Free Modules With More Than One Basis - Bland - Example 5, page 56

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am trying to understand Section 2.2 on free modules and need help with Example 5 showing a module with two bases ... ...

Example 5 reads as follows:View attachment 3569I am having trouble understanding the notation and meaning of $$M = \bigoplus_{ \mathbb{N} } \mathbb{Z}$$ ... ...

Further I am having considerable trouble seeing how/why $$M \cong M \bigoplus M$$ ... ...

Now I am taking $$\bigoplus_{ \mathbb{N} } \mathbb{Z}$$ to be an external direct sum ... ...

Bland defines an external direct sum as follows:https://www.physicsforums.com/attachments/3570So ... ... following the above definition (at least I think I am correctly following it ...) we have:

$$M = \bigoplus_{ \mathbb{N} } \mathbb{Z}$$

$$= \mathbb{Z} \bigoplus \mathbb{Z} \bigoplus \mathbb{Z} \bigoplus \ \ ... \ ... \ ... \ \bigoplus \mathbb{Z} \ \ ... \ ... \ ... ( \mathbb{N} \text{ copies } )$$

$$= \{ (z_\alpha ) \in \prod_{ \mathbb{N} } \mathbb{Z}_i \text{ where } i \in \mathbb{N} \}$$

$$= ?$$Can someone please confirm my analysis so far ... as far as it goes, anyway ...

Can someone also please explain and clarify the meaning of $$\bigoplus_{ \mathbb{N} } \mathbb{Z}$$, and further, demonstrate how Bland deduces that $$M \cong M \bigoplus M$$?

Needless to say, I do not follow the rest of the example ... ... so any help with that will also be appreciated ...

Peter

 

[caffeinemachine]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am trying to understand Section 2.2 on free modules and need help with Example 5 showing a module with two bases ... ...

Example 5 reads as follows:View attachment 3569I am having trouble understanding the notation and meaning of $$M = \bigoplus_{ \mathbb{N} } \mathbb{Z}$$ ... ...

Further I am having considerable trouble seeing how/why $$M \cong M \bigoplus M$$ ... ...

Now I am taking $$\bigoplus_{ \mathbb{N} } \mathbb{Z}$$ to be an external direct sum ... ...

Bland defines an external direct sum as follows:https://www.physicsforums.com/attachments/3570So ... ... following the above definition (at least I think I am correctly following it ...) we have:

$$M = \bigoplus_{ \mathbb{N} } \mathbb{Z}$$

$$= \mathbb{Z} \bigoplus \mathbb{Z} \bigoplus \mathbb{Z} \bigoplus \ \ ... \ ... \ ... \ \bigoplus \mathbb{Z} \ \ ... \ ... \ ... ( \mathbb{N} \text{ copies } )$$

$$= \{ (z_\alpha ) \in \prod_{ \mathbb{N} } \mathbb{Z}_i \text{ where } i \in \mathbb{N} \}$$

$$= ?$$Can someone please confirm my analysis so far ... as far as it goes, anyway ...

Can someone also please explain and clarify the meaning of $$\bigoplus_{ \mathbb{N} } \mathbb{Z}$$, and further, demonstrate how Bland deduces that $$M \cong M \bigoplus M$$?

Needless to say, I do not follow the rest of the example ... ... so any help with that will also be appreciated ...

Peter

You are correct in interpretting $\bigoplus_{\mathbf N}\mathbf Z$ as
$$\mathbf Z\oplus\mathbf Z\oplus\mathbf Z\oplus\cdots$$But you make a mistake when you write $$\bigoplus_{\mathbf N}\mathbf Z=\{(z_\alpha)\in\prod_{\mathbf N}\mathbf Z_i\text{ where } i\in \mathbf N \}$$What you ought to write is:$$\bigoplus_{\mathbf N}\mathbf Z=\{(z_j)_{j\in \mathbf N}\in \prod_{i\in \mathbf N}\mathbf Z | \ z_j=0 \text{ for almost all } $j$\}$$As for showing that $M$ and $M\oplus M$ are isomorphic, where $M=\bigoplus_{\mathbf N}\mathbf Z$, note that the bases of $M$ and $M\oplus M$ have the same cardinality.

 

[Math Amateur]

You are correct in interpretting $\bigoplus_{\mathbf N}\mathbf Z$ as
$$\mathbf Z\oplus\mathbf Z\oplus\mathbf Z\oplus\cdots$$But you make a mistake when you write $$\bigoplus_{\mathbf N}\mathbf Z=\{(z_\alpha)\in\prod_{\mathbf N}\mathbf Z_i\text{ where } i\in \mathbf N \}$$What you ought to write is:$$\bigoplus_{\mathbf N}\mathbf Z=\{(z_j)_{j\in \mathbf N}\in \prod_{i\in \mathbf N}\mathbf Z | \ z_j=0 \text{ for almost all } $j$\}$$As for showing that $M$ and $M\oplus M$ are isomorphic, where $M=\bigoplus_{\mathbf N}\mathbf Z$, note that the bases of $M$ and $M\oplus M$ have the same cardinality.

Thanks caffeinemachine,

Appreciate the help ... yes, see my error ... thanks ...

However, you write:

"As for showing that $M$ and $M\oplus M$ are isomorphic, where $M=\bigoplus_{\mathbf N}\mathbf Z$, note that the bases of $M$ and $M\oplus M$ have the same cardinality"Can you explain further regarding demonstrating an isomorphism between $M$ and $M\oplus M$?

Can you, for example, indicate the one-to-one and onto function or mapping underlying the isomorphism?

Thanks again for the help ...

Peter

 

[caffeinemachine]

Thanks caffeinemachine,

Appreciate the help ... yes, see my error ... thanks ...

However, you write:

"As for showing that $M$ and $M\oplus M$ are isomorphic, where $M=\bigoplus_{\mathbf N}\mathbf Z$, note that the bases of $M$ and $M\oplus M$ have the same cardinality"Can you explain further regarding demonstrating an isomorphism between $M$ and $M\oplus M$?

Can you, for example, indicate the one-to-one and onto function or mapping underlying the isomorphism?

Thanks again for the help ...

Peter

We prove a a general fact that if $M$ and $N$ are free $R$-modules with bases $B$ and $C$ respectively such that the cardinality of $B$ and the cardinality of $C$ are the same, then $M\cong N$.

Proof:
Let $f:B\to C$ be a bijection and say $B=\{x_\alpha\}_{\alpha\in J}$.

Define $$\varphi\left(\sum_{\alpha\in J}x_\alpha r_\alpha\right)=\sum_{\alpha\in J} f(x_\alpha) r_\alpha$$

Can you show that $\varphi$ is an $R$-module isomorphism?

 

[Math Amateur]

We prove a a general fact that if $M$ and $N$ are free $R$-modules with bases $B$ and $C$ respectively such that the cardinality of $B$ and the cardinality of $C$ are the same, then $M\cong N$.

Proof:
Let $f:B\to C$ be a bijection and say $B=\{x_\alpha\}_{\alpha\in J}$.

Define $$\varphi\left(\sum_{\alpha\in J}x_\alpha r_\alpha\right)=\sum_{\alpha\in J} f(x_\alpha) r_\alpha$$

Can you show that $\varphi$ is an $R$-module isomorphism?

Thanks caffeinemachine ...

OK, I will take up your suggestion and I will try to construct a proof of the following proposition ...

If

$$M,N$$ are free $$R$$-modules with bases $$B$$ and $$C$$ respectively such that the cardinality of $$B$$ equals the cardinality of $$C$$

then

$$M \cong N$$(***NOTE*** I cannot find this Theorem - are you able to give me a reference to it in some text?)
Now let $$f$$ be a bijection $$f \ : \ B \longrightarrow C$$ and let $$B = \{ x_\alpha \}_{ \alpha \in J }$$ and $$C = \{ y_\alpha \}_{ \alpha \in J }$$Define $$\phi \ : \ M \longrightarrow N$$ as follows:

$$ \phi ( \sum_{ \alpha \in J } x_\alpha r_\alpha ) = \sum_{ \alpha \in J } f(x_\alpha) r_\alpha
$$Then ... we need to show that $$\phi \ : \ M \longrightarrow N$$ is an R-module homomorphism ...So, we need to show that

(a) $$\phi ( x + y ) = \phi ( x ) + \phi ( y ) \ \ \text{ for all } x,y \in M$$

and

(b) $$\phi ( x r ) = \phi ( x ) r \text{ for all } x \in M \text{ and for all } r \in R$$
Now, consider $$x, y \in M$$ such that

$$x = \sum_{ \alpha \in J } x_\alpha a_\alpha \ , \ y = \sum_{ \alpha \in J } x_\alpha b_\alpha \ \ \text{ where } \ a_\alpha , b_\alpha \in R$$Then

$$\phi ( x + y )$$$$= \phi ( \sum_{ \alpha \in J } x_\alpha a_\alpha + \sum_{ \alpha \in J } x_\alpha b_\alpha )$$$$= \phi ( \sum_{ \alpha \in J } x_\alpha ( a_\alpha + b_\alpha) )$$$$= \sum_{ \alpha \in J } f (x_\alpha) ( a_\alpha + b_\alpha)
$$$$= \sum_{ \alpha \in J } f (x_\alpha) a_\alpha + \sum_{ \alpha \in J } f (x_\alpha) b_\alpha$$$$= \phi ( x ) + \phi ( y )$$and $$\phi ( x r )$$$$= \phi ( ( \sum_{ \alpha \in J } x_\alpha a_\alpha ) r )$$$$= \phi ( \sum_{ \alpha \in J } x_\alpha (a_\alpha r ) )$$$$= ( \sum_{ \alpha \in J } f(x_\alpha) (a_\alpha r ) )
$$$$= ( \sum_{ \alpha \in J } f(x_\alpha) (a_\alpha ) r $$$$= \phi ( x ) r$$
Thus $$\phi $$ is an R-module homomorphism.
Now we need to show that $$\phi $$ is injective and surjective.Injectivity of $$\phi \ : \ M \longrightarrow N$$


We have to show that if $$\phi ( x ) = \phi ( y ) \text{ then } x = y $$.

That is we have to show that if

$$\phi ( \sum_{ \alpha \in J } x_\alpha a_\alpha ) = \phi ( \sum_{ \alpha \in J } x_\alpha b_\alpha )$$

then

$$\sum_{ \alpha \in J } x_\alpha a_\alpha = \sum_{ \alpha \in J } x_\alpha b_\alpha
$$
Now

$$\phi ( ( \sum_{ \alpha \in J } x_\alpha a_\alpha ) = \phi ( ( \sum_{ \alpha \in J } x_\alpha b_\alpha ) \Longrightarrow \sum_{ \alpha \in J } f(x_\alpha) a_\alpha = \sum_{ \alpha \in J } f(x_\alpha) b_\alpha$$

Thus it must be that $$a_\alpha = b_\alpha $$

so

$$\sum_{ \alpha \in J } x_\alpha a_\alpha = \sum_{ \alpha \in J } x_\alpha b_\alpha
$$

and so $$\phi $$ is injective.Surjectivity of $$\phi \ : M \ \longrightarrow N $$

We need to show that if $$w \in N \text{ then } \exists \ x \in M \text{ such that } \phi (x) = w $$

Let $$c = \{ y_\alpha \}_{ \alpha \in J }$$

then

$$w = \sum_{ \alpha \in J } y_\alpha c_\alpha \text{ where } c_\alpha \in R $$

Since $$f$$ is a bijection, for each $$y_\alpha$$ these is an $$x_\alpha$$ such that $$f ( x_\alpha ) = y_\alpha$$

So we can write

$$w = \sum_{ \alpha \in J } f( x_\alpha ) c_\alpha $$

BUT ... ... then for

$$ x = \sum_{ \alpha \in J } x_\alpha c_\alpha \in M$$

we have

$$\phi (x) = \sum_{ \alpha \in J } f( x_\alpha ) c_\alpha = w
$$

Thus $$\phi$$ is surjective ... ... which completes the proof ...Can you please confirm the above is correct and critique the proof ...

Such help would be much appreciated ... ...

 

[caffeinemachine]

Thanks caffeinemachine ...

OK, I will take up your suggestion and I will try to construct a proof of the following proposition ...

If

$$M,N$$ are free $$R$$-modules with bases $$B$$ and $$C$$ respectively such that the cardinality of $$B$$ equals the cardinality of $$C$$

then

$$M \cong N$$(***NOTE*** I cannot find this Theorem - are you able to give me a reference to it in some text?)
Now let $$f$$ be a bijection $$f \ : \ B \longrightarrow C$$ and let $$B = \{ x_\alpha \}_{ \alpha \in J }$$ and $$C = \{ y_\alpha \}_{ \alpha \in J }$$Define $$\phi \ : \ M \longrightarrow N$$ as follows:

$$ \phi ( \sum_{ \alpha \in J } x_\alpha r_\alpha ) = \sum_{ \alpha \in J } f(x_\alpha) r_\alpha
$$Then ... we need to show that $$\phi \ : \ M \longrightarrow N$$ is an R-module homomorphism ...So, we need to show that

(a) $$\phi ( x + y ) = \phi ( x ) + \phi ( y ) \ \ \text{ for all } x,y \in M$$

and

(b) $$\phi ( x r ) = \phi ( x ) r \text{ for all } x \in M \text{ and for all } r \in R$$
Now, consider $$x, y \in M$$ such that

$$x = \sum_{ \alpha \in J } x_\alpha a_\alpha \ , \ y = \sum_{ \alpha \in J } x_\alpha b_\alpha \ \ \text{ where } \ a_\alpha , b_\alpha \in R$$Then

$$\phi ( x + y )$$$$= \phi ( \sum_{ \alpha \in J } x_\alpha a_\alpha + \sum_{ \alpha \in J } x_\alpha b_\alpha )$$$$= \phi ( \sum_{ \alpha \in J } x_\alpha ( a_\alpha + b_\alpha) )$$$$= \sum_{ \alpha \in J } f (x_\alpha) ( a_\alpha + b_\alpha)
$$$$= \sum_{ \alpha \in J } f (x_\alpha) a_\alpha + \sum_{ \alpha \in J } f (x_\alpha) b_\alpha$$$$= \phi ( x ) + \phi ( y )$$and $$\phi ( x r )$$$$= \phi ( ( \sum_{ \alpha \in J } x_\alpha a_\alpha ) r )$$$$= \phi ( \sum_{ \alpha \in J } x_\alpha (a_\alpha r ) )$$$$= ( \sum_{ \alpha \in J } f(x_\alpha) (a_\alpha r ) )
$$$$= ( \sum_{ \alpha \in J } f(x_\alpha) (a_\alpha ) r $$$$= \phi ( x ) r$$
Thus $$\phi $$ is an R-module homomorphism.
Now we need to show that $$\phi $$ is injective and surjective.Injectivity of $$\phi \ : \ M \longrightarrow N$$


We have to show that if $$\phi ( x ) = \phi ( y ) \text{ then } x = y $$.

That is we have to show that if

$$\phi ( \sum_{ \alpha \in J } x_\alpha a_\alpha ) = \phi ( \sum_{ \alpha \in J } x_\alpha b_\alpha )$$

then

$$\sum_{ \alpha \in J } x_\alpha a_\alpha = \sum_{ \alpha \in J } x_\alpha b_\alpha
$$
Now

$$\phi ( ( \sum_{ \alpha \in J } x_\alpha a_\alpha ) = \phi ( ( \sum_{ \alpha \in J } x_\alpha b_\alpha ) \Longrightarrow \sum_{ \alpha \in J } f(x_\alpha) a_\alpha = \sum_{ \alpha \in J } f(x_\alpha) b_\alpha$$

Thus it must be that $$a_\alpha = b_\alpha $$

so

$$\sum_{ \alpha \in J } x_\alpha a_\alpha = \sum_{ \alpha \in J } x_\alpha b_\alpha
$$

and so $$\phi $$ is injective.Surjectivity of $$\phi \ : M \ \longrightarrow N $$

We need to show that if $$w \in N \text{ then } \exists \ x \in M \text{ such that } \phi (x) = w $$

Let $$c = \{ y_\alpha \}_{ \alpha \in J }$$

then

$$w = \sum_{ \alpha \in J } y_\alpha c_\alpha \text{ where } c_\alpha \in R $$

Since $$f$$ is a bijection, for each $$y_\alpha$$ these is an $$x_\alpha$$ such that $$f ( x_\alpha ) = y_\alpha$$

So we can write

$$w = \sum_{ \alpha \in J } f( x_\alpha ) c_\alpha $$

BUT ... ... then for

$$ x = \sum_{ \alpha \in J } x_\alpha c_\alpha \in M$$

we have

$$\phi (x) = \sum_{ \alpha \in J } f( x_\alpha ) c_\alpha = w
$$

Thus $$\phi$$ is surjective ... ... which completes the proof ...Can you please confirm the above is correct and critique the proof ...

Such help would be much appreciated ... ...

Your proof is perfectly fine.

I don't know of a book where this theorem is mentioned. It is a sort of obvious fact which book fail to mention. But I should mention that I am just a beginner at module theory and am not aware of a good book on the subject. At least the ones I've seen I didn't like.