Ytdyklly

More precisely, why is it that all rings are required by the axioms to have commutativity in addition, but are not held to the same axiom regarding multiplication? I know that we have commutative and non-commutative rings depending on whether or not they are commutative in multiplication, but I am wondering why it is that the axioms were defined that way, providing us with this option.

I am using this list of axioms, from David Sharpe’s Rings and factorization:

The first rings that were considered were generally commutative; polynomial rings, then from work of Dedekind in number fields. The properties were then abstracted Fraenkel and Noether, who still dealt mostly with commutative rings.

However, it soon became apparent that there were too many instances where commutativity of multiplication did not hold. You had famously the quaternions, of course, but you also had matrices and, more generally, the endomorphism ring of an abelian group (where “multiplication” is composition of functions). So that we have two different, related, notions: commutative rings and noncommutative rings, just like we have noncommutative groups and commutative/abelian groups.

Now, why do this with multiplication and not with addition? Well, if you take your definition of ring above, which includes a unity, but you drop condition (3) (that is, you require everything except you do not require that addition be commutative), it turns out that the other eight axioms force commutativity of addition.

Indeed, suppose you have a structure $(R,+,cdot,0,1)$ that satisfies axioms (1), (2), and (4)-(9) above. I claim that one can deduce (3). Indeed, let $a,bin R$. Then using associativity on the left first, and associativity on the right second, we have
$$begin{align*}
(1+1)(a+b) &= 1(a+b) + 1(a+b) = a+b+a+b\
(1+1)(a+b) &= (1+1)a + (1+1)b = a+a+b+b.
end{align*}$$
From this we get that $a+b+a+b = a+a+b+b$. Now add the inverse of $a$ on the left and the inverse of $b$ on the left on both sides to get
$$begin{align*}
(-a) + a + b + a + b + (-b) &= 0+b+a+0 = b+a\
(-a) + a + a + b + b + (-b) &= 0+a+b+0 = a+b
end{align*}$$
Thus, we conclude that $a+b=b+a$. That is, commutativity of addition is a consequence of the other eight axioms.

The reason we include it is two-fold: one, is that it is much nicer to say that the first few axioms force $(R,+)$ to be a commutative/abelian group. The second is that it is also common to consider rings without unity, and if we do that, then it is no longer true that addition is forced to be commutative. To see this, note that if $(G,cdot)$ is any group with identity element $e_G$, and we define a multiplication on $G$ by letting $a*b=e_G$ for all $a,bin G$, then $(G,cdot,*)$ satisfies axioms (1)-(8) given above. But if the original group is not commutative, then the “addition” in this ring is not commutative. So if we want to consider rings without unity, we do want to explicitly require addition to be commutative.