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

In summary: Thanks caffeinemachine ...OK, I will take up your suggestion and I will try to construct a proof of the following proposition ...IfM,N are free R-modules with bases B and C respectively such that the cardinality of B equals the cardinality of CthenM \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
  • #1
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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 \(\displaystyle M = \bigoplus_{ \mathbb{N} } \mathbb{Z}\) ... ...

Further I am having considerable trouble seeing how/why \(\displaystyle M \cong M \bigoplus M\) ... ...

Now I am taking \(\displaystyle \bigoplus_{ \mathbb{N} } \mathbb{Z}\) 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:

\(\displaystyle M = \bigoplus_{ \mathbb{N} } \mathbb{Z}\)

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

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

\(\displaystyle = ?\)Can someone please confirm my analysis so far ... as far as it goes, anyway ...

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

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

Peter
 
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  • #2
Peter said:
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 \(\displaystyle M = \bigoplus_{ \mathbb{N} } \mathbb{Z}\) ... ...

Further I am having considerable trouble seeing how/why \(\displaystyle M \cong M \bigoplus M\) ... ...

Now I am taking \(\displaystyle \bigoplus_{ \mathbb{N} } \mathbb{Z}\) 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:

\(\displaystyle M = \bigoplus_{ \mathbb{N} } \mathbb{Z}\)

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

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

\(\displaystyle = ?\)Can someone please confirm my analysis so far ... as far as it goes, anyway ...

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

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.
 
  • #3
caffeinemachine said:
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
 
  • #4
Peter said:
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?
 
  • #5
caffeinemachine said:
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

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

then

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

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

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

and

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

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

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


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

That is we have to show that if

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

then

\(\displaystyle \sum_{ \alpha \in J } x_\alpha a_\alpha = \sum_{ \alpha \in J } x_\alpha b_\alpha
\)
Now

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

Thus it must be that \(\displaystyle a_\alpha = b_\alpha \)

so

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

and so \(\displaystyle \phi \) is injective.Surjectivity of \(\displaystyle \phi \ : M \ \longrightarrow N \)

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

Let \(\displaystyle c = \{ y_\alpha \}_{ \alpha \in J }\)

then

\(\displaystyle w = \sum_{ \alpha \in J } y_\alpha c_\alpha \text{ where } c_\alpha \in R \)

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

So we can write

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

BUT ... ... then for

\(\displaystyle x = \sum_{ \alpha \in J } x_\alpha c_\alpha \in M\)

we have

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

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

Such help would be much appreciated ... ...
 
Last edited:
  • #6
Peter said:
Thanks caffeinemachine ...

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

If

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

then

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

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

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

and

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

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

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


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

That is we have to show that if

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

then

\(\displaystyle \sum_{ \alpha \in J } x_\alpha a_\alpha = \sum_{ \alpha \in J } x_\alpha b_\alpha
\)
Now

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

Thus it must be that \(\displaystyle a_\alpha = b_\alpha \)

so

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

and so \(\displaystyle \phi \) is injective.Surjectivity of \(\displaystyle \phi \ : M \ \longrightarrow N \)

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

Let \(\displaystyle c = \{ y_\alpha \}_{ \alpha \in J }\)

then

\(\displaystyle w = \sum_{ \alpha \in J } y_\alpha c_\alpha \text{ where } c_\alpha \in R \)

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

So we can write

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

BUT ... ... then for

\(\displaystyle x = \sum_{ \alpha \in J } x_\alpha c_\alpha \in M\)

we have

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

Thus \(\displaystyle \phi\) 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.
 

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

1. What is a free module?

A free module is a type of module in abstract algebra that has a basis, meaning it can be generated by a linearly independent set of elements.

2. Can a free module have more than one basis?

Yes, a free module can have more than one basis. This is known as a free module with multiple bases.

3. What is the significance of multiple bases in a free module?

The presence of multiple bases in a free module allows for more flexibility and different ways of expressing the elements in the module. It also allows for different perspectives and interpretations of the module structure.

4. How are free modules with multiple bases represented?

Free modules with multiple bases can be represented using a matrix, where the rows correspond to the different bases and the columns correspond to the elements in each basis.

5. What is an example of a free module with multiple bases?

An example of a free module with multiple bases is the set of polynomials with coefficients in a field, where it can be generated by multiple bases such as {1, x, x^2} or {x, x^2, x^3}.

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