MHB Free modules, M_n(R)-modules and M_N(R)-R-bimodules - Berrick and Keating

[Math Amateur]

I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).

In Chapter 1: Basics under "1.2.3 Bimodules" B&K introduce some modules that they say play a key role in the text.

To ensure I understood these I wrote out some details of each example module (using $$R^2$$ instead of $$R^n$$ for ease of presentation) ... BUT ... I am unsure of my interpretation and need someone to critique my analysis ... and, hopefully confirm it is correct.

Section "1.2.3 Bimodules" in B&K reads as follows:https://www.physicsforums.com/attachments/3353In the above text B&K write:

" ... ... Let $$R^n$$ be the free right $$R$$-module of rank $$n$$, that is, the set of $$n \times 1$$ matrices over $$R$$. The usual addition and multiplication of matrices makes $$R^n$$ into a left $$M_n(R)$$-module and an $$M_n(R)-R$$-bimodule ... ... "

I will now write out these examples (cases) for $$R^2$$ ... ...

Case 1 $$R^2$$ as the free right module of rank $$2$$ Now, $$R^2$$ as the free right module of rank $$2$$ consists of the set of $$n \times 1$$ matrices over $$R$$, such as:

$$a = \begin{bmatrix}a_1\\a_2 \end{bmatrix}$$, $$ \ b = \begin{bmatrix}b_1 \\ b_2 \end{bmatrix}$$ where $$a_1, a_2, b_1, b_2 \in R$$

Briefly, addition, as in:

$$a + b = \begin{bmatrix}a_1\\a_2 \end{bmatrix} + \begin{bmatrix}b_1 \\ b_2 \end{bmatrix} = \begin{bmatrix}a_1 + b_1 \\a_2 + b_2 \end{bmatrix}$$

gives an additive abelian group and the right module action (scalar multiplication) defined by:

$$ar = \begin{bmatrix}a_1\\a_2 \end{bmatrix} r = \begin{bmatrix}a_1 r \\a_2 r \end{bmatrix}$$ for $$a \in R^2, r \in R$$

creates (together with the above abelian group) a right R-module.
Case 2 Make $$R^2$$ into a left $$M_n(R)$$-module

Define $$a + b$$ as in Case 1 above to make $$R^2$$ an additive abelian group ...

Define a left action $$ra$$, where $$r \in M_2(R)$$ and $$a \in R^2$$ ... so typically, for:

$$a = \begin{bmatrix}a_1\\a_2 \end{bmatrix}$$ and $$ r = \begin{bmatrix}c_{11} & c_{12} \\ c_{21} & c_{22} \end{bmatrix}$$

we have

$$\begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{bmatrix} \begin{bmatrix} a_1 \\a_2 \end{bmatrix} = \begin{bmatrix} c_{11}a_1 + c_{12}a_2 \\ c_{21}a_1 + c_{22}a_2 \end{bmatrix}$$

which creates a left $$M_2(R)$$-module
Case 3 Make $$R^2$$ into an $$M_2(R)-R$$-bimodule Now make $$R^2$$ into an additive abelian group as in case 2.

For this case there are two actions:

(i) a left action $$ma$$ where $$m \in M_2(R)$$ and $$a \in R^2$$ as in Case 2 above

and

(ii) a right action $$ar$$ where $$r \in R$$ and $$a \in R^2$$ as in case 1 above ...

These two actions create a $$M_2(R)-R$$-bimodule where we must have:

$$m(ar) = (ma)r$$

Can someone please critique my analysis and point out any errors or misinterpretations OR confirm that the above analysis is correct ...

Peter

 

[Deveno]

So far, so good, but you must SHOW:

$(ma)r = m(ar)$.

But this can be deduced from the fact matrix multiplication (where defined) is associative. Or, explicitly, that both products evaluate to:

$\begin{bmatrix}(c_{11}a_1 + c_{12}a_2)r\\(c_{21}a_1 + c_{22}a_2)r\end{bmatrix}$

by invoking right-distributivity of $R$.

We have some interesting parallels vis-a-vis the left/right dilemma:

left/right
domain/co-domain
row/column

various interesting constructions arise in (module) algebra from "swapping" the two. You should try to see these kinds of things as "essentially alike" (for example, in linear algebra ($F$-modules), "dual vectors" are just row-vectors, instead of "column-vectors", and in matrices, we have the transpose of a matrix. Mathematicians like to use the same bag of tricks over and over again: hey, it worked once, why not use it again...?)