MHB Making a Right R-module Into a Left R-Module - Bland, page 26 ....

[Math Amateur]

I am reading Paul E. Bland's book: Rings and Their Modules and am currently focused on Section 1.4 Modules ... ...

On page 26 Bland defines a right $$R$$-module. After giving the definition, Bland discusses making a right $$R$$-module into a left $$R$$-module when $$R$$ is a non-commutative ring ... ... I need help with some aspects of that discussion ... ...

Bland's definition of a right $$R$$-module together with his discussion of making a right $$R$$-module into a left $$R$$-module when $$R$$ is a non-commutative ring reads as follows:
View attachment 4897
Now, in the above text Bland writes the following:

" ... ... If $$R$$ is a noncommutative ring, then a right $$R$$-module cannot be made into a left $$R$$-module by setting $$ a \cdot x = xa $$. ... ... "Now, I think Bland means that if we are given a right $$R$$-module over a noncommutative ring $$R$$, then we have conditions (1) to (4) holding as in Definition 1.4.1 above ...

We then try to make this right $$R$$-module into a left $$R$$-module by, essentially, defining a new 'scalar product' $$\cdot$$ by simply setting $$a \cdot x = xa$$ so that we alter (1) to (4) ...

So, as I understand it, the terms of (1) will alter as follows:

$$x(a + b)$$ becomes $$(a + b) \cdot x$$

and

$$xa$$ becomes $$a \cdot x$$

and

$$xb$$ becomes $$b \cdot x$$

so that (1) above now reads as follows:

$$(a + b) \cdot x = a \cdot x + b \cdot x$$ ... ... ... ... ... (*)

which reads as if we had used $$\cdot$$ as the binary operation $$R \times M \rightarrow M$$ such that $$(a, x) \mapsto a \cdot x$$ ... ... that is, as if we had defined a left action of $$R$$ on $$M$$ ... ... and ... ... (1) then reads correctly (as (*) ) for a left $$R$$-module ... ...

Bland then assures us that the same occurs for (2) and (4) ... but mentions that simply setting $$a \cdot x = xa$$ will not work in the same way for (3) ... ...Is my reasoning correct so far ... can someone confirm this and/or point out shortcomings or errors ... ...
Then Bland goes about demonstrating that this does not work for (3) by writing the following:

" ... ... If setting $$a \cdot x = xa$$ for all $$x \in M$$ and $$a \in R$$ were to make $$M$$ into a left $$R$$-module, then we would have:

$$ (ab) \cdot x = x(ab) = (xa)b = b \cdot (xa) = b \cdot (a \cdot x) = (ba) \cdot x $$

... ... for all $$x \in M$$ and $$a, b \in R$$. ... ... "

... BUT ... I cannot see how $$a \cdot x = xa$$ and the 'rule' (3) gives us

$$b \cdot (a \cdot x) = (ba) \cdot x$$
Can someone please show me how/why this is true given only a \cdot = xa and the 'rule' (3) ... ... ?
Bland then writes the following:

" ... ... Examples of left $$R$$-modules over a noncommutative ring abound, where $$(ab) \cdot x \neq (ba) \cdot x$$ ... ... "

Can someone please give an illustrative example (or two) of a left R-module over a noncommutative ring where $$(ab) \cdot x \neq (ba) \cdot x $$
Hope someone can help with the above questions ... ...

Peter

 

[steenis]

This is an old post, do you still need an answer?

 

[Math Amateur]

This is an old post, do you still need an answer?

HI Steenis ... yes, I still need an answer ... no one has ever replied to this post of mine ... and I have never resolved the issue ...

Peter

 

[steenis]

Question 1
Your reasoning is correct so far.
We have to prove (1), (2), (3), and (4) of definition 1.4.1 adapted for the new scalar product.
For instance we have to prove (1): $(a+b)\bullet x = a\bullet x + b\bullet x$.
Proof: $(a+b)\bullet x = x(a+b) = xa + xb = a\bullet x + b\bullet x$
(2) and (4) are also easy to prove.

Question 2
But (3) is causing problems.
(3) becomes $b\bullet (a\bullet x) = (ba)\bullet x$; we have to prove this.
Proof: $b\bullet (a\bullet x) = b\bullet (xa) = (xa)b = x(ab) = (ab)\bullet x = ? = (ba)\bullet x$
You see, we cannot prove this unless $R$ is commutative: $ab=ba$.
Therefore: if R is noncommutative, the right R-module M canNOT be made into a left R-module by setting $a\bullet x = xa$.

Question 3
I think Bland meant: "right R-modules"

Define $R=M_2 (\Bbb R)$ as a right R-module over itself.

And take:

$x = \left (\begin{matrix} 1&0 \\ 0&1 \end{matrix} \right )$

$a = \left (\begin{matrix} 1&0 \\ 0&0 \end{matrix} \right )$

$b = \left (\begin{matrix} 2&2 \\ 2&2 \end{matrix} \right )$

Then:

$(ab)\bullet x = x(ab) = \left (\begin{matrix} 2&2 \\ 0&0 \end{matrix} \right )$

$(ba)\bullet x = x(ba) = \left (\begin{matrix} 2&0 \\ 2&0 \end{matrix} \right )$