What happens when you don't require rings to be abelian groups?

Asked Dec 12 '16 at 05:46

Active Dec 12 '16 at 08:18

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My question is as stated in the title: What do we lose when we drop the condition that a ring $R$ be an abelian group (i.e. it may be the case that $a+b \neq b+a$ for some $a,b \in R$).

If we allow this condition then we may consider, for example, any group $G$ to be a (noncommutative) ring with multiplication being conjugation in $G$ and addition being multiplication in $G$. In such a "ring" $G$, we always have $0=1$, so I suspect things might get strange.

Has anyone thought at all about such objects? Are they entirely uninteresting?

edited Dec 12 '16 at 08:18

user26857

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asked Dec 12 '16 at 05:46

Exit path

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This answer may help you: http://math.stackexchange.com/a/346682/133781 – Xam Dec 12 '16 at 05:48
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@Xammm It definitely does, thank you! – Exit path Dec 12 '16 at 05:50
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http://math.stackexchange.com/questions/609364/why-is-ring-addition-commutative?noredirect=1&lq=1 – Pawel Dec 12 '16 at 05:54
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They (semirings) tend to be more interesting to computer scientists (who find them very interesting) than mathematicians. Computer scientists in general tend to be more interesting in monoids than groups. – Derek Elkins left SE Dec 12 '16 at 08:28
Well, since this question is about dropping commutativity of the underlying group (not inverses), semirings are not so relevant as nearrings. – rschwieb Dec 12 '16 at 17:40

What happens when you don't require rings to be abelian groups?

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