MHB Noetherian Modules - Bland, Example 3, Section 4.2 ....

[Math Amateur]

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

Currently I am focused on Section 4.2 Noetherian and Artinian Modules [FONT=MathJax_Main][FONT=MathJax_Math] ... ...

I need some help in order to fully understand Example 3, Section 4.2 ...

Example 3, Section 4.2 reads as follows:View attachment 8095
My questions are as follows:Question 1

Can someone please explain/illustrate the nature of $$V_1$$ ... ?
Question 2

Can someone please demonstrate exactly how $$V_1 \subseteq V_2 \subseteq V_3 \subseteq$$ ...
Help will be appreciated ...

Peter

 

[castor28]

Hi Peter,

Take $V=\mathbb{R}^\mathbb{N}$, the set of sequences of real numbers $(x_1,x_2,\ldots)$ under pointwise addition. $V_n$ is the subspace of sequences in which all elements after the nth are 0: $x_i=0$ for $i>n$. The first such subspaces are:

$V_1=\{(x_1,0,0,\ldots)\mid x_1\in\mathbb{R}\}$
$V_2=\{(x_1,x_2,0,0,\ldots)\mid x_,x_2\in\mathbb{R}\}$
$V_3=\{(x_1,x_2,x_3,0,0,\ldots)\mid x_1,x_2,x_3\in\mathbb{R}\}$
​

and so on.

I think the inclusions $V_1 \subset V_2 \subset V_3 \subset\ldots$ are rather obvious; these are strict inclusions, and each $V_n$ is isomorphic to $\mathbb{R}^n$

 

[Math Amateur]

Hi Peter,

Take $V=\mathbb{R}^\mathbb{N}$, the set of sequences of real numbers $(x_1,x_2,\ldots)$ under pointwise addition. $V_n$ is the subspace of sequences in which all elements after the nth are 0: $x_i=0$ for $i>n$. The first such subspaces are:

$V_1=\{(x_1,0,0,\ldots)\mid x_1\in\mathbb{R}\}$
$V_2=\{(x_1,x_2,0,0,\ldots)\mid x_,x_2\in\mathbb{R}\}$
$V_3=\{(x_1,x_2,x_3,0,0,\ldots)\mid x_1,x_2,x_3\in\mathbb{R}\}$
​

and so on.

I think the inclusions $V_1 \subset V_2 \subset V_3 \subset\ldots$ are rather obvious; these are strict inclusions, and each $V_n$ is isomorphic to $\mathbb{R}^n$

Thanks for the help, castor28 ...

Mind you ... although I can see your definition of the V_i works ... I am not sure how you derived it from the definitions given by Bland ...

Can you comment ...

Peter

 

[castor28]

Thanks for the help, castor28 ...

Mind you ... although I can see your definition of the V_i works ... I am not sure how you derived it from the definitions given by Bland ...

Can you comment ...

Peter

Hi Peter,

The idea was to show a concrete and intuitive example.

As any basis of $V$ is infinite, it contains a countable subset.
I choose the set of elements $\{e_1=(1,0\ldots), (e_2 = (0,1,0\ldots), \ldots\}$. As these elements are linearly independent, they can be completed to a basis of $V$. Note that any basis of $V$ will do, since we only need to exhibit an infinite ascending chain of subspaces, and that property does not depend on a particular choice of basis.

The subspace $V_n$ is the subspace generated by $\{e_1,\ldots,e_n,0\ldots\}$; its elements are the sequences $\{(x_1, x_2,\ldots,x_n) \mid x_i\in\mathbb{R}\}$.

Note that the $\{e_i\mid i\in\mathbb{N}\}$ do NOT constitute a basis of $V$; they only span the set of sequences with finitely many non-zero terms (the direct sum instead of the direct product). A basis of $V$ is uncountable, and the proof of its existence relies on the axiom of choice; in practice, that means that you cannot describe such a basis explicitly.

You can keep the argument at an abstract level if you prefer, without assuming any form for a "given" basis $\{e_\alpha\}$ of $V$. You can still extract a countable sequence of basis elements $(e_1,e_2,\ldots)$ (although you can no longer write them down explicitly as sequences), and, for each $n$, define $V_n$ as the subspace generated by $\{e_1,\ldots,e_n\}$, which has dimension $n$. These subspaces satisfy the stated strict inclusions, since you get $V_{n+1}$ by adding $e_{n+1}$ to $V_n$.

 

[Math Amateur]

Hi Peter,

The idea was to show a concrete and intuitive example.

As any basis of $V$ is infinite, it contains a countable subset.
I choose the set of elements $\{e_1=(1,0\ldots), (e_2 = (0,1,0\ldots), \ldots\}$. As these elements are linearly independent, they can be completed to a basis of $V$. Note that any basis of $V$ will do, since we only need to exhibit an infinite ascending chain of subspaces, and that property does not depend on a particular choice of basis.

The subspace $V_n$ is the subspace generated by $\{e_1,\ldots,e_n,0\ldots\}$; its elements are the sequences $\{(x_1, x_2,\ldots,x_n) \mid x_i\in\mathbb{R}\}$.

Note that the $\{e_i\mid i\in\mathbb{N}\}$ do NOT constitute a basis of $V$; they only span the set of sequences with finitely many non-zero terms (the direct sum instead of the direct product). A basis of $V$ is uncountable, and the proof of its existence relies on the axiom of choice; in practice, that means that you cannot describe such a basis explicitly.

You can keep the argument at an abstract level if you prefer, without assuming any form for a "given" basis $\{e_\alpha\}$ of $V$. You can still extract a countable sequence of basis elements $(e_1,e_2,\ldots)$ (although you can no longer write them down explicitly as sequences), and, for each $n$, define $V_n$ as the subspace generated by $\{e_1,\ldots,e_n\}$, which has dimension $n$. These subspaces satisfy the stated strict inclusions, since you get $V_{n+1}$ by adding $e_{n+1}$ to $V_n$.

Hi castor28 ... thanks again!

Still reflecting on your post ...

Note that I still have trouble interpreting what Bland meant by the notation $$V_i = \bigoplus_{ i = 1 }^n x_{ \alpha_i } D$$ ... ... and the problem of determining what it means for $$n = 1,2,3, ... ...$$

Thanks again for your thoughts and your help ...

Peter

 

[castor28]

Hi castor28 ... thanks again!

Still reflecting on your post ...

Note that I still have trouble interpreting what Bland meant by the notation $$V_i = \bigoplus_{ i = 1 }^n x_{ \alpha_i } D$$ ... ... and the problem of determining what it means for $$n = 1,2,3, ... ...$$

Thanks again for your thoughts and your help ...

Peter

Hi Peter,

In this case, we have $D=\mathbb{R}$. Let us look first at the case $n=1$. We have:
$$V_1 = e_1\mathbb{R} = \{e_1x\mid x\in\mathbb{R}\}$$

This is a one-dimensional vector space (it is the free $\mathbb{R}$-module on $e_1$)

For $n=2$, we have the direct sum of two such subspaces (this is also a subspace):
$$\begin{align*}
V_2 &= e_1\mathbb{R}\oplus e_2\mathbb{R}\\
&= \{e_1x\mid x\in\mathbb{R}\} \oplus \{e_2y\mid y\in\mathbb{R}\}\\
&= \{e_1x + e_2y\mid x,y\in\mathbb{R}\}
\end{align*}$$
Note that the sum is direct because $e_1$ and $e_2$ are independent. This shows that $V_2$ is the subspace generated by $e_1$ and $e_2$, and the argument can be continued by induction.

 

[steenis]

Q1:
$V_1 = x_{\alpha_1} D \cong D$, with $x_{\alpha_1} \in \Gamma = \{\alpha_1, \alpha_2, \cdots \} $. It is a one-dimensional subspace of $V$ with basis $\{ x_{\alpha_1} \}$.

$V_2 = x_{\alpha_1} D + x_{\alpha_2} D$, with $x_{\alpha_1} \neq x_{\alpha_2}$ (why?).
thus $x_{\alpha_1} D \cap x_{\alpha_2} D = 0$ thus $V_2 = x_{\alpha_1} D \oplus x_{\alpha_2} D$.
So $V_2$ is a two-dimensional vectorspace with basis $\{x_{\alpha_1}, x_{\alpha_2} \}$
And so on.

Q2:
Take two elements $x_{\alpha_i}$ and $x_{\alpha_j}$, $i \neq j$, from $\Gamma \subset \Delta$, $\Delta$ is a basis of $V$, thus $x_{\alpha_i} \neq x_{\alpha_j}$.

The double indices are annoying so I write $x_\alpha$ for $x_{\alpha_i}$ and $x_\beta$ for $x_{\alpha_j}$.

Clear is that $x_\alpha D \leq x_\alpha D + x_\beta D $, you can prove that.

However, because $x_\beta \notin x_\alpha D$, we have $x_\alpha D \neq x_\alpha D + x_\beta D $

and so $x_\alpha D < x_\alpha D + x_\beta D $:

$x_\alpha D$ is a proper subspace of $x_\alpha D + x_\beta D $

You can apply this to $V_1$, $V_2$ ...