MHB Commutative rings are IBN-rings - Bland Proposition 2.2.11

[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 2.2.11.

Proposition 2.2.11 and its proof read as follows:View attachment 3588
Proposition 2.2.11 relies on the definition of an IBN-ring so I am providing Bland's definition of an IBN-ring which reads as follows:
View attachment 3589
Now in the proof of Proposition 2.2.11, Bland defines $$R$$ as a commutative ring ... ... and then has to show (see above definition of an IBN-ring) that for every free $$R$$-module $$F$$, any two bases of $$F$$ have the same cardinality.Bland then considers a free module $$F$$ ... ... ... ... BUT ... instead of considering bases of $$F$$ ... ... ... ... Bland, instead, shows that $$\{ x_\alpha + F \mathscr{m} \}_\Delta$$ is a basis for the vector space $$F/F \mathscr{m}$$ ... ... and also shows that $$\text{dim}_{R/ \mathscr{m}} ( F/F \mathscr{m} ) = \text{ card } ( \Delta )$$

... ... and then shows that

$$\text{ card } ( \Delta ) = \text{ card } ( \Gamma )$$

for any other basis $$\Gamma$$ of the vector space $$F/F \mathscr{m}$$
BUT ... ... we should be showing that all bases of $$F$$ (and NOT $$F/F \mathscr{m}$$ ) have the same cardinality ? !... ... so then ... ... how has Bland shown that every free $$R$$-module $$F$$ of the ring $$R$$ has bases of the same cardinality?Could someone explain the logic of Bland's proof of Proposition 2.2.11 ...

I would really appreciate help ...

Peter

 

[Fallen Angel]

Hi Peter,Bland considers two bases with set of indices $$\nabla, \Gamma $$ of the free module $$F$$ and then , in order to show they have the same cardinality, he constructs a vector space over $$R$$ in such a way that two basis of the vector space has cardinality $$card(\nabla),card(\Gamma)$$.
Hence $$card(\nabla)=card(\Gamma)$$ (Becasuse it is proved that two bases of a vector space has the same cardinality) and then the cardinality of the two initial bases of $F$ is the same.

 

[Math Amateur]

Hi Peter,Bland considers two bases with set of indices $$\nabla, \Gamma $$ of the free module $$F$$ and then , in order to show they have the same cardinality, he constructs a vector space over $$R$$ in such a way that two basis of the vector space has cardinality $$card(\nabla),card(\Gamma)$$.
Hence $$card(\nabla)=card(\Gamma)$$ (Becasuse it is proved that two bases of a vector space has the same cardinality) and then the cardinality of the two initial bases of $F$ is the same.

Hi Fallen Angel,

Thanks so much for your help ... really appreciate it ...

Now ... ... just reflecting on your argument ...

So you are saying that $$\{ x_\alpha + F \mathfrak{m} \}_{ \Delta }$$ and $$\{ x_\beta + F \mathfrak{m} \}_{ \Gamma }$$ are not only bases for the vector space $$F/ F \mathfrak{m}$$ ... ... BUT ... are also bases for $$F$$ as well?

But I do not see how they are bases for $$F$$.

Are you able to explain how they are bases for $$F$$ ... ?

Sorry to be slow ... and not follow you exactly ...

PeterPS By the way, thanks for you help with the symbol $$\mathfrak{m}$$***EDIT*** ***EDIT*** ***EDIT***oh! thanks to you, I think I now see the logic ... I was not reading the proof carefully enough ...

We start with a basis $$ \{x_\alpha \}_\Delta $$of cardinality $$\Delta$$ and then construct a basis (of the same cardinality) $$\{ x_\alpha + F \mathfrak{m} \}_{ \Delta }$$ of the vector space $$F/ F \mathfrak{m}$$.Then we consider another basis $$ \{ \overline{ x}_\beta \}_\Gamma $$ ... ... and through the same argument show that

$$\{ \overline{x}_\beta + F \mathfrak{m} \}_{ \Gamma }$$

is also a basis of the vector space $$F/ F \mathfrak{m}$$

Then since the dimension of a vector space is unique, we have

$$\text{ card } ( \Delta ) = \text{ card } ( \Gamma )$$

Is that correct?

 

[Fallen Angel]

Yeah, you got it! :D

 

[Math Amateur]

Yeah, you got it! :D

Yes, thanks to you!

Peter