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November 20th, 2010, 10:27 PM  #1 
Newbie Joined: Nov 2010 Posts: 10 Thanks: 0  Need some help with a Dedekind domain proof
Hi! Firstly I will just put in some theory that we will use in the proof: Def.: A Dedekind domain is an integral domain R such that: (1) Every ideal is finitely generated; (2) Every nonzero prime ideal is a maximal ideal; (3) R is integrally closed in its field of fractions K = {a/b: a,b ? R, b ? 0} And futheremore (1) is equivalent to the conditions: (1') Every increasing sequence of ideals is eventually constant. (1'') Every nonempty set S of ideals has a (not necessarily unique) maximal member "M". Lemma: In a Dedekind domain every ideal contains a product of prime ideals Cor.2: (cancellation law) If A,B,C are ideals in a Dedekind domain, and AB = AC, then B = C Cor.3: If A and B are ideals in a Dedekind domain R, then AB iff A ? B And here come the theorem that I want to prove: Thm.1: Every ideal in a Dedekind domain is uniquely representable as a product of prime ideals in the proof, I just say "ok", if I understand the step and a comment or "?" if I do not understand it. Proof of thm.1: PART1: Every ideal is representable as a product of primes: i) If this was not true, then the set of ideals that could not be represented in this way would be nonempty and have a maximal member "M" (by (1'')). ok ii) M is different from R (we just consider proper ideals). ok iii) M is contained in a prime ideal (This come from a result that say that every proper ideal is contained in a maximal ideal, which must also be a prime ideal). ok iv) Then M = PI, where I is an ideal (Cor.3: AB iff A contain B). ok v) I contain M. ? (why is that?) vi) From (cor.2.) show that the containment is strict. If I = M then RM = PM,then R = P, which is absurd? (I don`t quite see this) vii) Then I is strictky larger then M and consequently I is a product of primes. But then so is M, contrary to the assumption I would really appreciate it if some one could lend me a hand with this proof. Thanks for all your effort. (The proof can be found in Daniel A.Marcus`s Number Fields page 59) 
November 21st, 2010, 02:35 PM  #2 
Senior Member Joined: Aug 2010 Posts: 195 Thanks: 5  Re: Need some help with a Dedekind domain proof
v) If and are ideals of a ring , then an element in is of the form with and . Since an ideal is closed under multiplication from any element in R, is both in and in . Similarly, since ideals are closed under addition, is an element of both and of . Hence we have that and . vi) If we have then becomes . For any ideal of , we have that because so for all . Hence we have and using cancellation, we have but is prime and hence a proper ideal, so this is impossible. 
November 22nd, 2010, 03:53 AM  #3 
Newbie Joined: Nov 2010 Posts: 10 Thanks: 0  Re: Need some help with a Dedekind domain proof
Ah, ok I see. Thank you 

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