TITLE: Why is ring addition commutative?
QUESTION [52 upvotes]: What is the motivation behind axiomatically forcing the underpinning group of a ring to be abelian? Noncommutative rings are vastly more complex than commutative ones, so I am assuming that allowing the additive operation to be noncommuting would just make matters worse. Is there something deeper here, or is this a restriction for the sake of convenience and simplicity?

REPLY [4 votes]: Its not just for convenience and simplicity; its actually totally fundamental. The problem with replacing abelian groups with general groups (or general monoids, for that matter) is that neither of these are the models of a commutative algebraic theory. Compared to this, the fact that commutativity follows from the other ring axioms pales to insignificance.
Let me explain.
First, a problem.
Exercise 1. Let $X$ and $Y$ denote commutative semigroups written additively, and let $f,g : X \rightarrow Y$ denote homomorphisms. Show that $f+g$ is a homomorphism.
I highly recommend you solve the above (fairly easy) exercise before going on. In doing so, you'll have noticed that both associativity and commutativity were used in the proof. However, with a bit of thought, we see that they weren't really fundamental to the proof. Basically, we have the expression
$$(f(x)+f(x'))+(g(x)+g(x'))$$
and we use associativity to drop the brackets, obtaining
$$f(x)+f(x')+g(x)+g(x')$$
then we switch the two inner terms, obtaining
$$f(x)+g(x)+f(x')+g(x')$$
and then we put the brackets back in:
$$(f(x)+g(x))+(f(x')+g(x'))$$
Thus, our proof ultimately comes down to the fact that $Y$ satisfies the following identity.
$$(a+b)+(c+d) \equiv (a+c)+(b+d)$$
So we get an idea: call magmas satisfying the above identity medial, prove the result for medial magmas rather than commutative semigroups, and then simply observe that every commutative semigroup is in fact a medial magma. Of course there might be a niggling doubt in the back of our minds; is the mediality condition actually a "natural" thing to consider? It looks just like a warped form of commutativity. The answer is: "Yes, it is a natural property to consider," but the justification for this claim will have to wait for later. So, lets  temporarily adopt the position that mediality is interesting and natural, and lets take some time to consider its consequences.
Exercise 2. Verify that every commutative semigroup is indeed a medial magma (I suggest using additive notation). Conversely, prove that every medial magma with a two-sided identity element is both associative and commutative.
The above exercise begs the question; do there exist magmas that are medial, but which aren't commutative semigroups? The answer is a resounding "yes!"
Example 1. If $(X,+)$ is an abelian group, then $(X,-)$ is a medial magma that is neither commutative, nor associative.
Example 2. If $I \subseteq \mathbb{R}$ and $f : I^2 \rightarrow I$ is defined by $f(x,y) = (x+y)/2$, then $(I,f)$ is a medial magma that is commutative, but not associative.
The best way to verify Example 1 is with the result of the following exercise; Example 1 is then obtained as the special case in which $*$ is taken as addition, $f$ denotes the identity function on $X,$ and $g$ represents unary negation.
Exercise 3. Let $(X,*)$ denote a medial magma, and suppose $f,g : X \rightarrow X$ are commutative endomorphisms, in the sense that $f \circ g = g \circ f$. Defining $x \diamond y = f(x) * g(y)$, show that $(X,\diamond)$ is also medial.
Finally, since we've come this far, lets check for a moment that we're not fooling ourselves, by verifying the result that motivated our interest in mediality in the first place:
Exercise 4. Let $X$ and $Y$ denote medial magmas written with an asterisk, and let $f,g : X \rightarrow Y$ denote homomorphisms. Verify that $f*g$ is a homomorphism.
So to recap, if $f$ and $g$ are parallel homomorphisms in the category of medial magmas, then so too is $f*g$. But, is this really a big deal? I think this is a huge deal; here's one reason why.
Let $(X,*)$ denote a medial magma and suppose $A$ is the set of all endomorphisms of $X$. Then we observe that $A$ has the following properties.

$(A,*)$ is a medial magma, where $*$ on $A$ is defined pointwise from $*$ on $X$.
$(A,\circ)$ is a monoid
$\circ$ distributes on both sides over $*$

Thus, $A$ is kind of like a ring. This motivates the following definition: a ringlike-entity appropriate for medial magmas is a triple $(A,*,\circ)$ satisfying the aforementioned axioms. Now suppose $A$ is a ringlike-entity appropriate for medial magmas and that $X$ is a medial magma. I leave it as an exercise for the reader to define the notion of "multiplication of a scalar in $A$ with a vector in $X$"; that is, a function $f : A \times X \rightarrow X$ that satisfies all the reasonable compatibility conditions. As inspiration; let $X$ denote an arbitrary medial magma, suppose $A$ is the set of all endomorphisms of $X$, and think of $f : A \times X \rightarrow X$ as the evaluation function.
Therefore, since we have notion of scalar multiplication, for any ringlike-entity appropriate for medial magmas $A$, we have a notion of $A$-module.
Lets go picture for a moment. Commutative algebra (think: abelian groups, rings, modules and algebras) is built on the theory of "abelian groups." However, what I'm trying to say here, is that we can replace this theory with others (e.g. the theory of commutative semigroups, or medial magmas) in order to obtain different versions of commutative algebra, each with their own version of "ring", "module" etc. Most generally, I think, is that we can replace the theory of "abelian groups" with any commutative algebraic theory. Examples of such theories include:

The theory of abelian groups, with addition and unary negation.
As above, except with with addition and binary subtraction this time.
The theory of medial magmas.
The theory of commutative semigroups.

Note that a theory with a single binary operation $*$ is a commutative algebraic theory if and only if it proves that $*$ is medial; that is why I argued earlier that mediality is a natural condition to consider. Furthermore, not every group is medial, and not every monoid is medial, and not every semigroup is medial, so it wouldn't make sense to try to build commutative algebra over these structures.
Anyway, I have much, MUCH more to say about this stuff, some of it quite abstract, but I don't want to give away too much of what I've been toying with, since it might be publishable someday.
In conclusion though, its not just for convenience and simplicity. For everything to work, we have to build commutative algebra over some base commutative algebraic theory.