I wonder about the following: Let $R$ be a noncommutative ring with $\alpha, \beta \in R$ such that $\alpha \beta \neq \beta \alpha$. Let $M,N,P$ be left-$R$-modules. Why does it make problems to define a map $B: M \times N \to P$ to be bilinear if $B(\lambda m,n) = \lambda B(m,n)$ and analogously for the second argument (and $B$ additive in both components)? That is, why does it make problems to allow pulling out scalars? We would get $\alpha \beta B(m,n) = \beta \alpha B(m,n)$. Why does that make the definition useless? In general, from $(\alpha \beta - \beta \alpha) B(m,n) = 0$ it doesnt follow $B(m,n) = 0$ (when $(\alpha \beta - \beta \alpha)$ is not a zero divisor in $R$), so i dont see the problem. I've read the "correct definition" of a tensorproduct/bilinear map over noncommutative rings begins like this: take a right-$R$-module $M$ and a left-$R$-module $N$ and an abelian group $P$. Then it makes sense to talk about bilinear maps $B: M \times N \to P$: we want them to satisfy $B(\alpha m,n) = B(m, \alpha n)$, i.e., instead of allowing to pull out scalars, we only are allowed to push them from one argument to the other. Why?

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    $\begingroup$ It might not be useless, but it "forgets" the fact that $R$ was not commutative by in fact becoming bilinear maps for modules over the commutative ring $R/I$ where $I$ is the ideal generated by all $ab - ba$. $\endgroup$ Nov 18 '14 at 12:35
  • $\begingroup$ @TobiasKildetoft That should be an answer. $\endgroup$ Nov 19 '14 at 14:23
  • $\begingroup$ @OscarCunningham I was hoping someone would answer who also had a better idea of why the "correct" version is as it is (since then I would not need to work it out myself). $\endgroup$ Nov 20 '14 at 8:03

Usually it's wiser to write the right module actions on the right of the module elements, so the last equation you have should be

$B( m\alpha,n) = B(m, \alpha n)$ since $M$ is a right module. You can look upon it as a sort of associativity condition. Remember that the prototype for bilinear operations is ring multiplication. Things are a different than a ring here, but still the basic idea of "I know how to do $m(\alpha n)$ and $(m\alpha)n$. In general these could be different, but if they are the same then I don't need parentheses and I just have $m\alpha n$."

This particular compatibility ingredient is usually called (in my experience) being "balanced." So by way of illustrating a balanced multiadditive map on the modules $M_R$, $_SP$ and the bimodule $_RN_S$, it would be a map from $T:M\times N\times P\to G$ where $G$ is an abelian group, such that the map is additive in $M$ and $N$ and $P$ independently, and the following "balancedness" requirements are met:



Of course, this can be generalized to any number of (bi)modules.

You mentioned "pulling out scalars," which I guess means something like $\alpha B(m,n)=B(\alpha m,n)$. This is also possible, but for that to be true you'd ask for a left bimodule structure on $M$. So if $_RM_S$ and $_SN$ are (bi)modules, you can show that the set of balanced maps defined with just $M_S$ and $_SN$ have a natural left $R$ module structure defined by $rB(m,n)=B(rm,n)$. Similarly, if $N$ is a bimodule, you can pull-in/push-out a scalar on the far right hand side: $B(m,nt)=B(m,n)t$.

  • $\begingroup$ The question asks about bilinear maps out of two left modules, but you say $m$ is a right module. Does $M$ have to be a right module? What stops working otherwise? $\endgroup$ Nov 20 '14 at 20:57
  • $\begingroup$ @OscarCunningham I was addressing the last paragraph where it says take a right-R-module M and the overall consistency of the "right definition." But you're right, I could try harder to address the first part about "why doesn't this other definition work." $\endgroup$
    – rschwieb
    Nov 20 '14 at 21:16

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