endomorphisms of a set. Then a unital semigroup is to a set as a unital ring is to an abelian group, so taking all of the endomorphisms of that group is again the same kind of theorem.

Answer by Grétar Amazeen for Why is there no Cayley's Theorem for rings?

2010-06-12T23:07:16Z

You can think of Cayley's theorem as the special case of Yoneda's lemma, where the category has only one object. If you take the additive version of Yoneda's lemma, and you plug in an additive category with one object then you get the desired statement for rings that's in Jack's answer. Viewing Cayley's theorem like this let's you generalize it to many other structures then groups.