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I have come across this paper. The authors prove that magmas satisfying the identity $$(xy)z=y(zx)\tag1$$ are nearly both associative and commutative. To be precise, they show that in such magmas, products of at least five elements are independent of the placement of brackets and the order of elements. They call this property five-niceness.

Then they say that if a non-associative ring (which means not necessarily associative) has multiplication satisfying this identity and the ring is semiprime, then it is both associative and commutative. Semiprimality means that for an ideal $I,$ we have $I^2=0\implies I=0.$

I have no experience with non-associative rings whatsoever. In particular, I have never seen a definition of an ideal in such a ring. I tried to find one on the internet but in vain. So my main question is

(a) What are ideals in non-associative rings?

And where can I read about them? Are they as useful in the theory of non-associative rings as ideals in associative rings? It's very difficult for me to start "thinking non-associatively" and I'm having trouble seeing what the problems could be.

Finally, I would like to ask about the identity. I probably have little chance of receiving an answer to these questions, but there's no harm in trying.

(b) Is there an example of a ring that satisfies $(1)$, is associative, but isn't commutative?

(c) Is there an example of a ring that satisfies $(1)$, is commutative, but isn't associative?

(d) Is there an example of a ring that satisfies $(1)$, but is neither commutative nor associative?

I do not require the rings to have identities.

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An ideal for a non-associative ring $R$, also say two-sided ideal, is just an additive subgroup $I \subseteq R$ such that for each $r \in R$ and $i \in I$ you have that $ri,ir \in I$. You can find the definition in the wikipedia page. –  Giorgio Mossa Apr 26 '12 at 12:50
    
It's easy to see that ideals are all the kernels of ring homomorphisms. –  Giorgio Mossa Apr 26 '12 at 12:51
    
Do your ring contain 1? If yes, then $z=1$ guarantees commutativity. If no, then sums of non-empty words with two letters subject to associativity and cyclic rotation of three factors should work for (b). –  Phira Apr 26 '12 at 12:57
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I'd like to see an explicit example of a magma satisfying that condition which isn't associative or commutative. A natural one if possible. What if this turns out like the phd student who wrote his paper on Holder continuous maps with exponent greater than $1$? –  Ragib Zaman Apr 26 '12 at 12:57
    
@RagibZaman Take a look at this paper. There's an example on page 201 of a magma that satisfies all possible identities in three variables but isn't commutative. –  user23211 Apr 26 '12 at 13:41
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I think you need to gain some general experience with nonassociative rings and algebras. Many of the algebras we encounter in classroom mathematics are associative,but there are several important types of nonassociative algebras that appear in both algebra and particle physics.There are very few general algebra texts that discuss them to any depth.The major exception is Nathan Jacobson's Basic Algebra-Jacobson,of course,was one of the most active developers of this area after A.A.Alberts , who was probably the first major algebraicist to do research in nonassociative rings and algebras. The most famous examples are, of course,the Lie rings and algebras.But there are several other important classes: the octonions, alternative algebras and the Jordan algebras. And believe it or not,one of the most famous examples is usually studied in high school calculus and is almost never described this way: Three dimensional Euclidean space is a nonassociative algebra under ordinary vector addition and the cross product.In fact, it is both noncommutative and nonassociative under the cross product. (Try and prove it-good exercise for a beginning algebra student!) For much more on nonassociative algebras, the standard text is Richard Schafer's An Introduction To Nonassociative Algebras. A more recent and comprehensive source is Kevin McCrimmon's A Taste Of Jordan Algebras . I think a lot of your questions will be answered after a deep study of these sources. Good luck!

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Thanks a lot! That's a lot of material so I'll have probably forgotten this question by the time I've studied enough to be prepared to confront them, but that's fine with me. :) –  user23211 Apr 26 '12 at 17:26
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