Definition of Ring

I'm studying Abstract Algebra right now, currently covering rings. In the introduction of the subject, I am curious as to why there is no need for there to be a multiplicative identity. I understand that in order to be a ring, we require the set to be an abelian group under addition operation and a monoid under multiplication. But what is the reason for the monoid, rather than group under multiplication--or lack of multiplication?

• You seem to be asking two questions - are you asking about the multiplicative identity or multiplicative inverses? For the former, see math.stackexchange.com/questions/48587/… . For the latter, well, that's what fields are for. Jul 24, 2011 at 23:18
• Related questions here and here. Jul 24, 2011 at 23:20
• It's the former. Thanks both for the links. Jul 24, 2011 at 23:35
• Jul 24, 2011 at 23:37
• See also this question on examples/motivation for non-unital rings. Note that the definition of a ring also requires that the additive and multiplicative structures are related - by the distributive law. Without such one would simply have a set with two completely unrelated structures. Jul 25, 2011 at 0:17

Basically, you want to encompass a large enough class of objects that the study is useful, while at the same time having "enough" structure to be able to say interesting things.

This leads to a difficult balance: the more requirements you put on the structure, the fewer structures that the definition will cover. For example, if you require that the structure be an abelian group under both addition and multiplication, then $\mathbb{Z}$, the integers, don't qualify. So, the fewer conditions (requirements), the more structures you expect to satisfy them.

On the other hand, in order to be able to say interesting things you usually need assumptions. That means that the more conditions you put on the structure, the more things you have to "play with", and the more likely you are to be able to say interesting (or far reaching) things. Consider for example "finite groups". We are very far from having a satisfactory answer to the question "What are all the finite groups?"; but throw in the simple condition that multiplication commute, and the question "What are all the finite abelian groups?" already has a very good answer (the structure theorem for finitely generated abelian groups). So, the more conditions you place on the objects you want to study, the more you expect to be able to say about them.

And so, we usually study semigroups and groups rather than magmas; why? Because groups and semigroups are prevalent enough that a lot of objects satisfy the conditions, and at the same time the requirements are strong enough to let us say lots of interesting things about them (we have a harder with semigroups than with groups, and an even harder with loops and magmas...)

For rings, a good balance turns out to be when we require the multiplicative structure to be either a semigroup or a monoid, rather than full-fledged group. (We do study the cases where you get a group: you get fields and division rings). A monoid structure (with a multiplicative identity) used to be prefered, but it turns out to exclude a lot of interesting cases (many of which arise in places like functional analysis). So an expansion of the class of structures to those in which you only have a semigroup structure has been prefered, though many authors still assume all rings have an identity and that homomorphisms between rings map identities to identities (e.g., Lam's books all make this assumption).

So... it's a balancing act between trying to "cover" a lot and at the same time being able to "say" a lot. Asking for the multiplicative structure to be a semigroup or a monoid is a good balance. There are other, weaker structures (such as near-rings and semi-rings) as well, just like with groups you have semigroups, quasigroups, loops, and magmas.

The point of algebraic structures is not just to have algebraic structures, but rather to have algebraic structures that reflect the things that are already out there.

One of the simplest algebraic objects out there is the collection of integers, $\mathbb{Z}$. We have two operations, multiplication and addition, and they satisfy all sorts of properties. However, if we demanded that we have multiplicative inverses, then $\mathbb{Z}$ would not fit into our mold, and we want some sort of definition that describes the structure present in $\mathbb{Z}$.

What if we want to consider $2\mathbb{Z}\subset \mathbb{Z}$, the even integers? We still have multiplication and addition, but we no longer have a multiplicative identity. Should this be considered a ring? Or something else? This depends on who you ask, as a lot of people require rings to have multiplicative identities, and they will explicitly say "ring without unit" otherwise. Still, regardless what we call it, it is clear that we should have some sort of algebraic structure that models the properties of $2\mathbb{Z}$.

If, when we ignore $0$, we have a group under multiplication, we get the notion of a division ring, and if the group is abelian we get this notion of a field, which is very important. The rational numbers, the real numbers, the complex numbers are all fields. Some of the most beautiful mathematics comes from working over fields, as there are things you can do over fields that you cannot do over arbitrary rings. However, as we have seen, it is important that we consider rings, because otherwise there would be basic algebraic objects out there begging for a name.

• "If, when we ignore 0, we have a group under multiplication, we get the notion of a field" I think you mean an abelian group. Otherwise you get a division ring. Jul 25, 2011 at 0:41
• @Jacob Yes, for some reason I was thinking about commutative rings. Although, division rings are also called "skew fields", and share many similar properties. If history had been slightly different, there would be fields and commutative fields. Still, making the change. Jul 25, 2011 at 0:56
• @Aaron: This is something that I've thought about with regard to requiring or not requiring a multiplicative identity in rings: requiring a multiplicative identity doesn't allow $n\mathbb{Z} \subset \mathbb{Z}$, $n \ge 2$ to be an ideal (since an ideal is a subring), which doesn't seem very convenient or natural. Do you agree that this is a good motivator for not requiring a multiplicative identity? Jan 17, 2013 at 18:06
• @AlexPetzke: No, I think language usage trumps this consideration. Most of the non-unital rings I see are $k$-algebras for some (unital, often a field) ring $k$. Moreover, this additional $k$-algebra structure ends up being important, so "$k$-algebra" couldn't really be omitted. For people in algebraic geometry or rep theory, explicitly calling them unital rings every single time would be cumbersome. Papers would have blanket disclaimers that "every ring is assumed unital unless noted" and most algebaists would start assuming that rings are unital. Jan 17, 2013 at 21:42

Taking your "rather than group under multiplication" phrase:

It is interesting to consider things which have some properties similar to the integers under addition and multiplication. So you want a group under addition but not under multiplication.

Many mathematicians only assume that the set is a semigroup not monoid under multiplication for example Nathan Jacobson,I. N. Herstein,Seth Warner,N. McCoy, and Van der Werden. There are some theorems on rings which require a multiplicative (right or left or an) identity however. For an interesting nonunital ring see "Planetmath.org" - Klein 4 ring. See Hopkins theorem (an Artinian ring with identity is a Noetherian ring) for a need for identity element for a ring. Also rings with an identity element have a maximal ideal,a maximal right ideal, and a maximal left ideal using Zorn's lemma.Many early results in Ring Theory did not need the asumption that a ring was unital. Many fascinating non unital rings and subrings are found in matrix rings over a non unital ring.

If lots of interesting stuff can be proved without assuming there's a unit, then assuming there's a unit would merely complicate the proofs with an irrelevant assumption. So if you're getting this stuff from a textbook, see which results the book proves without assuming a unit.