Noetherian Rings and the Free R-module R^(n) .... Bland, Corollary 4.2.8 .... ....

[Math Amateur]

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

I am focused on Section 4.2: Noetherian and Artinian Modules and need some help to understand the proof of Corollary 4.2.8

Proposition 4.2.7 and its Corollary 4.2.8 read as follows:View attachment 8210Bland states but does not prove Corollary 4.2.8 ...

Can someone please help me to establish a proof for Corollary 4.2.8 ...

Help will be appreciated ...

Peter=================================================================================The above text by Bland refers to right Noetherian rings and to \(\displaystyle R^{ (n) }\) ... Bland's definitions for these entities follow ...Bland defines right Noetherian rings in Definition 4.2.1 which reads as follows:View attachment 8211Bland defines the free module \(\displaystyle R^{ (n) }\) on page 52 as follows:View attachment 8212
Hope the text above helps ...

Peter

 

[steenis]

Use the "Notation" on p.51, cor.2.2.4. and prop.4.2.7.

 

[Math Amateur]

Use the "Notation" on p.51, cor.2.2.4. and prop.4.2.7.

By definition (Notation page 51) ... where \(\displaystyle \{ M_\alpha \}_{ i = 1 }^n\) is a family of modules such that \(\displaystyle M_\alpha = M \) ...

... we have that \(\displaystyle M^{ (n) } = M \times M \times \ ... \ ... \ \times M\) ... (n factors)

... in other words \(\displaystyle R^{ (n) } = R \times R \times \ ... \ ... \ \times R\) ... (n factors) ... ... ... ... ... (1)

... and ...

... since we are dealing with a finite direct product we have ...

\(\displaystyle R^{ (n) } \cong \bigoplus_{ i = 1 }^n R_\alpha\) where \(\displaystyle R_\alpha = R\) ... ... ... ... ... (2)But ... problem ... formulae (1) and (2) are dealing with rings not modules ... is this a problem ...?

Can you clarify ... ?
Note:

I know a ring can be considered as a "module over itself" ... but does this mean that for the proof we simply consider the rings as modules .. ... ? ... ... or does it have other effects ...

If we just simply consider the rings as modules then it seems to me we can just apply Proposition 4.2.7 with \(\displaystyle M_i = R_i\) ... but then we are not using Corollary 2.2.4 that you suggested was necessary ...

Can you help ...

Peter

 

[steenis]

Yes, consider $R$ as a right $R$-module, see definition 4.2.1
prop. 4.2.7 will do, cor.2.2.4 is not really necessary

 

[Math Amateur]

Yes, consider $R$ as a right $R$-module, see definition 4.2.1
prop. 4.2.7 will do, cor.2.2.4 is not really necessary

Thanks for the help, Steenis ...

Peter

 

[Math Amateur]

Yes, consider $R$ as a right $R$-module, see definition 4.2.1
prop. 4.2.7 will do, cor.2.2.4 is not really necessary

Thanks again Steenis ...

Proof of Corollary 4.2.8 should then be as follows:

A ring \(\displaystyle R\) is Noetherian (and hence a Noetherian module)

\(\displaystyle \Longleftrightarrow \bigoplus_{ i = 1 }^n R_i\) is a Noetherian module by Proposition 4.2.7

\(\displaystyle \Longleftrightarrow R^{ (n) }\) is a Noetherian module by definition of \(\displaystyle R^{ (n) }\) ... Is that correct?

Peter

 

[steenis]

Yes, that is correct

(mention that $R_i = R$)