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Is there a specific terminology for rings in which we do not require associativity on the multiplication ? Like, for example we say semiring for a ring where we do not require the additive inverses. Any references are appreciated.

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    $\begingroup$ There are Non-associative algebras, like Lie algebras or Jordan algebras. There are also Lie rings and non-associative rings, see here, or here. $\endgroup$ – Dietrich Burde Oct 23 '18 at 15:29
  • $\begingroup$ Thank you, I already know a little bit about non-associative algebras. I was wondering if the specific case of rings has a terminology, and deserves attention. $\endgroup$ – Sov Oct 23 '18 at 15:34
  • $\begingroup$ Yes, it does. Already Lie rings deserve attention. $\endgroup$ – Dietrich Burde Oct 23 '18 at 15:41
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Generically, this is normally just called a nonassociative ring.

I don't think there is a common term like "semiring," or at least it is not commonly used.

It is just like "noncommutative ring." There is no cute pithy term, just a straightforward adjective.

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  • $\begingroup$ But both of the accepted definitions of the term 'ring', although differing as to whether the existence of a multiplicative identity is or isn't required, agree in that both require associativity and neither requires commutativity, so the parallel between 'nonassociative ring' and 'noncommutative ring' is not exact. The former (as the referenced quotation makes clear) generalises the term 'ring', but the latter does not. (I'm sorry to quibble! But the ambiguity of the term 'ring' almost obliges one to be pedantic.) $\endgroup$ – Calum Gilhooley Oct 23 '18 at 17:19
  • $\begingroup$ @CalumGilhooley Standard practice is that "noncommutative ring" means "not necessarily commutative" and I apply it the same way with "non associative ring" meaning "not necessarily associative ring", so both terms 'generalize' ring for me. But I do not know for sure if people treat "nonassociative ring" like I do. But yes, as you mentioned, there is no universal consensus on terms, so one can't really say with certainty that the dichotomy you mention exists. $\endgroup$ – rschwieb Oct 23 '18 at 17:26
  • $\begingroup$ I meant that the unadorned term 'ring' means 'noncommutative ring' (in precisely your sense, i.e. 'not necessarily commutative ring'), but it does not mean 'nonassociative ring' (again in precisely your sense), therefore the two longer terms are not 'just like' one another. (I hope I may stop quibbling now!) $\endgroup$ – Calum Gilhooley Oct 23 '18 at 17:37

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