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?

  • 1
    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 $ab - ba$. – Tobias Kildetoft Nov 18 '14 at 12:35
  • @TobiasKildetoft That should be an answer. – Oscar Cunningham Nov 19 '14 at 14:23
  • @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). – Tobias Kildetoft 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:

$T(mr,n,p)=T(m,rn,p)$

$T(m,ns,p)=T(m,n,sp)$

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: $B(m,nt)=B(m,n)t$.

  • 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? – Oscar Cunningham Nov 20 '14 at 20:57
  • @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." – rschwieb Nov 20 '14 at 21:16

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.