Tensor Products of Modules - Bland - Remark, Page 65

In summary, Bland is discussing the difficulties in proving the well-definedness of a map g in Section 2.3 of his book "Rings and Their Modules". This is due to the large size of cosets and the need to ensure that g remains constant on cosets. Bland also mentions the R-balanced map h = \rho' ( f \times id_N ) and demonstrates its R-balanced property by showing it satisfies the necessary conditions.
  • #1
Math Amateur
Gold Member
MHB
3,998
48
I am reading Paul E. Bland's book "Rings and Their Modules ...

Currently I am focused on Section 2.3 Tensor Products of Modules ... ...

I need some help in order to fully understand the Remark that Bland makes on Pages 65- 66

Bland's remark reads as follows:View attachment 5648
View attachment 5651
Question 1

In the above text by Bland we read the following:

"... ... but when \(\displaystyle g\) is specified in this manner it is difficult to show that it is well defined ... ... "

What does Bland mean by showing \(\displaystyle g\) is well defined and why would this be difficult ... ...Reflection on Question 1Reflecting ... ... I suspect that when Bland talks about \(\displaystyle g\) being "well defined" he means that if we choose a different element ... say, \(\displaystyle \sum_{ i = 1}^m n'_i ( x'_i \otimes y'_i )\) in the same coset as \(\displaystyle \sum_{ i = 1}^m n_i ( x_i \otimes y_i )\) ... ... then \(\displaystyle g\) still maps onto \(\displaystyle \sum_{ i = 1}^m n_i ( f(x_i) \otimes y_i ) \) ... ... is that correct ...
Question 2

In the above text by Bland we read the following:

"... ... Since the map \(\displaystyle h = \rho' ( f \times id_N )\) is an R-balanced map ... ... "Why is \(\displaystyle h = \rho' ( f \times id_N )\) an R-balanced map ... can someone please demonstrate that this is the case?
Hope someone can help ... ...

Peter==================================================================================The following text including some relevant definitions may be useful to readers not familiar with Bland's textbook... note in particular the R-module in Bland's text means right R-module ...
View attachment 5650
 
Last edited:
Physics news on Phys.org
  • #2
Yes, one has to ensure that if $x_i \otimes y_i = x'_i \otimes y'_i$ that $f(x_i) \otimes y_i = f(x'_i) \otimes y'_i$, which can be hard to do, since the coset containing $(x_i,y_i)$ is often incredibly large (the sums can then be put together "by extending by linearity"). So your answer about $g$ is correct, it must be constant on cosets (and not depend on the representative pairs $(x_i,y_i)$).

Let's just demonstrate $h$ is $R$-balanced:

$h(x_1+x_2,y) = [\rho' \circ (f \times 1_N)](x_1 + x_2,y)$

$= \rho'(f(x_1+x_2),y) = \rho'(f(x_1) + f(x_2),y)$ (since $f$ is a module homomorphism)

$= [f(x_1)+f(x_2)]\otimes y$ (since $\rho'$ is the canonical tensor map)

$= f(x_1)\otimes y + f(x_2)\otimes y$ (since the canonical tensor map is, by construction, $R$-balanced)

$=\rho'(f(x_1),y) + \rho'(f(x_2),y)$ (definition of what $\rho'$ is)

$ =[\rho'\circ(f\times1_N)](x_1,y) + [\rho'\circ(f \times 1_N)](x_2,y)$

$= h(x_1,y) + h(x_2,y)$.

Similarly,

$h(x,y_1+y_2) = [\rho'\circ(f \times 1_N)](x,y_1+y_2)$

$= \rho'(f(x),y_1+y_2) = f(x)\otimes(y_1+y_2) = f(x)\otimes y_1+f(x)\otimes y_2$

$= [\rho'(f \times1_n)](x,y_1) + [\rho'\circ(f \times 1_N)](x,y_2)$

$= h(x,y_1) + h(x,y_2)$.

Finally, for $a \in R$:

$h(xa,y) = [\rho'\circ(f\times 1_N)](xa,y)$

$= \rho'(f(xa),y) = \rho'(f(x)a,y) = f(x)a\otimes y$

$= f(x) \otimes ay = \rho'(f(x),ay) = [\rho'\circ(f \times 1_N)](x,ay) = h(x,ay)$

Note that we are "wedding" a right-module to a left-module "over the middle". If $R$ is commutative, we don't have to take such precautions, and can just use the more standard term $R$-bilinear.
 

FAQ: Tensor Products of Modules - Bland - Remark, Page 65

1. What is a tensor product of modules?

A tensor product of modules is a way of combining two modules over a ring into a single module. It is denoted by the symbol ⊗ and has properties similar to those of a vector space.

2. How is a tensor product of modules defined?

A tensor product of modules is defined as the quotient of the free module generated by the Cartesian product of the two given modules, divided by a certain submodule generated by elements of the form (am, b) - (a, mb), where a and b are elements of the respective modules and m is an element of the ring.

3. What is the significance of tensor products of modules in mathematics?

Tensor products of modules are important in mathematics because they allow us to extend the concept of linear combinations and linear transformations to modules over a ring, which may not have a multiplication operation defined.

4. Can tensor products of modules be extended to other algebraic structures?

Yes, tensor products can be extended to other algebraic structures such as algebras, groups, and Lie algebras. However, the definition and properties may vary slightly depending on the specific structure.

5. Are there any applications of tensor products of modules in real-world problems?

Tensor products of modules have various applications in fields such as physics, engineering, and computer science. For example, they are used in quantum mechanics to describe the entangled states of multiple particles, and in signal processing to analyze multi-dimensional signals.

Back
Top