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Forum index » Science and Technology » Math » Research
Definition for UFD
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Pete Klimek
science forum beginner


Joined: 15 May 2005
Posts: 7

PostPosted: Fri Jun 09, 2006 3:02 am    Post subject: Re: Definition for UFD Reply with quote
Quote:
...
To prove the second part of the claim, we'll show that the quotient field
K
of R is Q(X) (i.e., all rational functions of X with coefficients in Q).
Since
Q(X) clearly satisfies Ribenboim's UFD condition, this will complete the
proof.
...

No, this part doesn't work. It works if we write Ribenboim's condition
using the irreducibles in Z[X], which of course is a UFD. But there are
other irreducibles in R. For instance, clearly X+1 and X-2 are irreducible
in R. Their product is X^2 - 2X - 2. But that has value 2 at X=0 and X=1,
so (1/2)X^2 - X - 1 is in R and is easily seen to be another irreducible.


William C. Waterhouse
Penn State


Thanks for the correction. I don't see any obvious way to fix this argument,
so the original question remains unanswered.

Pete Klimek
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Hagen
science forum beginner


Joined: 28 Apr 2005
Posts: 42

PostPosted: Wed Jun 14, 2006 8:13 am    Post subject: Re: Definition for UFD Reply with quote
So, the original question is still open!
Hello ring theorists - any suggestions?

What we know so far is that Ribenboim's
definition coincides with the original one
if every element of the ring can be factored
into irreducibles.

A potential counter example thus must be
non-noetherian or more precisely must possess
an infinite strictly ascending chain of principal
ideals.

H
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