Directly Finite and Directly Infinite R-Modules - Bland S2.2

[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:

Question 1

In the above Example from Bland's text we read the following:" ... ... If $M = \bigoplus_\mathbb{N} \mathbb{Z}$, then it follows that $M \cong M \oplus M$ ... ... "How ... exactly ... do we know that it follows that $M \cong M \oplus M$ ... ... ?

Question 2In the above Example from Bland's text we read the following:" ... ...$R = \text{ Hom}_\mathbb{Z} (M, M) \cong \text{ Hom}_\mathbb{Z} (M, M \oplus M )$$\cong \text{ Hom}_\mathbb{Z} (M, M) \oplus \text{ Hom}_\mathbb{Z} (M, M)$$\cong R \oplus R$ ... ... "Although the above relationships look intuitively reasonable ... how do we know ... formally and rigorously that:

$\text{ Hom}_\mathbb{Z} (M, M) \cong \text{ Hom}_\mathbb{Z} (M, M \oplus M )$

$\cong \text{ Hom}_\mathbb{Z} (M, M) \oplus \text{ Hom}_\mathbb{Z} (M, M)$Hope someone can help ...

Peter

 

[fresh_42]

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:

Question 1

In the above Example from Bland's text we read the following:" ... ... If $M = \bigoplus_\mathbb{N} \mathbb{Z}$, then it follows that $M \cong M \oplus M$ ... ... "How ... exactly ... do we know that it follows that $M \cong M \oplus M$ ... ... ?

Question 2In the above Example from Bland's text we read the following:" ... ...$R = \text{ Hom}_\mathbb{Z} (M, M) \cong \text{ Hom}_\mathbb{Z} (M, M \oplus M )$$\cong \text{ Hom}_\mathbb{Z} (M, M) \oplus \text{ Hom}_\mathbb{Z} (M, M)$$\cong R \oplus R$ ... ... "Although the above relationships look intuitively reasonable ... how do we know ... formally and rigorously that:

$\text{ Hom}_\mathbb{Z} (M, M) \cong \text{ Hom}_\mathbb{Z} (M, M \oplus M )$

$\cong \text{ Hom}_\mathbb{Z} (M, M) \oplus \text{ Hom}_\mathbb{Z} (M, M)$Hope someone can help ...

Peter

If you split $ℕ$ in even and odd numbers you get two copies which are as big as $ℕ$ is. Thus you can "double" your basis index set without gaining or losing something. Set up the bijections between $ℕ$ and $2ℕ$ and $ℕ$ and $2ℕ + 1$ and expand them to the isomorphism $⊕_{ℕ} ℤ ≅ (⊕_{ℕ} ℤ) ⊕ (⊕_{ℕ} ℤ)$.

The first isomorphism in question 2 is clear, because it is simply the substitution of $M$ by $M⊕M$ as developed in question 1.
So it remains to show that $\text{ Hom}_\mathbb{Z} (M, M \oplus M ) \cong \text{ Hom}_\mathbb{Z} (M, M) \oplus \text{ Hom}_\mathbb{Z} (M, M).$
Take a homomorphism $φ \in \text{ Hom}_\mathbb{Z} (M, M \oplus M )$. Then every image $φ(m)$ splits into two components $φ_i(m) \; : \; φ(m) = φ_1(m) + φ_2(m)$ or as a vector $φ(m) = (φ_1(m) , φ_2(m)).$
Then each $φ_i$ is a homomorphism of $M$ and $φ = φ_1 + φ_2 = (φ_1 , φ_2)$ which establishes the required isomorphism.

 

[Math Amateur]

If you split $ℕ$ in even and odd numbers you get two copies which are as big as $ℕ$ is. Thus you can "double" your basis index set without gaining or losing something. Set up the bijections between $ℕ$ and $2ℕ$ and $ℕ$ and $2ℕ + 1$ and expand them to the isomorphism $⊕_{ℕ} ℤ ≅ (⊕_{ℕ} ℤ) ⊕ (⊕_{ℕ} ℤ)$.

The first isomorphism in question 2 is clear, because it is simply the substitution of $M$ by $M⊕M$ as developed in question 1.
So it remains to show that $\text{ Hom}_\mathbb{Z} (M, M \oplus M ) \cong \text{ Hom}_\mathbb{Z} (M, M) \oplus \text{ Hom}_\mathbb{Z} (M, M).$
Take a homomorphism $φ \in \text{ Hom}_\mathbb{Z} (M, M \oplus M )$. Then every image $φ(m)$ splits into two components $φ_i(m) \; : \; φ(m) = φ_1(m) + φ_2(m)$ or as a vector $φ(m) = (φ_1(m) , φ_2(m)).$
Then each $φ_i$ is a homomorphism of $M$ and $φ = φ_1 + φ_2 = (φ_1 , φ_2)$ which establishes the required isomorphism.

Thanks for the help, fresh_42 ... appreciate your help ...

Sorry to be slow in replying ... been traveling ...

You write:

" ... ... If you split $ℕ$ in even and odd numbers you get two copies which are as big as $ℕ$ is. Thus you can "double" your basis index set without gaining or losing something. Set up the bijections between $ℕ$ and $2ℕ$ and $ℕ$ and $2ℕ + 1$ and expand them to the isomorphism $⊕_{ℕ} ℤ ≅ (⊕_{ℕ} ℤ) ⊕ (⊕_{ℕ} ℤ)$ ... ... "So ... you define two functions $\psi_1$ and $\psi_2$ as follows:

$\psi_1 \ : \ \mathbb{N} \longrightarrow 2 \mathbb{N} \ \text{ where } \psi_1 (x) = 2x$

and

$\psi_2 \ : \ \mathbb{N} \longrightarrow 2 \mathbb{N} + 1 \ \text{ where } \psi_2 (x) = 2x + 1$These are both isomorphisms ... BUT ... how exactly (details?) do we expand them to the isomorphism $⊕_{ℕ} ℤ ≅ (⊕_{ℕ} ℤ) ⊕ (⊕_{ℕ} ℤ)$ ... ... ?Can you help ... ?

Peter

 

[fresh_42]

Hi Peter!

You can settle an isomorphism $⊕_{ℕ} ℤ ≅ (⊕_{2ℕ} ℤ) ⊕ (⊕_{2ℕ+1} ℤ)$ by the following map:

If $a_n$ denotes an element $a$ at the $n$-th position, i.e. an element of $⊕_{ℕ} ℤ$, then we map it to $(a_{2k},0)$ if $n=2k$ is even and to $(0,a_{2k+1})$ if $n=2k+1$ is odd. Again the index only shows the position of $a$ and I wrote the elements of $(⊕_{2ℕ} ℤ) ⊕ (⊕_{2ℕ+1} ℤ)$ as pairs. It doesn't matter, you could write it as well as a sum. I simply felt that $a_{2k}+ 0$ might be confusing.

So it remains to show that, e.g. $⊕_{ℕ} ℤ ≅ ⊕_{2ℕ} ℤ$. This could be done by mapping $a_n$ to $a_{2n}$.

 

[Math Amateur]

Hi Peter!

You can settle an isomorphism $⊕_{ℕ} ℤ ≅ (⊕_{2ℕ} ℤ) ⊕ (⊕_{2ℕ+1} ℤ)$ by the following map:

If $a_n$ denotes an element $a$ at the $n$-th position, i.e. an element of $⊕_{ℕ} ℤ$, then we map it to $(a_{2k},0)$ if $n=2k$ is even and to $(0,a_{2k+1})$ if $n=2k+1$ is odd. Again the index only shows the position of $a$ and I wrote the elements of $(⊕_{2ℕ} ℤ) ⊕ (⊕_{2ℕ+1} ℤ)$ as pairs. It doesn't matter, you could write it as well as a sum. I simply felt that $a_{2k}+ 0$ might be confusing.

So it remains to show that, e.g. $⊕_{ℕ} ℤ ≅ ⊕_{2ℕ} ℤ$. This could be done by mapping $a_n$ to $a_{2n}$.

Thanks for the help, fresh_42 ...

Peter