Recently I have encountered an exercise from the book A survey of modern algebra by Birkhoff and MacLane, which states:
Let R be a set equipped with two operations (addition and multiplication) that satisfies:
(i) Associative laws of addition and multiplication
(ii) Distributive law: a(b+c) = ab + ac
(iii) Zero: R contains 0 such that: a+0 = a for all a.
(iv) Unity: R contains 1 (not equal 0) such that: a1 = a for all a
(iv) For all a, the equations a+x = 0 and y+a = 0 have solutions x and y in R.
Prove that a+b = b+a for a,b in R.
I came up with the following argument:
a + (b + a) + b = (a+b) + (a+b) = (a+b)(1+1) = a(1+1) + b(1+1) = (a+a) + (b+b) = a + (a+b) + b
Then b+a = a+b, since left and right cancellation are possible. However, this solution doesn't impress me much, since it employs both the left and right distributivity, which are not provided by (iii).
Can anyone provide a formal proof for this exercise? Or the exercise itself is not well stated?
Near rings provide examples that show the necessity of both distributive laws (besides having an identity).
The simplest near ring is built from a group $G$ (not necessarily abelian), written additively, and considering the group $M(G)$ of all mappings from $G$ to $G$ (with pointwise operation). As multiplication take composition of maps. Then $M(G)$ is a near ring with identity which is not a ring (the distributive law is only one-sided) and the additive group structure is not commutative unless $G$ is abelian to begin with.