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Topic: Algebraic Over a Noetherian Domain (Read 692 times) |
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ChetN
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Posts: 6
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Algebraic Over a Noetherian Domain
« on: Sep 13th, 2006, 3:47pm » |
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I would appreciate any clues to this one::
Let E be a field extension of a Noetherian unique factorization domain D. Suppose that E = QD[a1,...,an] for some a1,...,an, and QD is quotient field of D. Show that E is algebraic over D.
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| « Last Edit: Sep 13th, 2006, 3:47pm by ChetN » |
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Deedlit
Senior Riddler
Posts: 476
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Re: Algebraic Over a Noetherian Domain
« Reply #1 on: Sep 13th, 2006, 4:27pm » |
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Does "field extension of a domain D" mean something other than an extension of QD?
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ChetN
Newbie
Posts: 6
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Re: Algebraic Over a Noetherian Domain
« Reply #2 on: Sep 14th, 2006, 2:51pm » |
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no...it is the same thing
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Michael Dagg
Senior Riddler
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Posts: 500
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Re: Algebraic Over a Noetherian Domain
« Reply #3 on: Sep 15th, 2006, 11:45am » |
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Very Nice problem!
Just as good as solving this problem itself is the discovery
that it is one of several theorems that are closely related,
but this one is not as visible as its relatives (but is one in a
different form).
Getting to this point with this particular problem says
that you already know that the homomorphic image of a
Noetherian ring is itself Noretherian (that's certainly
textbook nevertheless) and, yes, apart from the problem
statement we really are really talking about rings in the
general sense and here we also know about R-modules and
that a quotient field is the smallest field extension into
which an integral domain may be embedded.
The main result is about the existence of zeros in the
algebraic closure of D .
You can show this by contradiction if you can prove:
Let E = D(Z1,...,Zn) be a rational function field over
D with n >= 1 indeterminates, then E is not
finitely generated as ring over Q_D.
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| « Last Edit: Sep 15th, 2006, 11:48am by Michael Dagg » |
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Regards,
Michael Dagg
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Deedlit
Senior Riddler
Posts: 476
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Re: Algebraic Over a Noetherian Domain
« Reply #4 on: Sep 15th, 2006, 1:48pm » |
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I'm afraid I'm not following. It doesn't look like we're given any information at all about E, beyond that it is a field extension of D. So why not just take QD [a], with a transcendental?
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ChetN
Newbie
Posts: 6
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Re: Algebraic Over a Noetherian Domain
« Reply #5 on: Sep 16th, 2006, 8:26am » |
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it is not that easy. i know that D can be obtained from QD by a ring adjunction. i think i do and also I think I don't know what he means by n > 1 indetermines.
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| « Last Edit: Sep 16th, 2006, 8:45am by ChetN » |
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ChetN
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Posts: 6
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Re: Algebraic Over a Noetherian Domain
« Reply #6 on: Sep 16th, 2006, 8:47am » |
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ah never mind the indeterminates...
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ChetN
Newbie
Posts: 6
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Re: Algebraic Over a Noetherian Domain
« Reply #7 on: Sep 16th, 2006, 2:22pm » |
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ok, without proof and assuming that statement is true "does" give me a tool to work with namely in the sense that E is generated as an S module with S=QD, as i am talking about S not being a generated ring over QD
as for the transcendental suggestion...I have to ask if you know or not if you do "know" that a transcendental extension of rationals in a field/set by itself iis/does make any sense here over an integral domain and.... hardly not workable here and humm well at least to my knowledge?
please enlighten me otheriwise
actually a transcendental field is not an extension of a rational field because there is no rationals in this kind of extension
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ChetN
Newbie
Posts: 6
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Re: Algebraic Over a Noetherian Domain
« Reply #9 on: Sep 18th, 2006, 7:27am » |
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i am messed up
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Michael Dagg
Senior Riddler
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Re: Algebraic Over a Noetherian Domain
« Reply #10 on: Sep 27th, 2006, 2:52pm » |
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OK, well, for those who may be interested in this problem, it is the
field version of the Hilbert Nullstellensatz theorem. You may not be
able to google this particular form but you will get the others.
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Regards,
Michael Dagg
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