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The beginner of Ring Theory may have came across following question:

Let $(R,+,\cdot)$ be a ring (may be non-commutative) with unity. Show that the condition (axiom) $a+b=b+a$ for all $a,b$ in defition of ring can be deduced from other axioms of ring.

Just expand $(1+1).(a+b)$ in two ways to deduce above axiom.

I was considering if this is still valid for ring $R$ without unity; is this true?

Q. If $(R,+,\cdot)$ is a ring without unity, can we deduce axiom $a+b=b+a$ ($a,b\in R$)from other axioms in definition of ring?

Beginner
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  • Not sure about the definitions right now, but: Is every abelian group $(G, +, 0)$ with trivial multiplication (i.e. $ab = 0$ for all $a,b \in G$) a commutative ring without identity? If yes, the case for $G$ not abelian should do it? –  Jun 12 '17 at 08:38
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    @PaulK Seems that we were thinking the same thing at the same time. – badjohn Jun 12 '17 at 08:40
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    See also https://math.stackexchange.com/questions/609364/why-is-ring-addition-commutative (the most upvoted answer mentions this construction). – Arnaud D. Jun 12 '17 at 08:40
  • See also https://math.stackexchange.com/questions/1336199/smallest-two-sided-nearring. There are nontrivial examples there. – Batominovski Jun 12 '17 at 08:41

1 Answers1

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Take any non-Abelian group $G$ and call its operation addition and its identity 0. Define multiplication by $\forall a, b \in G: ab = 0$. This satisfies the axioms of a ring without identity except for the commutativity of addition.

badjohn
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  • I went on to try to think of more interesting examples but the links provided by the others have done that job well. It is worth remembering this trivial multiplication as a test case for hypotheses. – badjohn Jun 12 '17 at 09:31