# Tensor Products of Modules - Bland - Remark, Page 65

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• Math Amateur
In summary, in Section 2.3 of Paul E. Bland's book "Rings and Their Modules," Bland discusses tensor products of modules and makes a remark on pages 65-66 about the difficulty in showing that a specific map, g, is well defined. This means that for any element in the same coset, g must still map onto the same output. Bland also mentions that the map h = ρ'(f × id_N) is an R-balanced map. This means that for any elements a and c in M and b and d in N, if (a,b) and (c,d) are in the same coset, then (f(a),b) and (f(c
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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

Question 1

In the above text by Bland we read the following:

"... ... but when ##g## is specified in this manner it is difficult to show that it is well defined ... ... "

What does Bland mean by showing ##g## is well defined and why would this be difficult ... ...Question 2

In the above text by Bland we read the following:

"... ... Since the map ##h = \rho' ( f \times id_N )## is an R-balanced map ... ... "Why is ##]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 ...

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Good morning MA. I wonder if you are affected by the wild weather this morning from the East Coast Low in the Tasman Sea.

In the example given, the purported definition of $g$ is that $g(x\otimes y)=f(x)\otimes y$. If we think about what this means, it is saying that

$$g\Big(\chi_{(x,y)}+K\Big)=\chi_{(f(x),y)}+K'$$

where $K$ and $K'$ are the submodules of $Z$ and $Z'$ respectively that have the required 'zero' properties inherited from the modules $M,M',N$; and $Z$ and $Z'$ are the free R-modules generated by ${\{\chi_{(u,v)}\ :\ u\in M\wedge v\in N\}}$ and ${\{\chi_{(u,v)}\ :\ u\in M'\wedge v\in N\}}$ respectively.

For this to be well-defined, we require that, for any $a,c\in M$ and $b,d\in N$ such that $(a,b)$ and $(c,d)$ are in the same coset of $K$, it will be the case that $(f(a),b))$ and $(f(c),d)$ are in the same coset of $K'$. Otherwise, $g$ will not be a function because $g\Big(\chi_{(a,b)}+K\Big)\neq g\Big(\chi_{(c,d)}+K\Big)$ even though $\chi_{(a,b)}+K=\chi_{(c,d)}+K$. Hence we have two different output values for a single input.

The argument involving the diagram appears to be a proof that $g$, thus defined, is indeed well-defined.

Math Amateur
Hi Andrew ... thanks for helping ///

I am currently in regional Victoria ... back in Tasmania soon ... weather wet and cold here but not wild like around Sydney and the NSW coastline ...

Regarding you post ... yes, understand ... so what you are saying is ... ... that when Bland talks about ##g## being "well defined" he means that if we choose a different element ... say, ##\sum_{ i = 1}^m n'_i ( x'_i \otimes y'_i )## in the same coset as ##\sum_{ i = 1}^m n_i ( x_i \otimes y_i )## ... ... then ##g## still maps onto ##\sum_{ i = 1}^m n_i ( f(x_i) \otimes y_i ) ## ... ... is that correct ...?

Yes, that's right.

Math Amateur
andrewkirk said:
Yes, that's right.
Thanks Andrew ...

Peter

## 1. What is a tensor product of modules?

A tensor product of modules is a mathematical operation that combines two modules to create a new module. It is denoted by the symbol ⊗ and is defined by a set of rules that describe how elements of the two modules interact with each other.

## 2. How is a tensor product of modules calculated?

The tensor product of two modules M and N is calculated by taking the direct sum of all possible combinations of elements from M and N, and then taking the quotient space by the submodule generated by certain elements that satisfy the bilinearity condition.

## 3. What is the significance of tensor products of modules?

Tensor products of modules have many applications in mathematics, physics, and engineering. They are particularly useful in linear algebra and representation theory, and can also be used to study algebraic varieties and geometric objects.

## 4. Can tensor products of modules be commutative?

No, tensor products of modules are not commutative. This means that M ⊗ N is not necessarily equal to N ⊗ M. However, they are associative, meaning that (M ⊗ N) ⊗ P is isomorphic to M ⊗ (N ⊗ P) for any three modules M, N, and P.

## 5. Are there any special properties of tensor products of modules?

Yes, there are several special properties of tensor products of modules. These include properties related to exact sequences, projective and injective modules, and flat modules. Tensor products also satisfy the universal property, meaning that they can be uniquely defined by certain bilinear maps.

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