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Consider a noncommutative ring with unity $R$, three left $R$-modules $M,N,P$ and a map $f\colon\;M\times N\to P$ such that:

$ f(m+m',n)=f(m,n)+f(m',n)\\ f(m,n+n')=f(m,n)+f(m,n')\\ f(rm,n)=rf(m,n)=f(m,rn) $

A map like this does exist: the null one. What about others? Can exist a non-null map satisfying those axioms? From the third point, given two scalars $r,s\in R$, it is evident that $(rs-sr)f(m,n)=0$ for all $m,n$ and this could be a problem given that $R$ is non-commutative.

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Such a map can exist, but not for every noncommutative ring. For example, no such map can exist over a division ring. This is because the annihilator of the left tensor product module is a nonzero left ideal, hence the left tensor product is always trivial in that case.

For an example where the map does exist, let $T$ be the tensor algebra over a field, and let $I$ be the two sided ideal generated by $rs-sr$. The quotient is the symmetric algebra $S$ over the field. Let $f:S\times S\to S$ be given by multiplication. Then this is a nonzero left bilinear map over $T$.

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  • $\begingroup$ +1 very natural. The last paragraph is just a convenient illustration of a ring where the commutator ideal is proper, so (moppio89) if you have any other such examples in mind, they'll work too. $\endgroup$
    – rschwieb
    Dec 15 '14 at 15:35
  • $\begingroup$ For example, the commutator ideal in the upper triangular matrix ring $T_2(F)$ is just the strictly upper triangular matrices. The quotient is $F\times F$, and that would work as another example. $\endgroup$
    – rschwieb
    Dec 15 '14 at 15:38

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