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Noetherian Rings

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Dummit and Foote in Chapter 9 - Polynomial Rings define Noetherian rings as follows:

Definition. A commutative ring R with 1 is called Noetherian if every ideal of R is finitely generated.

Question: Does this mean that R itself must be finitely generated since R is an ideal of R?

This question is important in the context of fields since D&F go on to say that every field is Noetherian. In the case of fields the only ideals are the trivial ideal {0} and the field itself. But this would mean every field is finitely generated which does not seem to be corect.

Can anyone clarify these issues for me?

Peter

[This has also been posted on MHF]
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Question: Does this mean that R itself must be finitely generated since R is an ideal of R?
Take into account that for every commutative and unitary ring $R$ (noetherian or not), we have $R=R\cdot 1=(1)$.

But this would mean every field is finitely generated which does not seem to be corect.
Why? In such a case, $R$ is an $R$-vector space and a basis is $B=\{1\}$.