Why is an integral domain a commutative ring with unity?

I was wondering why an integral domain is required to be commutative and have a multiplicative neutral element.

For example the quaternions would be a non-commutative integral domain.

The interesting property is that it has no zero divisors, so why require it to be commutative and have a $1$?

In other words: what consequences follow from having a one vs. not having one when investigating rings without zero divisors?

Solution 1:

Yes, people do study rings without unity (see for example this previous answer for some comments on rings without unity). So people do study commutative-rings-without-unit-that-have-no-zero-divisors.

Likewise, people do study non-commutative rings (with or without identity), and study such rings that have no zero divisors (though from what I can tell from my colleagues, zero divisors don't worry them quite as much as nilpotent elements do, so "reduced rings" seem to play a bigger role). So certainly, people study things that are "almost domains except for one of the properties" (nobody drops the $1\neq 0$ condition, though, because that makes it too easy).

A ring with 1 which is not the zero ring and has no zero divisors is called a domain (at least in Lam's First Course in Noncommutative Rings). They are defined on page 3 of the book, so pretty early. They can get pretty hairy: if you have a domain, it may not be embeddable in a division ring (in contrast to the integral domain case, where you always have the field of fractions); this was proven by Mal'cev, and is presented in Lam's Lectures on Rings and Modules, section 9B.

So really your question is more along the lines of "Why do we distinguish precisely integral domains, and not some of these other objects?" Maybe we should study rings without zero divisors, give them a special name (say, "wuzeds", for "without zero divisors", and then call integral domains "commutative wuzeds"?)

Partly, history. Ring theory was born from ideas of Dedekind, and later Noether and Artin, which arose from number theory considerations; they all took place in the commutative setting, and most of what Dedekind worked with were in fact (a special family of examples of) what we now call "integral domains". Integral domains were simply studied more. Also, a lot of ring theory originally arose as a way to try to abstract these ideas, generalize the results; a lot of the results were based on polynomials, on matrices, and other objects constructed from the rings, which people were familiar with in the case of the integers and other prototypes. Most of these objects become very hard when you switch to rings without identity (for example, whereas every ring with identity can be embedded in both its polynomial and its matrix rings in canonical ways, rings without identity are far more difficult to embed), or when you switch to noncommutative rings.

So people were studying special cases of commutative rings before they were studying noncommutative rings. That's also why we call them fields and skew-fields (or division rings), and not "commutative division rings" and "division rings". (Bourbaki attempted to change this, by defining "field" to be a ring with identity, $1\neq 0$, in which every nonzero element has an inverse, i.e. what we usually call a division ring, and then talking about fields and commutative fields; it did not stick outside of France).

Solution 2:

Why do you say that the interesting property of an integral domain is that it not have zero divisors? This is giving short shrift to the other properties, i.e. having 1, and being commutative.

The question of rings without 1, and their relevance within mathematics, has been discussed in various places here and on MO: see e.g. here, here, and here.

The differences between commutative and non-commutative rings are immense. Essentially, in commutative rings one can localize, and in non-commutative rings one cannot. See e.g. this answer for more discussion of this.

The theory of integral domains is based on all three properties (commutativity, existence of 1, no zero-divisors).

To give a concrete example, the enveloping algebra of a Lie algebra over a field has no zero divisors. If the Lie algebra is of finite dimension, it is Noetherian. (One sees this because the enveloping algebra admits a filtration, the filtration by degree, whose associated graded is a polynomial ring --- so in particular a domain --- in as many variables as the dimension of the Lie algebra --- and so Noetherian for a finite dimensional Lie algebra.)

Now lets suppose our Lie algebra $\mathfrak g$ is finite dimensional and semi-simple over a field of characteristic zero, say $\mathbb C$ just to fix ideas. The enveoping algebra $U(\mathfrak g)$ contains an augmentation ideal (the kernel of the action on the trivial representation, if you like), call it $I$; so there is a short exact sequence $$0 \to I \to U(\mathfrak g) \to \mathbb C \to 0.$$ Now because $\mathfrak g$ is semi-simple, any extension of the trivial representation $\mathbb C$ by itself splits, and so is again trivial. This implies that $I^2 = I$.

On the other hand, in a Noetherian integral domain (commutative!), one has the result that $I^2 = I$ implies $I = 0$. (See e.g. this answer.)

Geometrically, the idea is that if $A$ is a commutative ring with one and $I$ is an ideal such that $I = I^2$, then Spec $A/I$ is a closed subscheme of Spec $A$ which has no non-trivial normal directions into Spec $A$. If one pictures this, you will see that the only way this seems possible, intuitively, is if Spec $A/I$ is open as well as closed in Spec $A$ (otherwise there would be some directions pointing out of Spec $A/I$ into the rest of Spec $A$). One can then actually prove that if $A$ is furthermore Noetherian, and if $I = I^2$, then $A$ factors as a product $A/I \times B$, so Spec $A$ is the disjoint union of Spec $A/I$ and Spec $B$. In particular, if $A$ is a domain (so that Spec $A$ is irreducible, and so connected), one finds that either $I = A$ or $I = 0$.

The example of the enveloping algebra shows that the non-commutative situation is completely different: we have the ring $U(\mathfrak g)$ without zero divisors, and Noetherian, but with a non-trivial idempotent ideal. This is (at least for me) quite hard to interpret geometrically, and certainly suggests that it is reasonable to separate the study of (commutative) integral domains from the more general study of (possibly non-commutative) rings without zero divisors.

Final remark: I have no objection to using the term integral domain or domain in the non-commutative context. (I should note that, like many commutative algebraists, I frequently say and write domain to mean commutative integral domain.) My answer here is not intended to advocate any position on terminology, but just to explain why the commutative context is quite different from the more general non-commutative one, and hence why it makes sense to make a special study of the commutative case.

Incidentally, my answer could reasonably be taken in a more general sense as (at least partly) explaining why we make a special study of commutative rings (the subject of commutative algebra) rather than just studying all rings at once.