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

  1. Oct 23, 2016 #1
    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:


    ?temp_hash=12e20c959a0b3d5f9994e048d34f062d.png


    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
     

    Attached Files:

  2. jcsd
  3. Oct 23, 2016 #2

    fresh_42

    Staff: Mentor

    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.
     
  4. Oct 23, 2016 #3

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

    Yes, hopefully someone can enlighten us further ...

    Peter
     
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