I 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 ... ...

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

Example 3, Section 4.2 reads as follows:

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

 

[andrewkirk]

$$V_1\triangleq x_{\alpha_1}D\triangleq \{x_{\alpha_1}d\ :\ d\in D\}$$
and recall that $x_{\alpha_1}$ is the first element of the set $\Gamma$ and 'scalar' multiplication of module elements by division-ring elements occurs on the right.

The direct sum referred to is an internal direct sum.

For $n\geq 1$ we have:

$$V_n \triangleq \bigoplus_{k=1}^n x_{\alpha_k} D
\triangleq \left\{\sum_{k=1}^n x_{\alpha_k}d_k\ :\ \forall k\ (d_k\in D)\right\}$$

To see that $V_{n}\subset V_{n+1}$ observe that $V_n$ is just the set of elements of $V_{n+1}$ with $d_{n+1}=0_D$.

I can't help adding that it strikes me as unfortunate that the author uses this (in my opinion) ugly method of double-indexing the elements of $\Gamma$, as $x_{\alpha_i}$. It is much cleaner and less confusing to make $\Delta$ the basis, rather than an index set for a basis, and just write $x$ for an arbitrary element of the basis, and $x_i$ for an element of the countable subset $\Gamma$ of the basis set. With the double-indexing it is harder to read, harder to conceptualise and it takes longer to write the latex code.

 

[Math Amateur]

$$V_1\triangleq x_{\alpha_1}D\triangleq \{x_{\alpha_1}d\ :\ d\in D\}$$
and recall that $x_{\alpha_1}$ is the first element of the basis and 'scalar' multiplication of module elements by division-ring elements occurs on the right.

The direct sum referred to is an internal direct sum.

For $n\geq 1$ we have:

$$V_n \triangleq \bigoplus_{k=1}^n x_{\alpha_k} D
\triangleq \left\{\sum_{k=1}^n x_{\alpha_k}d_k\ :\ \forall k\ (d_k\in D)\right\}$$

To see that $V_{n}\subset V_{n+1}$ observe that $V_n$ is just the set of elements of $V_{n+1}$ with $d_{n+1}=0_D$.

I can't help adding that it strikes me as unfortunate that the author uses this (in my opinion) ugly method of double-indexing his basis, as $x_{\alpha_i}$. It is much cleaner and less confusing to just write $x_i$ if the index set $\Delta$ is known to be countable, and $x_\alpha$ with $\alpha \in\Delta$ where $\Delta$ is not known to be necessarily countable. With the double-indexing it is harder to read, harder to conceptualise and it takes longer to write the latex code.

Thanks for the help and guidance Andrew ...

But ... just a clarification ...

You write:

" ... ... $x_{\alpha_1}$ is the first element of the basis ... ... "

But the basis is $\{ x_\alpha \}_\Delta$ not $\{ x_\alpha \}_\Gamma$ ... $\alpha_1$ is the first element of the index set $\Gamma$ ...

Isn't it possible that $\Delta$ is an uncountably infinite set with no first element ... but ... not sure what this would mean for the basis ...

Can you clarify ...?

Peter

 

[andrewkirk]

Yes, sorry, I should have written "first element of the set $\Gamma$", which is prescribed as a countable, indexed subset of $\Delta$. I'll go back and correct my post.

 

[Math Amateur]

Thanks Andrew ...

Peter