My Math Forum Need some help with a Dedekind domain proof

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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 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 dont 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.Marcuss 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 $A$ and $B$ are ideals of a ring $R$, then an element in $AB$ is of the form $a_1b_1 + \cdots + a_kb_k$ with $a_i \in A$ and $b_i \in B$. Since an ideal is closed under multiplication from any element in R, $a_ib_i$ is both in $A$ and in $B$. Similarly, since ideals are closed under addition, $a_1b_1 + \cdots + a_kb_k$ is an element of both $A$ and of $B$. Hence we have that $AB \subset A$ and $AB \subset B$. vi) If we have $I= M$ then $M=PI$ becomes $M=PM$. For any ideal $A$ of $R$, we have that $RA=A$ because $1 \in R$ so $1 \cdot a= a \in RA$ for all $a \in A$. Hence we have $RM=PM$ and using cancellation, we have $R=P$ but $P$ 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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