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 \(\displaystyle R\) as a commutative ring ... ... and then has to show (see above definition of an IBN-ring) that for every free \(\displaystyle R\)-module \(\displaystyle F\), any two bases of \(\displaystyle F\) have the same cardinality.Bland then considers a free module \(\displaystyle F\) ... ... ... ... BUT ... instead of considering bases of \(\displaystyle F\) ... ... ... ... Bland, instead, shows that \(\displaystyle \{ x_\alpha + F \mathscr{m} \}_\Delta\) is a basis for the vector space \(\displaystyle F/F \mathscr{m}\) ... ... and also shows that \(\displaystyle \text{dim}_{R/ \mathscr{m}} ( F/F \mathscr{m} ) = \text{ card } ( \Delta )\)

... ... and then shows that

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

for any other basis \(\displaystyle \Gamma\) of the vector space \(\displaystyle F/F \mathscr{m}\)
BUT ... ... we should be showing that all bases of \(\displaystyle F\) (and NOT \(\displaystyle F/F \mathscr{m}\) ) have the same cardinality ? !... ... so then ... ... how has Bland shown that every free \(\displaystyle R\)-module \(\displaystyle F\) of the ring \(\displaystyle R\) 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 \(\displaystyle \nabla, \Gamma \) of the free module \(\displaystyle F\) and then , in order to show they have the same cardinality, he constructs a vector space over \(\displaystyle R\) in such a way that two basis of the vector space has cardinality \(\displaystyle card(\nabla),card(\Gamma)\).
Hence \(\displaystyle card(\nabla)=card(\Gamma)\) (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 \(\displaystyle \nabla, \Gamma \) of the free module \(\displaystyle F\) and then , in order to show they have the same cardinality, he constructs a vector space over \(\displaystyle R\) in such a way that two basis of the vector space has cardinality \(\displaystyle card(\nabla),card(\Gamma)\).
Hence \(\displaystyle card(\nabla)=card(\Gamma)\) (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 \(\displaystyle \{ x_\alpha + F \mathfrak{m} \}_{ \Delta }\) and \(\displaystyle \{ x_\beta + F \mathfrak{m} \}_{ \Gamma }\) are not only bases for the vector space \(\displaystyle F/ F \mathfrak{m}\) ... ... BUT ... are also bases for \(\displaystyle F\) as well?

But I do not see how they are bases for \(\displaystyle F\).

Are you able to explain how they are bases for \(\displaystyle F\) ... ?

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

PeterPS By the way, thanks for you help with the symbol \(\displaystyle \mathfrak{m}\)***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 \(\displaystyle \{x_\alpha \}_\Delta \)of cardinality \(\displaystyle \Delta\) and then construct a basis (of the same cardinality) \(\displaystyle \{ x_\alpha + F \mathfrak{m} \}_{ \Delta }\) of the vector space \(\displaystyle F/ F \mathfrak{m}\).Then we consider another basis \(\displaystyle \{ \overline{ x}_\beta \}_\Gamma \) ... ... and through the same argument show that

\(\displaystyle \{ \overline{x}_\beta + F \mathfrak{m} \}_{ \Gamma }\)

is also a basis of the vector space \(\displaystyle F/ F \mathfrak{m}\)

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

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

Is that correct?

 

[Fallen Angel]

Yeah, you got it! :D

 

[Math Amateur]

Yeah, you got it! :D

Yes, thanks to you!

Peter