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Math Help - Noetherian Rings

  1. #1
    Super Member Bernhard's Avatar
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    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
    Last edited by Bernhard; July 19th 2013 at 08:19 PM.
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    Ant
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    Re: Noetherian Rings

    Quote Originally Posted by Bernhard View Post
    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!
    Thanks from Bernhard
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    Super Member Bernhard's Avatar
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    Re: Noetherian Rings

    Quote Originally Posted by Ant View Post
    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
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