# What happens if we remove the requirement that $\langle R, + \rangle$ is abelian from the definition of a ring?

Ever since I learned the definition of a ring, I've wondered why the additive group is required to be abelian. What happens if we allow $\langle R, + \rangle$ to be nonabelian as well as $\langle R, \cdot \rangle$? Is this impossible? If not, what is the this type of algebraic structure called?

Here is the definition of a ring that I am using:

• $\langle R , + \rangle$ is an abelian group with identity $0$.
• $\langle R , \cdot \rangle$ is associative.
• $a\cdot (b + c) = a\cdot b + a \cdot c$.
• $(a + b)\cdot c = a\cdot c + b \cdot c$.
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Typically addition is commutative by definition. I can't think of an example where addition is non commutative. In such cases, we typically call it multiplication. –  nigelvr Dec 19 '12 at 18:17
With the axioms you've given, you can make any group into the additive group of a ring by defining $x \cdot y = 0$ for all $x,y$. This is one reason it is useful to require a multiplicative identity, which most -- though not all -- contemporary mathematicians regard as part of the "standard" definition of a ring. In fact I have in my memory that commutativity of addition actually follows from the other axioms. Someone else will elaborate on that shortly, I presume. –  Pete L. Clark Dec 19 '12 at 18:30
@PeteL.Clark, I remember this in the context of vector space axioms, as in math.stackexchange.com/questions/479000/…. You can remove the requirement that vector addition is commutative; it'll follow from the the other axioms. –  lhf Sep 19 '13 at 5:47

If you require the distributive law on just one side, you obtain what is called a near ring.

If you require distributivity on both sides, this tends to force $+$ to be commutative.

Indeed, computing the product $(\alpha + \beta)(a + b)$ using distributivity on the left, and then on the right, and then doing it again in the opposite order, you deduce that $$\beta a + \alpha b = \alpha b + \beta a.$$ So at least on the additive subgroup of $R$ generated by elements which are products of two other elements, the operation $+$ is commutative. If e.g. you require in addition that $R$ contains a multiplicative identity, then $+$ will be commutative on all of $R$.

Edit: This argument is a variant of the Eckmann-Hilton argument.

It might also help to think about the basic example of a near-ring (as discussed in wikipedia), namely maps from a group $G$ to itself, with $+$ being the group operation (applied pointwise to maps) and $\cdot$ being composition of maps. The right distributive law is trivially true, but imposing the left distributive law would then say that we are looking at maps of a group that preserve the group operation, i.e. endomorphisms from $G$ to $G$. But for this to be a near-subring of the near-ring of all maps, then we would need that the pointwise sum of two homomorphisms is again a homomorphism, a condition which holds only if $G$ is abelian.

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Thus, if we take the ring axioms as you've listed them, and change the first one to only require that $\langle R,+\rangle$ is a group, we get the axioms for a right near ring that is also a left near ring under the same operations.