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

[Math Amateur]

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 $$g$$ 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 ... ...Reflection on Question 1Reflecting ... ... I suspect 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 ...
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 ...
View attachment 5650

 

[Deveno]

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.