Understanding the Standard Free Right Module on a Set X?

[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 the meaning of B&K's terminology ... ...

Section 2.1.16 reads as follows:View attachment 3384
In the above text B&K write the following:

" ... ... Note that the elements of \(\displaystyle \text{Fr}_R (x)\) are formal sums

\(\displaystyle m = \sum_{x \in X}x r_x (m)\)

with \(\displaystyle r_x (m) \in R\),

almost all \(\displaystyle r_x (m)\) being \(\displaystyle 0\),

and that \(\displaystyle m = n\) in \(\displaystyle \text{Fr}_R (x)\) if and only if \(\displaystyle r_x (m) = r_x (n)\) for all \(\displaystyle x \in X\). ... ... "



I do not understand the notation:

\(\displaystyle m = \sum_{x \in X}x r_x (m)\)

Indeed ... ... what is \(\displaystyle r_x (m)\)? ... ... What is the meaning of this notation? ... ... What are B&K trying to indicate by this notation?Since

\(\displaystyle \text{Fr}_R (x) = \bigoplus_ X xR\)

is an external direct sum, it seems to me that the elements of \(\displaystyle \text{Fr}_R (x)\) are sequences of the form \(\displaystyle (x_\alpha r)\) where \(\displaystyle x_\alpha \in X\) and \(\displaystyle r \in R \) ... ... Can someone please clarify this situation and explain what B&K mean by their notation ...Further, it would help if someone could briefly explain the canonical embedding ...Finally, can someone explain how the above definition of a free module matches or integrates with the definition in some texts (e.g M.E. Keating's undergraduate text on modules) of a free R-module as an R-module that has a basis?Peter

 

[Euge]

Every element $m\in \text{Fr}_R(X)$ has a unique representation as a sum $\sum_{x\in X} xr_x(m)$, where the $r_x(m)$ are $R$-scalars. That's why $m = n$ in $\text{Fr}_R(X)$ if and only if $r_x(m) = r_x(n)$ for all $x\in X$.

The canonical embedding $\iota_X : X \hookrightarrow \text{Fr}_R(X)$ is given by

\(\displaystyle \iota_X = \sum_{y\in X} y1_y.\)

It sends an element $x\in X$ to $x\cdot 1\in \text{Fr}_R(X)$.

 

[Math Amateur]

Thanks for the help Euge ... appreciate it ...

But ... I am still having trouble fully understanding B&K ...

If the \(\displaystyle r_x(m)\) are simply \(\displaystyle R\)-scalars, why do B&K not just use the notation '\(\displaystyle r\)' for them ... why use a complicated notation like \(\displaystyle r_x(m)\)? ... what are they trying to say/emphasize?

Also I am still trying to understand the canonical embedding ... how does \(\displaystyle \iota_X = \sum_{y\in X} y1_y.\) send \(\displaystyle x\) to \(\displaystyle x \cdot 1\)?

Hope that you can help further ...

Peter

 

[Math Amateur]

Thanks for the help Euge ... appreciate it ...

But ... I am still having trouble fully understanding B&K ...

If the \(\displaystyle r_x(m)\) are simply \(\displaystyle R\)-scalars, why do B&K not just use the notation '\(\displaystyle r\)' for them ... why use a complicated notation like \(\displaystyle r_x(m)\)? ... what are they trying to say/emphasize?

Also I am still trying to understand the canonical embedding ... how does \(\displaystyle \iota_X = \sum_{y\in X} y1_y.\) send \(\displaystyle x\) to \(\displaystyle x \cdot 1\)?

Hope that you can help further ...

Peter

Hi again Euge,

Your post above caused me to do some more reading and thinking ...

I went back to B&K Section 2.1.15 which reads as follows:View attachment 3470
https://www.physicsforums.com/attachments/3471Now ... ... I think I understand this section which is just previous to the Section 2.1.16 that confuses me ...

In Section 2.1.15, the direct sum of \(\displaystyle \Lambda\) copies of \(\displaystyle L\) is given as follows:

\(\displaystyle L^\Lambda = \{ ( l_\lambda ) \ | \ l_\lambda \in L, l_\lambda = 0 \text{ for almost all } \lambda \}\)Now, I take it that, in the above, \(\displaystyle \lambda\) is the index for the set \(\displaystyle \Lambda\), that is, \(\displaystyle \lambda \in \Lambda\) and \(\displaystyle \lambda\) runs through the ordered index set.So then ... when we come to Section 2.1.16 and the module \(\displaystyle Fr_R(X)\), we also have a direct sum ... this time of copies of \(\displaystyle R\) ... ... BUT ... ... this time the indexed set is unordered ... ?! ... ... whatever difference that makes ... ... ?

So ... ... based on Section 2.1.15 (see above) I would expect that we would have:

\(\displaystyle Fr_R(X) = R^X = \{ ( r_x ) \ | \ r_x \in R \text{ and } x \in X \text{ and } r_x = 0 \text{ for almost all } x \}
\)


Is that correct?If so ... ... then how does my analysis above square with B&K who state that the elements of \(\displaystyle Fr_R(X)\) are formal sums as follows:

\(\displaystyle m = \sum_{x \in X} x r_x(m)
\)

with \(\displaystyle r_x(m) \in R\), almost all \(\displaystyle r_x(m)\) being \(\displaystyle 0\), and that \(\displaystyle m = n\) in \(\displaystyle Fr_R(X)\) if and only if \(\displaystyle r_x(m) = r_x(n)\) for all \(\displaystyle x\) in \(\displaystyle X\).Further, I find it very peculiar indeed that members of the index set \(\displaystyle X\) are members of the module \(\displaystyle Fr_R(X)\)!I hope someone can clarify the above for me ...

Help will be appreciated ... ...

Peter***EDIT***

Just thinking some more ...

... ... in Section 2.1.16, are B&K actually taking a set \(\displaystyle X\) of elements of a module \(\displaystyle M\) and generating a submodule \(\displaystyle Fr_R(x)\) ... ... ? ... hmm ... still doesn't seem right ...
Peter

 

[Euge]

If the \(\displaystyle r_x(m)\) are simply \(\displaystyle R\)-scalars, why do B&K not just use the notation '\(\displaystyle r\)' for them ... why use a complicated notation like \(\displaystyle r_x(m)\)? ... what are they trying to say/emphasize?

The notation $r_x(m)$ indicates the dependence on $x$ and $m$. Writing an element $m \in \text{Fr}_R(X)$ as $\sum_{x\in X} xr$ is ambiguous -- is $r$ the same for every $x\in X$ or is it variable? However, I think it's fine in most contexts to write $m = \sum_{x\in X}xr_x$.

Also I am still trying to understand the canonical embedding ... how does \(\displaystyle \iota_X = \sum_{y\in X} y1_y.\) send \(\displaystyle x\) to \(\displaystyle x \cdot 1\)?
Peter

This is because $\iota_X(x) = \sum_{y\in X} y1_y(x)$, and $1_y(x) = 1$ when $y = x$ and $0$ otherwise.

 

[Math Amateur]

The notation $r_x(m)$ indicates the dependence on $x$ and $m$. Writing an element $m \in \text{Fr}_R(X)$ as $\sum_{x\in X} xr$ is ambiguous -- is $r$ the same for every $x\in X$ or is it variable? However, I think it's fine in most contexts to write $m = \sum_{x\in X}xr_x$.
This is because $\iota_X(x) = \sum_{y\in X} y1_y(x)$, and $1_y(x) = 1$ when $y = x$ and $0$ otherwise.

I am still trying to understand Berrick and Keating's description of the standard free right \(\displaystyle R\)-module on \(\displaystyle X\) ... ... (see text above in the initial post of this thread) ... ...

... ... (indeed, maybe I am overthinking it!)
I am going to try to give a simple example ... ... and hope that someone can critique my construction ... ... thus furthering my understanding of the matter ...Assume \(\displaystyle X = \{ x_1, x_2, x_3 \} \)

(where the subscripts \(\displaystyle 1,2,3\) do NOT indicate an ordering)

Then ...

\(\displaystyle Fr_R (X) = \oplus_X xR\)

\(\displaystyle = x_1R + x_2R + x_3R\)

\(\displaystyle = \{ x_1r_1 + x_2r_2 + x_3r_3 \ | \ r_1, r_2, r_3 \in R \}
\)

Further, since the external direct sum is isomorphic to the direct product we can write the elements of \(\displaystyle Fr_R (X)\) as \(\displaystyle \{ x_1r_1, x_2r_2, x_3r_3 \}\) ... ... ...

... ... indeed, this may be a convenient way to think about and work with the elements of \(\displaystyle Fr_R (X)\).Is my analysis above correct?

I would really appreciate someone confirming that my analysis above is essentially correct and/or critiquing my analysis and pointing out errors/weaknesses.

PeterFor the convenience of readers I am providing the relevant text again from the B&K text, as follows:View attachment 3524

 

[Euge]

Your initial description of $ Fr_R(X) $ with $X=\{x_1, x_2, x_3\}$ is correct, but the second description (i.e., where the elements are of the form $\{x_1 r_1, x_2 r_2, x_3 r_3\}$ is incorrect. Think about this carefully: How can $ Fr_R(X) $ be an $R$- module? Why do the elements of $ Fr_R(X)$ make sense? For starters, a multiplication between $ X $ and $R$ needs to be defined such that $ xr\in R$ for all $ x\in X $ and $ r\in R $. Then since $ R$ is a ring, closure under addition holds in $ R $; so all finite sums of terms of the form $ xr $ make sense.

If you want to properly identify $ Fr_R(X) $ with the direct sum of copies of $ R$ indexed by $ X $ (from your viewpoint), associate an element $\sum_X xr_x \in Fr_R(X)$ with the element $(r_x) \in R^X$.

 

[Math Amateur]

Your initial description of $ Fr_R(X) $ with $X=\{x_1, x_2, x_3\}$ is correct, but the second description (i.e., where the elements of the form $\{x_1 r_1, x_2 r_2, x_3 r_3\}$ is incorrect. Think about this carefully: How can $ Fr_R(X) $ be an $R$- module? Why do the elements of $ Fr_R(X)$ make sense? For starters, a multiplication between $ X $ and $R$ needs to be defined such that $ xr\in R$ for all $ x\in X $ and $ r\in R $. Then since $ R$ is a ring, closure under addition holds in $ R $; so all finite sums of terms of the form $ xr $ make sense.

If you want to properly identify $ Fr_R(X) $ with the direct sum of copies of $ R$ indexed by $ X $ (from your viewpoint), associate an element $\sum_X xr_X \in Fr_R(X)$ with the element $(r_x) \in R^X$.

Thank you Euge ... most helpful ...

Still reflecting on your point regarding the second description ... but I think I have understood your point ...

Thanks again,

Peter

 

[Math Amateur]

Your initial description of $ Fr_R(X) $ with $X=\{x_1, x_2, x_3\}$ is correct, but the second description (i.e., where the elements of the form $\{x_1 r_1, x_2 r_2, x_3 r_3\}$ is incorrect. Think about this carefully: How can $ Fr_R(X) $ be an $R$- module? Why do the elements of $ Fr_R(X)$ make sense? For starters, a multiplication between $ X $ and $R$ needs to be defined such that $ xr\in R$ for all $ x\in X $ and $ r\in R $. Then since $ R$ is a ring, closure under addition holds in $ R $; so all finite sums of terms of the form $ xr $ make sense.

If you want to properly identify $ Fr_R(X) $ with the direct sum of copies of $ R$ indexed by $ X $ (from your viewpoint), associate an element $\sum_X xr_x \in Fr_R(X)$ with the element $(r_x) \in R^X$.

Thanks so much Euge ... but I think I need some further help ...

In my post above, I state the following:

" ... ... Then ...

\(\displaystyle Fr_R (X) = \bigoplus_X xR\)

\(\displaystyle = x_1R + x_2R + x_3R\)

\(\displaystyle = \{ x_1r_1 + x_2r_2 + x_3r_3 \ | \ r_1, r_2, r_3 \in R \}
\) Further, since the external direct sum is isomorphic to the direct product we can write the elements of \(\displaystyle Fr_R (X)\) as \(\displaystyle \{ x_1r_1, x_2r_2, x_3r_3 \}\) ... ... ...

... ... indeed, this may be a convenient way to think about and work with the elements of \(\displaystyle Fr_R (X)\). ... ... "Then in your post you write:

" ... ... Your initial description of $ Fr_R(X) $ with $X=\{x_1, x_2, x_3\}$ is correct, but the second description (i.e., where the elements of the form $\{x_1 r_1, x_2 r_2, x_3 r_3\}$ is incorrect. ... ... "
I wish to fully understand why the second description is incorrect/mistaken ... ... so to make clear my thinking/argument I will try to make clear the thinking that led me
to believe that ... ... " ... Further, since the external direct sum is isomorphic to the direct product we can write the elements of \(\displaystyle Fr_R (X)\) as \(\displaystyle \{ x_1r_1, x_2r_2, x_3r_3 \}\) ... ... ... "

I will step back and use Dummit and Foote's (D&F's) description of the direct product and the direct sum together with Proposition 5 which indicates an isomorphism between the external direct sum (direct product) and the internal direct sum. I am also including D&F's definition of the internal direct sum and free R-modules as follows:https://www.physicsforums.com/attachments/3552View attachment 3553Now, ... if we examine:

\(\displaystyle Fr_R (X) = \bigoplus_X xR\)

\(\displaystyle = x_1R + x_2R + x_3R\)

\(\displaystyle = \{ x_1r_1 + x_2r_2 + x_3r_3 \ | \ r_1, r_2, r_3 \in R \}
\)

where

\(\displaystyle X = \{ x_1, x_2, x_3 \} \)... ... ... then we see that \(\displaystyle Fr_R (X)\) is the sum of three cyclic modules which, indeed, are submodules \(\displaystyle N_1, N_2, N_3\) of \(\displaystyle Fr_R (X)\) ... ... But according to D&F (see above text) we have:

\(\displaystyle N_1 + N_2 + N_3 \cong N_1 \times N_2 \times N_3\)

But then ... ...

The elements of \(\displaystyle N_1\) are \(\displaystyle x_1r_1\)

The elements of \(\displaystyle N_2\) are \(\displaystyle x_2r_2\)

The elements of \(\displaystyle N_3\) are \(\displaystyle x_3r_3\)Then ... surely we can write the the elements of \(\displaystyle Fr_R (X)\) (which up to an isomorphism are elements of \(\displaystyle N_1 \times N_2 \times N_3\) as follows:

\(\displaystyle \{ x_1r_1, x_2r_2, x_3r_3 \}\)
Now ... ... obviously there is something wrong with my analysis above ... ... but what exactly?

Can you please indicate where my thinking/analysis is in error?

I would very much appreciate a clarification of this matter which is causing me considerable confusion ...

Thanks again for your help to date on this matter ... ...

Peter

 

[Euge]

I think the confusion here lies in our different uses of notation. When you write $\{x_1r_1,x_2r_2,x_3r_3\}$, you mean the ordered triple $(x_1 r_1, x_2 r_2, x_3 r_3)$ in $x_1 R \times x_2 R \times x_3 R$, right? If so, then your assertions are correct and I misinterpreted your notation.

 

[Math Amateur]

I think the confusion here lies in our different uses of notation. When you write $\{x_1r_1,x_2r_2,x_3r_3\}$, you mean the ordered triple $(x_1 r_1, x_2 r_2, x_3 r_3)$ in $x_1 R \times x_2 R \times x_3 R$, right? If so, then your assertions are correct and I misinterpreted your notation.

Thanks so much for your help, Euge ...

Hmm... yes, understand your point ...

I used $\{x_1r_1,x_2r_2,x_3r_3\}$ instead of the ordered triple $(x_1 r_1, x_2 r_2, x_3 r_3)$ only because at the start of their Section 2.1.16 on The free module Fr_R (X) Berrick and Keating write:

"Suppose now that we wish to take a sum of copies of R indexed by an unordered set X. ... ... "So, to sum up what you have said (my current interpretation of what you have said, anyway ... ... ) is as follows:

My analysis is correct for an ordered set \(\displaystyle X = \{x_1,x_2,x_3 \}\) where the ordering is \(\displaystyle x_1 < x_2 < x_3\) ... if we change $\{x_1r_1,x_2r_2,x_3r_3\}$ to the ordered triple $(x_1 r_1, x_2 r_2, x_3 r_3)$.

Is that correct?

Can you please explain how we deal with an unordered set \(\displaystyle X = \{x_1,x_2,x_3 \} \)?

Thanks again for your help?

Peter

 

[Euge]

I mentioned the ordered triples (i.e., 3-tuples) since you identified $\text{Fr}_R(X)$ with a direct product. I didn't say that there was an ordering of $X$.

 

[Math Amateur]

I mentioned the ordered triples (i.e., 3-tuples) since you identified $\text{Fr}_R(X)$ with a direct product. I didn't say that there was an ordering of $X$.

Hi Euge,

You are indeed correct ... you did not mention an ordering ... ...

... ... BUT ... I am still puzzled as to how the analysis differs in the cases when X is ordered and when X is unordered ...

Can you help in this matter?

Peter

 

[Euge]

Hi Euge,

You are indeed correct ... you did not mention an ordering ... ...

... ... BUT ... I am still puzzled as to how the analysis differs in the cases when X is ordered and when X is unordered ...

Can you help in this matter?

Peter

You cannot always identity the free $R$-module on $X$ with the direct product (although you can if $X$ is finite). Elements of a direct product indexed by $X$ are $X$-tuples; they are undefined if $X$ is not ordered (such as when $X$ is the set of complex numbers). When $X$ is ordered though, $Fr_R(X)$ can be identified with $R^X$ (I showed you the identification in an earlier post).

 

[Math Amateur]

You cannot always identity the free $R$-module on $X$ with the direct product (although you can if $X$ is finite). Elements of a direct product indexed by $X$ are $X$-tuples; they are undefined if $X$ is not ordered (such as when $X$ is the set of complex numbers). When $X$ is ordered though, $Fr_R(X)$ can be identified with $R^X$ (I showed you the identification in an earlier post).

Thanks Euge ... very much appreciate the help ...

So the standard free right module on a set \(\displaystyle X \) always exists with elements being able to be expressed as formal sums ... but if \(\displaystyle X\) is unordered, its elements cannot be identified with a direct product ...

It has got me wondering ... so the construction of \(\displaystyle Fr_R (X)\) cannot be via an external direct sum since that is via (and identified with) a direct product?

Can you clarify?

Peter

 

[Euge]

Thanks Euge ... very much appreciate the help ...

So the standard free right module on a set \(\displaystyle X \) always exists with elements being able to be expressed as formal sums ... but if \(\displaystyle X\) is unordered, its elements cannot be identified with a direct product ...

It has got me wondering ... so the construction of \(\displaystyle Fr_R (X)\) cannot be via an external direct sum since that is via (and identified with) a direct product?

Can you clarify?

Peter

Direct sums are generally different from direct products. Also, to construct $Fr_R(X)$, you have to first define a multiplication between $X$ and $R$ so that the sums $\sum_X x r_x$ make sense and that the action involved makes $Fr_R(X)$ into an $R$-module.