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In the question linear algebra over a division ring vs. over a field is discussed the relationship between linear algebra over a field and linear algebra over a division ring. Roughly speaking the quation is: what happens if the scalars do not commute?

This led me to think about the following: what happens if the scalars do not commute and are also not associative?

Are there any theorems that hold for a module over an associative division ring but do not hold for a module over a nonassociative division ring?

Remark: Let $D$ be a nonassociative division ring. A left $D$-module $M$ consists of an abelian group $(M, +)$ and an operation $D × M → M$ such that for all $r, s \in D$ and $x, y \in M$, we have:

  1. r(x+y) = rx + ry

  2. (r+s)x = rx + sx

  3. 1.x = x.

A right $D$-module is defined similarly.

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    $\begingroup$ What is your definition of a module over a non-associative ring $R$? If you mimic the usual definition, then you will get $((rs)t)m = (rs)(tm) = r(s(tm))=r((st)m) = (r(st))m$. This shows that your module action factors over $R/((rs)t=r(st))_{r,s,t \in R}$, the free associative ring on $R$, which places you within usual module theory. $\endgroup$
    – Martin Brandenburg
    Mar 28, 2015 at 14:11
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    $\begingroup$ Good comment Martin. This is studied in "Osborn, J. Marshall Modules over nonassociative rings. Comm. Algebra 6 (1978), no. 13, 1297–1358", but I do not have access to this article . $\endgroup$
    – zacarias
    Mar 28, 2015 at 14:49

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