MHB Help with Paul E. Bland's Division Rings and IBN-rings Prop 2.2.10 Proof

[Math Amateur]

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 the proof of Proposition 2.2.10.

Proposition 2.2.10 and its proof read as follows:View attachment 3587My question/problem is concerned with Bland's proof of Proposition 2.2.10 above.

Bland asks us to consider two bases $$\mathscr{B}$$ and $$\mathscr{B}'$$ where

$$\mathscr{B} = \{ x_1,x_2, \ ... \ ... \ x_n \} $$

and

$$ \mathscr{B}' = \{ y_1,y_2, \ ... \ ... \ y_m \} $$ Bland then goes through a process whereby he starts with the basis

$$\mathscr{B} = \{ x_1,x_2, \ ... \ ... \ x_n \} $$

and replaces various $$x_i$$ with $$y_i$$ until he reaches the basis

$$\{ \text{ the } x_i \text{ not eliminated } \} \cup \{ y_1,y_2, \ ... \ ... \ y_m \}$$THEN ... ... Bland argues as follows:

" ... ... But $$\mathscr{B}'$$ is a maximal set of linearly independent vectors of V, so it cannot be the case the $$n \lt m$$. Hence $$n \ge m$$. Interchanging $$\mathscr{B}$$ and $$\mathscr{B}'$$ in the argument gives $$m \ge n$$ and this completes the proof. ... ... "HOWEVER ... ... we know (straight away, without going through the construction process), that since $$\mathscr{B}$$ and $$\mathscr{B}'$$ are both maximal sets of linearly independent vectors that we cannot have $$n \lt m$$, nor can we have $$m \gt n$$, so we must have $$m = n$$.My question is as follows:

What is the relevance of the construction process above that results in the basis:

$$\{ \text{ the } x_i \text{ not eliminated } \} \cup \{ y_1,y_2, \ ... \ ... \ y_m \}$$?

Why do we need this process/construction?

Indeed, the argument for the Proposition seems to follow straight from the fact that both $$\mathscr{B}$$ and $$\mathscr{B}'$$ are both maximal sets of linearly independent vectors ... ... ?

Can someone please clarify the above issue?

Peter

 

[Fallen Angel]

Hi Peter,

First, I don't have this book so I have no idea what it says at Prop 2.2.8

But, being a maximal set of linearly independent vectors means that every other vector that you could add will be linearly dependent, not that it's maximal in the number of elements.

In the case we are working, with vector spaces, the fact that given two bases, both have the same cardinality needs a prove.
He assumes that $$n >m$$ and then adds some linearly indepent vectors to one of the basis, what is a contradiction with the fact to be maximal, and the other inequality holds just by the symmetry of the proof.

Actually, this is not true for general modules.