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