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Let $R$ be a commutative and associative ring with unit. Can $R$ be embedded in a ring $\hat{R}$ wich is both non commutative and non associative ?

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closed as too localized by Qiaochu Yuan, Andreas Blass, Bruce Westbury, Pete L. Clark, Mark Sapir Nov 4 '11 at 17:10

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What do you mean by a nonassociative ring? –  MTS Nov 4 '11 at 16:34
    
And what do you mean by embedded? If I say that a ring $A$ is embedded in another ring $B$, I mean that there is an injective ring homomorphism from $A$ to $B$. What kind of morphism do you want from your associative ring to your nonassociative "ring"? –  MTS Nov 4 '11 at 16:36
    
I posted very similar comments, but then deleted them when I realized there is a wikipedia article on nonassociative rings, and it's the first hit from a google search. The second question is harder to answer...I don't know what subobjects are for these so-called non-associative rings. –  David White Nov 4 '11 at 16:51

1 Answer 1

Yes. The "can" question is not so interesting: think of the image of the integers inside any nonassociative algebra. But specific cases, and counting embeddings, are interesting. See work of Gross and Gan, "Commutative Subrings of Certain Non-associative Rings", Math. Ann. v.314 n.2, 1998.

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