# Thread: Noetherian Rings

1. ## Noetherian Rings

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

2. ## Re: Noetherian Rings

Originally Posted by Bernhard
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

Hi,

I believe the answer is yes. As you say, R is an ideal of R hence if R is noetherian then R must be finitely generated.

In general, when you consider R as an ideal of R then it is certainly finitely generated (simply take 1 as a generating set!).

So yes, fields only have two ideals both of which are finitely generated; {0} is an obvious generating set for the ideal {0}, and {1} is a generating set for R.

Of course R need not be finite in size!

It's been a little while since I've covered this material so don't take the above as gospel. Good luck!

3. ## Re: Noetherian Rings

Originally Posted by Ant
Hi,

I believe the answer is yes. As you say, R is an ideal of R hence if R is noetherian then R must be finitely generated.

In general, when you consider R as an ideal of R then it is certainly finitely generated (simply take 1 as a generating set!).

So yes, fields only have two ideals both of which are finitely generated; {0} is an obvious generating set for the ideal {0}, and {1} is a generating set for R.

Of course R need not be finite in size!

It's been a little while since I've covered this material so don't take the above as gospel. Good luck!
Thanks Ant, most helpful ... Appreciate your assistance ...

Peter