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In page 66 of Algebraic Number Theory Book by Neukirch,he defines $A(X)$ as the localization from integral domain $A$ with multiplicative subset $S$ as complement of $\bigcup_{P\in X} P $ where $P$ are prime ideals of $A$ over a set $X$.Hence prime ideals of $A(X)$ have $1-1 $ correspondence to a prime ideal containing in $\bigcup_{P\in X} P $.

1)Now it mentions that if $X$ is finite then only the prime ideals from $X$ survives in $A(X)$.This statement I couldn't prove except for special case when A is Dedekind domain where I used the proposition that if $I \subseteq \bigcup_{i=1}^{n} P_{i} $ where $I$ is any ideal in a commutative unital ring then $I \subseteq P_{i} $for some $i $.

I have further doubt in Pg-70 where

2)$X$ is defined as a set of non zero prime ideals of dedekind domain $\mathcal{O}$ which contains almost all prime ideals of it.Now what is this almost all?

3)Also how the only prime ideals in $\mathcal{O}(X)$ are just the images of $P \in X$ ,I mean $P_X = P\mathcal{O}(X)$, for $P \in X$ are the non zero prime ideals of $\mathcal{O}(X)$ and not for any other prime ideal containing in $\bigcup_{P\in X} P $.

4)Also further it uses this fact in proof of $(11.6)$ which is for $P \notin X$ we have $P\mathcal{O}(X) = (1)$.Here how can I proof that for $P^{'}\notin X \implies P^{'}\nsubseteq \bigcup_{P\in X} P $ so that it can prove the above statement?

No pie
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  • Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. – Community May 09 '22 at 14:00
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    The statement that $I \subset \bigcup\limits_{i=1}^n P_i$ implies $I \subset$ some $P_j$ is true in any commutative unital ring, not just Dedekind domains. – D_S May 09 '22 at 14:05
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    "Almost all" means "all but finitely many." I can't tell exactly what else you want to know as your question is a bit jumbled, please clarify. – D_S May 09 '22 at 14:07
  • Ok i edited it and bulleted the question.I understood the 2nd question as author introduce it into next section. – No pie May 09 '22 at 16:01
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    I was just reading this fragment and stumbled upon the same thing.
    Ad 1)
    As D_S wrote, it is true in general (see 'prime avoidance lemma')
    Ad 3) and 4)
    This is actually not true, so it's a slight mistake in a book (see https://math.stackexchange.com/questions/51362/can-a-prime-in-a-dedekind-domain-be-contained-in-the-union-of-the-other-prime-id for a counterexample and in-depth discussion). But if we set $Y={p\neq 0:p\subset \bigcup_{q\in X}q}$ then $X\subset Y$ so $Y$ omits only finitely many prime ideals and for $Y$ the statement from 4) is true (by the definition of $Y     – filipux Jun 07 '22 at 15:12

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