Understanding Bland's Example: Free Modules & Directly Finite/Infinite R-Modules

[Math Amateur]

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

Currently I am focused on Section 2.2 Free Modules ... ...

I need some help in order to fully understand Bland's Example on page 56 concerning directly finite and directly infinite R-modules ... ...

Bland's Example on page 56 reads as follows:View attachment 5615
Question 1

In the above Example from Bland's text we read the following:

" ... ... If \(\displaystyle M = \bigoplus_\mathbb{N} \mathbb{Z}\), then it follows that \(\displaystyle M \cong M \oplus M\) ... ... "How ... exactly ... do we know that it follows that \(\displaystyle M \cong M \oplus M\) ... ... ?

Question 2In the above Example from Bland's text we read the following:" ... ...\(\displaystyle R = \text{ Hom}_\mathbb{Z} (M, M) \cong \text{ Hom}_\mathbb{Z} (M, M \oplus M )\)\(\displaystyle \cong \text{ Hom}_\mathbb{Z} (M, M) \oplus \text{ Hom}_\mathbb{Z} (M, M)\) \(\displaystyle \cong R \oplus R\) ... ... "

Although the above relationships look intuitively reasonable ... how do we know ... formally and rigorously that:\(\displaystyle \text{ Hom}_\mathbb{Z} (M, M) \cong \text{ Hom}_\mathbb{Z} (M, M \oplus M )\)\(\displaystyle \cong \text{ Hom}_\mathbb{Z} (M, M) \oplus \text{ Hom}_\mathbb{Z} (M, M) \)
Hope someone can help ...

Peter

 

[Deveno]

Set $\phi: M \oplus M \to M$ by

$\phi[(m_1,m_2,m_3,\dots) \oplus (m'_1,m'_2,m'_3,\dots)] = (m_1,m'_1,m_2,m'_2,m_3,m'_3,\dots)$.

Verify $\phi$ is an isomorphism.

In short, twice a countable infinity is another countable infinity, we have the bijection:

$k \mapsto 2k$ of $\Bbb N$ with $2\Bbb N$ which can be used to create a bijection of $\Bbb N \times \Bbb N$ with $\Bbb N$ by "interlacing pairs into a single stream".

Elements of $R = \text{Hom}_{\Bbb Z}(M,M)$ can be represented as "infinite matrices with entries in $\Bbb Z$ that have a 'hard start' at the upper left corner". If we call such a matrix $A$, then applying $\phi$ to $A$ gives us an element of $\text{Hom}_{\Bbb Z}(M,M\oplus M)$, and you can verify this is another isomorphism:

$\text{Hom}_{\Bbb Z}(M,M) \to \text{Hom}_{\Bbb Z}(M,M \oplus M)$

Such bizarre things happen when the indexing set of a direct sum is infinite.

 

[Math Amateur]

Set $\phi: M \oplus M \to M$ by

$\phi[(m_1,m_2,m_3,\dots) \oplus (m'_1,m'_2,m'_3,\dots)] = (m_1,m'_1,m_2,m'_2,m_3,m'_3,\dots)$.

Verify $\phi$ is an isomorphism.

In short, twice a countable infinity is another countable infinity, we have the bijection:

$k \mapsto 2k$ of $\Bbb N$ with $2\Bbb N$ which can be used to create a bijection of $\Bbb N \times \Bbb N$ with $\Bbb N$ by "interlacing pairs into a single stream".

Elements of $R = \text{Hom}_{\Bbb Z}(M,M)$ can be represented as "infinite matrices with entries in $\Bbb Z$ that have a 'hard start' at the upper left corner". If we call such a matrix $A$, then applying $\phi$ to $A$ gives us an element of $\text{Hom}_{\Bbb Z}(M,M\oplus M)$, and you can verify this is another isomorphism:

$\text{Hom}_{\Bbb Z}(M,M) \to \text{Hom}_{\Bbb Z}(M,M \oplus M)$

Such bizarre things happen when the indexing set of a direct sum is infinite.

Thanks Deveno ... appreciate your help ...

Just reflecting on what you have written ...

Peter

 

[Math Amateur]

Set $\phi: M \oplus M \to M$ by

$\phi[(m_1,m_2,m_3,\dots) \oplus (m'_1,m'_2,m'_3,\dots)] = (m_1,m'_1,m_2,m'_2,m_3,m'_3,\dots)$.

Verify $\phi$ is an isomorphism.

In short, twice a countable infinity is another countable infinity, we have the bijection:

$k \mapsto 2k$ of $\Bbb N$ with $2\Bbb N$ which can be used to create a bijection of $\Bbb N \times \Bbb N$ with $\Bbb N$ by "interlacing pairs into a single stream".

Elements of $R = \text{Hom}_{\Bbb Z}(M,M)$ can be represented as "infinite matrices with entries in $\Bbb Z$ that have a 'hard start' at the upper left corner". If we call such a matrix $A$, then applying $\phi$ to $A$ gives us an element of $\text{Hom}_{\Bbb Z}(M,M\oplus M)$, and you can verify this is another isomorphism:

$\text{Hom}_{\Bbb Z}(M,M) \to \text{Hom}_{\Bbb Z}(M,M \oplus M)$

Such bizarre things happen when the indexing set of a direct sum is infinite.

Hi Deveno,

Thanks again for the help ...

You write:

"... ... $k \mapsto 2k$ of $\Bbb N$ with $2\Bbb N$ which can be used to create a bijection of $\Bbb N \times \Bbb N$ with $\Bbb N$ by "interlacing pairs into a single stream".Can you please help by demonstrating exactly how this bijection works ... ...
You also write:

"... ... Elements of $R = \text{Hom}_{\Bbb Z}(M,M)$ can be represented as "infinite matrices with entries in $\Bbb Z$ that have a 'hard start' at the upper left corner". ... ... Can you please help by giving more details on these matrices (epecially their left corner) and explain how the isomorphism you mention is established ...Thank you again for your invaluable help ...Peter

 

[Deveno]

Hi Deveno,

Thanks again for the help ...

You write:

"... ... $k \mapsto 2k$ of $\Bbb N$ with $2\Bbb N$ which can be used to create a bijection of $\Bbb N \times \Bbb N$ with $\Bbb N$ by "interlacing pairs into a single stream".Can you please help by demonstrating exactly how this bijection works ... ...
You also write:

"... ... Elements of $R = \text{Hom}_{\Bbb Z}(M,M)$ can be represented as "infinite matrices with entries in $\Bbb Z$ that have a 'hard start' at the upper left corner". ... ... Can you please help by giving more details on these matrices (epecially their left corner) and explain how the isomorphism you mention is established ...Thank you again for your invaluable help ...Peter

The usual way of creating such a bijection is via Cantor's Diagonal Method:

$(1,1) \mapsto 1$
$(1,2) \mapsto 2$
$(2,1) \mapsto 3$
$(1,3) \mapsto 4$
$(2,2) \mapsto 5$
$(3,1) \mapsto 6$
$(1,4) \mapsto 7$
$(2,3) \mapsto 8$
$(3,2) \mapsto 9$
$(4,1) \mapsto 10$
$(1,5) \mapsto 11$
$(2,4) \mapsto 12$
$(3,3) \mapsto 13$

and so on

 

[Math Amateur]

The usual way of creating such a bijection is via Cantor's Diagonal Method:

$(1,1) \mapsto 1$
$(1,2) \mapsto 2$
$(2,1) \mapsto 3$
$(1,3) \mapsto 4$
$(2,2) \mapsto 5$
$(3,1) \mapsto 6$
$(1,4) \mapsto 7$
$(2,3) \mapsto 8$
$(3,2) \mapsto 9$
$(4,1) \mapsto 10$
$(1,5) \mapsto 11$
$(2,4) \mapsto 12$
$(3,3) \mapsto 13$

and so on

oh! OK ... ... thanks Deveno ... most helpful ...At the risk of trying your patience ... ... are you able to help with my other question ... ...

That is ...You also write:

"... ... Elements of $R = \text{Hom}_{\Bbb Z}(M,M)$ can be represented as "infinite matrices with entries in $\Bbb Z$ that have a 'hard start' at the upper left corner". ... ... Can you please help by giving more details on these matrices (epecially their left corner) and explain how the isomorphism you mention is established ...Thanks again for you help ...

Peter

 

[Deveno]

oh! OK ... ... thanks Deveno ... most helpful ...At the risk of trying your patience ... ... are you able to help with my other question ... ...

That is ...You also write:

"... ... Elements of $R = \text{Hom}_{\Bbb Z}(M,M)$ can be represented as "infinite matrices with entries in $\Bbb Z$ that have a 'hard start' at the upper left corner". ... ... Can you please help by giving more details on these matrices (epecially their left corner) and explain how the isomorphism you mention is established ...Thanks again for you help ...

Peter

I'll also remark here that another way of showing a bijection is by using the Cantor-Schroeder-Bernstein Theorem, which states that if there exists an injection $A \to B$ and an injection $B \to A$, then there exists a bijection between them (the link gives the details of its construction), so we have the injections:

$\Bbb N \times \Bbb N \to \Bbb N$ given by

$(k,m) \mapsto 2^k3^m$ (injective by the uniqueness of prime factorization)

and $\Bbb N \to \Bbb N \times \Bbb N$ given by

$k \mapsto (k,1)$

Recall also that in $\bigoplus\limits_{\Bbb N} \Bbb Z$, each sequence of integers is "eventually zero", so that in our "infinite matrices" the action of our infinite matrix is equivalent to an action of a finite matrix.

We can represent elements of $M$ as finite sums of $e_i$ where:

$e_i = (0,\dots,0,1,,0,\dots,0,\dots)$ where the sole non-zero entry 1 is in the $i$-th position.

These form a $\Bbb Z$-basis for $M$, and so we can compute the finite sequence, for each $i$, of $A(e_i)$ for $A \in \text{Hom}_{\Bbb Z}(M,M)$.

This gives us an infinite number of "columns", each of which is only finitely long, but which may be of arbitrary length.

We can (in theory) perform matrix multiplication with two such matrices, because although each row is infinite, after we reach the end of non-zero entries in a column we can ignore the row-entries in the sum:

$\sum a_{ik}b_{kj}$

since for some value of $k = k_0$, we have $b_{kj} = 0$ for all $k > k_0$.

If we regard the basis $\{e_i\}$ as consisting of two identical copies (one tagged with primes), then instead of writing:

$A = (A(e_1),A(e_2),A(e_3),\dots)$ and

$A' = (A'(e'_1),A'(e'_2),A'(e'_3),\dots)$ with:

$A \oplus A' = (A(e_1),A(e_2),A(e_3),\dots)\oplus(A'(e'_1),A'(e'_2),A'(e'_3),\dots)$

we instead consider $A \oplus A'$ as being:

$A \oplus A' = (A(e_1),A'(e'_1),A(e_2),A'(e'_2),\dots)$

both of which are clearly infinitely-rowed matrices with eventually terminating columns.

It is not possible to create such representations for elements of $\text{Hom}_{\Bbb Z}(\prod\limits_{\Bbb N} \Bbb Z,\prod\limits_{\Bbb N} \Bbb Z)$ precisely because the sequences may never terminate.

Unfortunately, being a being with only finite resources, I cannot properly display an infinite object.

 

[Math Amateur]

I'll also remark here that another way of showing a bijection is by using the Cantor-Schroeder-Bernstein Theorem, which states that if there exists an injection $A \to B$ and an injection $B \to A$, then there exists a bijection between them (the link gives the details of its construction), so we have the injections:

$\Bbb N \times \Bbb N \to \Bbb N$ given by

$(k,m) \mapsto 2^k3^m$ (injective by the uniqueness of prime factorization)

and $\Bbb N \to \Bbb N \times \Bbb N$ given by

$k \mapsto (k,1)$

Recall also that in $\bigoplus\limits_{\Bbb N} \Bbb Z$, each sequence of integers is "eventually zero", so that in our "infinite matrices" the action of our infinite matrix is equivalent to an action of a finite matrix.

We can represent elements of $M$ as finite sums of $e_i$ where:

$e_i = (0,\dots,0,1,,0,\dots,0,\dots)$ where the sole non-zero entry 1 is in the $i$-th position.

These form a $\Bbb Z$-basis for $M$, and so we can compute the finite sequence, for each $i$, of $A(e_i)$ for $A \in \text{Hom}_{\Bbb Z}(M,M)$.

This gives us an infinite number of "columns", each of which is only finitely long, but which may be of arbitrary length.

We can (in theory) perform matrix multiplication with two such matrices, because although each row is infinite, after we reach the end of non-zero entries in a column we can ignore the row-entries in the sum:

$\sum a_{ik}b_{kj}$

since for some value of $k = k_0$, we have $b_{kj} = 0$ for all $k > k_0$.

If we regard the basis $\{e_i\}$ as consisting of two identical copies (one tagged with primes), then instead of writing:

$A = (A(e_1),A(e_2),A(e_3),\dots)$ and

$A' = (A'(e'_1),A'(e'_2),A'(e'_3),\dots)$ with:

$A \oplus A' = (A(e_1),A(e_2),A(e_3),\dots)\oplus(A'(e'_1),A'(e'_2),A'(e'_3),\dots)$

we instead consider $A \oplus A'$ as being:

$A \oplus A' = (A(e_1),A'(e'_1),A(e_2),A'(e'_2),\dots)$

both of which are clearly infinitely-rowed matrices with eventually terminating columns.

It is not possible to create such representations for elements of $\text{Hom}_{\Bbb Z}(\prod\limits_{\Bbb N} \Bbb Z,\prod\limits_{\Bbb N} \Bbb Z)$ precisely because the sequences may never terminate.

Unfortunately, being a being with only finite resources, I cannot properly display an infinite object.

Thanks for the help Deveno ... I really appreciate it ...

Now your knowledge of mathematics is excellent ... and your teaching ability is great ... but fully displaying an infinite object is bridge too far ... even for you :(Just working through your post in detail now ...

Peter