Isn't commutative property of $+$ in a ring obvious from other axioms?

Asked Aug 04 '19 at 04:20

Active Aug 04 '19 at 12:13

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Suppose $R$ be a nonempty set and $+,.$ are binary compositions on $R$ such that (R,+,.) satisfies all the properties of a ring except the commutative property of $+.$ Then

$[(a+b)-(b+a)]=[(a+b)+(-b+(-a))]$ (using distributive law) =$[a+(b-b)+(-a)]=0.$

So $a+b=b+a,\forall~a,b\in R$

Thus we see that commutative property of $+$ is obvious from other axioms.

Why do we need still mention it in the definition of a ring separately?

asked Aug 04 '19 at 04:20

Jave

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4

How do you prove that $-(b+a)=(-b)+(-a)$? – azif00 Aug 04 '19 at 04:23
ohk, I got it! I need unity in the ring. – Jave Aug 04 '19 at 04:25
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Jave, careful there. Without commutativity you would have $-(b+a)=(-a)+(-b)$. Remember that in a group (swithcing to multiplicative notation as I predict you to have seen it written this way) $(xy)^{-1}=y^{-1}x^{-1}$. I'm fairly sure this was Azif00's point. – Jyrki Lahtonen Aug 04 '19 at 04:41
See here. – Bill Dubuque Aug 04 '19 at 04:43
See https://math.stackexchange.com/questions/94350/why-do-we-assume-ab-ba-in-a-ring-with-1-also-is-it-true-with-rings-without – Gerry Myerson Aug 04 '19 at 05:57

Isn't commutative property of $+$ in a ring obvious from other axioms?

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