# What is the “right” definition of a ring?

This is somewhat related to Greg's question about groups and abelian groups. Suppose you met someone who was well-acquainted with groups, but who was unwilling to accept rings as a meaningful object of study. How would you describe rings to them in a natural way given that they like talking about groups?

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All the answers so far have motivated noncommutative ring theory. I'd be interested in seeing motivation for commutative rings (in the spirit of mathoverflow.net/questions/2551/…). – Eric Wofsey Oct 27 '09 at 3:09
A strange question (that's good :-). – Włodzimierz Holsztyński Apr 8 '13 at 6:32
Actually, algebraists of the past ... E Artin, ... Jacobson ... already answered this question a long time ago -- that's how they have approached the structure theory of rings. – Włodzimierz Holsztyński Apr 8 '13 at 6:36
(I mean that they provided at least a significant partial answer). – Włodzimierz Holsztyński Apr 8 '13 at 6:39

Well rings are naturally the objects which act on abelian groups - indeed composition always endows the endomorphisms of an abelian group with the structure of a ring. So if one is interested in the endomorphisms of groups one is actually interested in rings.

One can make this analogy more precise especially if one picks a particular ring and looks at the forgetful functor to abelian groups from its module category. This analogy can then be used again for instance to motivate the definition of plethory which are the natural objects which act on rings.

To address Eric's comment about commutative rings there is an analogue in this case of something which was mentioned in the discussion of groups versus abelian groups. Indeed one can obtain commutative rings by considering the identity in an additive symmetric monoidal category. In this case the endomorphisms of the tensor unit are endowed with an abelian group structure via the augmentation over abelian groups and the Eckmann-Hilton argument applied to tensoring endomorphisms and composing endomorphims forces the composition to be abelian. So from this point of view commutative rings are the gadgets which naturally act on the hom-sets of additive symmetric monoidal categories.

Since I mentioned this one can take this slightly further. If one considers such a category together with an autoequivalence (for instance if we take a tensor triangulated category) then one can consider the graded endomorphism ring of the identity. This naturally gives rise to an integer graded ring which is commutative up to some unit which squares to the identity and which has a natural action on the category.

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Here is some more support for rings as objects that act on abelian groups, as already mentioned by Greg Stevenson.

Someone well-acquainted with groups would likely know that representations of groups are important to their study. Given a group G, a representation of G over a field k is an action of G on a k-vector space V as a group of linear automorphisms. Familiarity with rings allows us to realize that this is the same as a ring homomorphism from the group ring kG into Endk(V). Then one can immediately begin to investigate group actions by asking questions about the structure of the group ring kG. In fact, one can even show that the category of G-modules (representations of G) is equivalent to the category of (say left) modules over the ring kG.

From this perspective, rings are important because they act on modules. In this vein, every ring can be realized as an endomorphism ring of a module: for a ring R, the right module RR satisfies R ≅ End(RR). (In analogy with the terminology for group representations, RR is sometimes referred to as the regular representation of R.)

To go one step further, the endomorphism ring of an object in any abelian (or even preadditive) category is a ring. (Though from Greg's post, it sounds as if one can go even further than this!) So we see that rings greatly generalize the notion of groups acting on objects with additive structure.

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Suppose this hypothetical person understood on an intuitive level the notion of symmetries of objects in Euclidean space but was unwilling to accept the formal definition of a vector space (more precisely the part where a field is necessary)... – Qiaochu Yuan Oct 27 '09 at 6:47

Much like abelian groups and groups, commutative rings and non-commutative rings have different motivations in my mind.

As people have said, non-commutative rings are naturally endomorphisms of abelian groups. The first non-commutative ring people should have in their head should be M_n, I think.

However, I'm surprised no one has brought up that commutative rings are naturally the set of functions on something. Granted, it takes a couple of semesters of algebraic geometry to make this true, but its the idea that motivates the theory. The first commutative ring people should have in their head is C[x], with Z as a second example where it seems tantalyzingly bizarre that everything still works (prime factorization, ideals, etc). Certainly, when I try to convince a skeptic that rings are awesome (which I have done a couple of times now), I wave my hands wildly and talk about how cool rings of functions on things are.

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Well-- the set of ring-valued functions. – Tim Campion Dec 7 '10 at 6:45

I'll offer another "explanation" for rings:

a ring (see here) is a monoid in the monoidal category of abelian groups (with respect to the standard tensor product of abelian groups).

This perspective is useful in that it shows what the right generalizations and categorifications of rings are. This is a general phenomenon: when you want to know which of several equivalent definitions is the fundamental or right one, check for which of these you can find natural oo-categorical versions. The more natural a concept, the easier it generalizes this way.

For rings, we notice that the category of abelian groups is the archetypical abelian category. The oo-version of an abelian category is a stable (oo,1)-category. The archetypical one is the (oo,1)-category of spectra - Spec. A commutative monoid in Spec is an "commutative oo-ring" usually called an E-oo ring. If it is non-commutative it is called an A-oo ring.

This is the story about rings and their vertical categorifciation. There is also insight into the nature of rings to be gained from their horizontal categorification:

a monoid in Ab, hence a ring, is equivalently an an enriched category with a single object over the category of abelian groups:

write pt for the single object of an Ab-enriched category, then Hom(pt,pt) is an abelian group equipped with a homomorphism of abelian groups Hom(pt,pt)otimesHom(pt,pt) --> Hom(pt,pt) that is associative and unital. So Hom(pt,pt) is some ring, and every ring works.

So a general Ab-enriched category may be thought of as a ringoid, if you wish.

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If they agreed that the representation theory of groups was interesting (and if they didn't agree to this, I might contest their claim to be well-acquainted with groups...) I would argue that thinking about modules for the group ring C[G] is a very clean way to do representation theory. (On preview, Manny Reyes wrote a more complete answer along the same lines while I started this, so I'll stop here.)

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I'll add in another little piece of the puzzle to help motivate the study of commutative rings once we believe that we are interested in endomorphisms of modules.

There's an exercise in Rotmans Homological Algebra book that if R and S are (non-commutative) rings, such that the module categories R-mod and S-mod are equivalent, then Z(R) and Z(S) are isomorphic as (commutative) rings. Here, of course, Z(R) means the center of R.

It follows from this that if R and S are commutative rings then R = S if and only if R-mod = S-mod. So if modules are what we want to study, it makes sense to single out those rings which have such a close relationship with their module categories.

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Of course, an different lesson from this is that Morita equivalence is not as interesting if you restrict yourself to the commutative world. – Chris Phan Jun 14 '10 at 9:36

One simple answer is that rings are modeled after the ring of integers. Admittedly, it is a commutative ring with many special properties, but it's a simple intuitive model.

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This is the usual explanation, but as far as this hypothetical person is concerned, Z is the free group on one element. What's this multiplication nonsense? – Qiaochu Yuan Oct 27 '09 at 14:25

Today I came across an expository paper which reminded me of this particular question. The paper is STANDARD DEFINITIONS CONCERNING RINGS by KEITH CONRAD. This is an answer to the title and clearly not to the body of the question.

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This explains why I sometimes forget to read the body only to find that(after I have typed up an answer)the body was a different story – Unknown Feb 17 '11 at 18:07
This could have better gone here:mathoverflow.net/questions/22579/… – Unknown Feb 17 '11 at 18:15