Standard Free Right R-Module on X - B&K Section 2.1.16

[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 Chapter2: Direct Sums and Short Exact Sequences in Section 2.1.16 B&K deal with the standard free right ##R##-module on a set ##X##. I need some help with understanding an aspect of the authors' discussion ... ...

Section 2.1.16 reads as follows:

In the above text by B&K they construct the standard free module on ##X## as the module##\text{Fr}_R (x) = \bigoplus_X xR##so that the elements of ##\text{Fr}_R (x)## are formal sums ##m = \sum_{x \in X}x r_x (m)##

with ##r_x (m) \in R## ... ...OK ... so far so good ... BUT ...

... can someone please explain to me how this enables B&K to say ..." ... ... It is clear that if ##\Lambda## is any convenient ordering of ##X##, we have


##\text{Fr}_R (x) = R^{ \Lambda }## ... ... "---------------------------------------------------------------------------------------------------------------------------------

To make my question more explicit ... suppose that ##X## is a finite ordered set ##X = \{ x_1, x_2, x_3 \}##

Then ##\text{Fr}_R (x) = \bigoplus_X xR = x_1 R + x_2 R + x_3 R##

... ... and elements of ##\text{Fr}_R (x)## would be of the form ##m = x_1 r_1 + x_2 r_2 + x_3 r_3##BUT ... to repeat my question ... with elements of this form how can we argue that
if ##\Lambda## is any convenient ordering of ##X##, we have ##\text{Fr}_R (x) = R^{ \Lambda }## ... ?
Note: I suspect that B&K may be expecting the reader to identify ##\sum x r_x(m)## with ##\sum r_x(m)## ...

... BUT ... if this is the case ... why introduce ##X## into a construction only to "identify" it away ... why not just define the standard free right ##R##-module as ##\text{Fr}_R (x) = R^{ \Lambda }## ?

Hope someone can help ...

Peter

 

[fresh_42]

I'm not quite sure, if I can follow the authors correctly. To me it seems a bit more complicated than necessary, or at least overly precise. As I understand them, say construct their free ##R##-module over ##X## as ##R^X=R^{|X|}##, i.e. the direct sum of ##|X|## copies of ##R##.

Now here comes the ordering into play. To define a sum in it, one has to make sure, that the components match. I mean, e.g. ##(f_1,f_2)+(f'_1,f'_2)## has to be ##(f_1+f'_1,f_2+f'_2)## and not ##(f_1+f'_2,f_2+f'_1)## in ##Fr_R(X)##. Looks pretty obvious here, but not anymore in arbitrary many copies of ##R## with an unordered bunch of explicit numbers of ##R##.

I would have written ##Fr_R(X)=\bigoplus_{x\in X}R## or ##Fr_R(X)=\bigoplus_{x\in X}R_x## or ##Fr_R(X)=\bigoplus_{x\in X}(x,R)## instead. The notation as multiplication ##xr## is somehow artificial and I would have preferred pairs. However, it might as well be, that I'm overlooking a logic trap here. Until now I lived well upon indexing ##R## to ##R_x## or ##(x,R)## instead of the ##xR## notation. Important is only that the components can be matched.

Note: I suspect that B&K may be expecting the reader to identify ##\sum x r_x(m)## with ##\sum r_x(m)## ...

I agree. Maybe someone else can enlighten both of us.

 

[Math Amateur]

I'm not quite sure, if I can follow the authors correctly. To me it seems a bit more complicated than necessary, or at least overly precise. As I understand them, say construct their free ##R##-module over ##X## as ##R^X=R^{|X|}##, i.e. the direct sum of ##|X|## copies of ##R##.

Now here comes the ordering into play. To define a sum in it, one has to make sure, that the components match. I mean, e.g. ##(f_1,f_2)+(f'_1,f'_2)## has to be ##(f_1+f'_1,f_2+f'_2)## and not ##(f_1+f'_2,f_2+f'_1)## in ##Fr_R(X)##. Looks pretty obvious here, but not anymore in arbitrary many copies of ##R## with an unordered bunch of explicit numbers of ##R##.

I would have written ##Fr_R(X)=\bigoplus_{x\in X}R## or ##Fr_R(X)=\bigoplus_{x\in X}R_x## or ##Fr_R(X)=\bigoplus_{x\in X}(x,R)## instead. The notation as multiplication ##xr## is somehow artificial and I would have preferred pairs. However, it might as well be, that I'm overlooking a logic trap here. Until now I lived well upon indexing ##R## to ##R_x## or ##(x,R)## instead of the ##xR## notation. Important is only that the components can be matched.I agree. Maybe someone else can enlighten both of us.

Thanks for the thoughts and the help, fresh_42 ...

Yes, hopefully someone can enlighten us further ...

Peter