Axioms of algebraic structure - ring

Question

If in the definition of ring $(R,+,\times$) we insist that it has unit element $1$. Then we can show that addition $(+)$ is commutative operation. However, most of the proof which I've seen in MSE use the following trick: $$x+x+y+y=(1+1)x+(1+1)y=(1+1)(x+y)=x+y+x+y$$ which lead to $x+y=y+x$.

But what if we use another approach, namely: Suppose $a=x+y$ and $b=y+x$. Then consider the expression: $$a-b=a+(-b)=(x+y)+(-(y+x))=(x+y)+((-y)+(-x))$$ and then using associativity of $+$ we get that: $a-b=0$ which leads to desired result.

Note that the crucial moment in my proof is the property $-b=(-1)\cdot b$ which can be proven easily without any reliance on abelian of addition.

I suppose that this approach is also correct. But what do you think about it?

Answer 1

Yes it is correct and much a more natural approach to think of . Good job. smn

Answer 2

One step in your proof actually assumes commutativity of addition, namely

$$-(y+x)=-y-x$$

Without establishing commutativity beforehand, we can only say that

$$-(y+x)=-x-y$$

To see why this is the case, consider how one would prove the last equality:

$$(y+x) + ((-x) + (-y)) = y + (x + (-x)) + (-y) = y + 0 + (-y) = 0$$

where we relied heavily on associativity.