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November 20th, 2010, 10:27 PM   #1
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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 non-empty 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 A|B 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 non-empty
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: A|B 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)
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November 21st, 2010, 02:35 PM   #2
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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.
Turgul is offline  
November 22nd, 2010, 03:53 AM   #3
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Re: Need some help with a Dedekind domain proof

Ah, ok I see. Thank you
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