In abstract algebra, we like to define various algebraic structures that generalize some very concrete object. For example, a field is defined using properties that enable addition, multiplication, subtraction, and division in a way analogous to the real or rational numbers, and a group is defined using properties that generalize composition of invertible functions.
Now, a ring, which generalizes the integers, is defined based on addition, subtraction, and multiplication (we can’t divide integers in general, since e.g. 1/2 is not an integer). More precisely, they’re defined by the following properties:
1. For all a and b, a + b = b + a
2. For all a, b, and c, a + (b + c) = (a + b) + c
3. There exists a number 0 such that a + 0 = a = 0 + a for all a
4. For each a, there exists an element -a such that a + (-a) = (-a) + a = 0
5. For all a, b, and c, a * (b * c) = (a * b) * c
6. For all a, b, and c, a * (b + c) = a * b + a * c = (b + c) * a
That a + b and a * b are always elements of the ring is implied in the definition. Some people require the ring to have an element 1 such that a * 1 = 1 * a = a for all a; the definition I learned doesn’t, and I think this way it works better; when the element 1 exist, I just call it a ring with 1. In similar vein, if for all a and b, a * b = b * a, the ring is called commutative.
Now, so far, so good. But what if we drop property 1? The ring will still form an additive group (groups are defined by properties 2-4 of rings; if they satisfy #1, they’re called abelian; why it is “abelian” and not “commutative” is beyond me), and multiplication will still distribute over addition.
However, Graeme Taylor of Modulo errors commented on Good Math, Bad Math, linking to a Usenet post showing that non-abelian rings are fairly restricted: in particular, every ring with 1 is abelian. This is because (a+a)+(b+b) = (1+1)(a+b) = (a+b)+(a+b) by distribution, so that a+b = b+a by cancellation.