MHB 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 $$\text{Fr}_R (x)$$ are formal sums

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

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

almost all $$r_x (m)$$ being $$0$$,

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



I do not understand the notation:

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

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

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

is an external direct sum, it seems to me that the elements of $$\text{Fr}_R (x)$$ are sequences of the form $$(x_\alpha r)$$ where $$x_\alpha \in X$$ and $$r \in R $$ ... ... 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

$$ \iota_X = \sum_{y\in X} y1_y.$$

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

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

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

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

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 $$\Lambda$$ copies of $$L$$ is given as follows:

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

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


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

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

with $$r_x(m) \in R$$, almost all $$r_x(m)$$ being $$0$$, and that $$m = n$$ in $$Fr_R(X)$$ if and only if $$r_x(m) = r_x(n)$$ for all $$x$$ in $$X$$.Further, I find it very peculiar indeed that members of the index set $$X$$ are members of the module $$Fr_R(X)$$!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 $$X$$ of elements of a module $$M$$ and generating a submodule $$Fr_R(x)$$ ... ... ? ... hmm ... still doesn't seem right ...
Peter

 

[Euge]

If the $$r_x(m)$$ are simply $$R$$-scalars, why do B&K not just use the notation '$$r$$' for them ... why use a complicated notation like $$r_x(m)$$? ... 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 $$ \iota_X = \sum_{y\in X} y1_y.$$ send $$x$$ to $$x \cdot 1$$?
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 $$R$$-module on $$X$$ ... ... (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 $$X = \{ x_1, x_2, x_3 \} $$

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

Then ...

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

$$= x_1R + x_2R + x_3R$$

$$= \{ x_1r_1 + x_2r_2 + x_3r_3 \ | \ r_1, r_2, r_3 \in R \}
$$

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

... ... indeed, this may be a convenient way to think about and work with the elements of $$Fr_R (X)$$.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 ...

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

$$= x_1R + x_2R + x_3R$$

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

... ... indeed, this may be a convenient way to think about and work with the elements of $$Fr_R (X)$$. ... ... "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 $$Fr_R (X)$$ as $$\{ x_1r_1, x_2r_2, x_3r_3 \}$$ ... ... ... "

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:

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

$$= x_1R + x_2R + x_3R$$

$$= \{ x_1r_1 + x_2r_2 + x_3r_3 \ | \ r_1, r_2, r_3 \in R \}
$$

where

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

$$N_1 + N_2 + N_3 \cong N_1 \times N_2 \times N_3$$

But then ... ...

The elements of $$N_1$$ are $$x_1r_1$$

The elements of $$N_2$$ are $$x_2r_2$$

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

$$\{ x_1r_1, x_2r_2, x_3r_3 \}$$
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 $$X = \{x_1,x_2,x_3 \}$$ where the ordering is $$x_1 < x_2 < x_3$$ ... 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 $$ X = \{x_1,x_2,x_3 \} $$?

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 $$X $$ always exists with elements being able to be expressed as formal sums ... but if $$X$$ is unordered, its elements cannot be identified with a direct product ...

It has got me wondering ... so the construction of $$Fr_R (X)$$ 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 $$X $$ always exists with elements being able to be expressed as formal sums ... but if $$X$$ is unordered, its elements cannot be identified with a direct product ...

It has got me wondering ... so the construction of $$Fr_R (X)$$ 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.