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When defining tensor products $M\otimes_R N$ over a commutative ring $R$ one can use a universal property with respect to bilinear maps $M\times N\rightarrow P$.

On the other hand, in the general case, for noncommutative rings one has to use balanced maps $M \times N \rightarrow Z$ instead of bilinear. Of course, in the first case $P$ is an $R$-module while in the second case $Z$ is just an abelian group.

Remember $f$ is biliniar if $f(mr, n)=rf(m, n)=f(m, nr)$ while $f$ is balanced if and only if $f(mr,n)=f(m, rn)$.

I have the following two natural questions:

  1. Why the two definitions coincide?
  2. Is there an example of a balanced map $M\times N \rightarrow P$ which is not bilinear? I cannot construct one by myself. Here I assume R commutative so I can speak about bilinear maps.
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The right bilinear maps would yield a tensor product of right modules if we construct the corresponding universal object, where $mr\otimes n=m\otimes nr$. The tensor product of two right modules is again a right module. The corresponding construction can also be done for left modules, which is a left module, and we can also take the tensor product of a right module and a left module, which is where the balanced maps come in. In general the tensor product of a right and a left module is only an abelian group as you mentioned, though it can be made into a module if there are module structures on the side not being "boxed in" by the tensor product. For this construction the coefficient ring can also be commutative, as it is possible for a module to have different actions by a commutative ring on the left and the right, which happens for example when we consider the skew group ring over a commutative ring that the group acts on.

If $R$ is any noncommutative ring, the multiplication map $(r,s)\mapsto rs$ is a balanced (also called middle linear) map that is not bilinear on the left or the right.

Now let $R$ be commutative of characteristic not equal to 2. Let $S$ be the free left $R[x]$ module with basis $\{1,y,y^2,\ldots\}$ and define a right action of $R[x]$ by $yx=-xy$. $S$ has the structure of a ring, essentially a polynomial ring in variables that do not commute. The multiplication map $S\times S\to S$ is then an $R[x]$-balanced map that is not bilinear.

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  • $\begingroup$ thanks! I would like an example in the case R is commutative; M, N, P three R-modules and an example of a balanced map $M\times N\rightarrow P$ which is not bilinear. $\endgroup$ – mathuser Dec 10 '14 at 8:35
  • $\begingroup$ @mathuser please see my edit. $\endgroup$ – Matt Samuel Dec 10 '14 at 14:14

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