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Mar 26, 2013 at 21:04 vote accept Martin Brandenburg
Mar 24, 2013 at 4:12 comment added user30180 Dear Martin: The advantage of being at a department as strong as Muenster is that it has a lot of people who know number theory quite well both at this level and way beyond, so you can also ask any of them in person. Proceeding in that way you may give more understanding of the reasoning than by waiting for someone on MO to declare "Yes, I agree with this argument" (assuming I have not made a blunder, which I do not believe I have).
Mar 24, 2013 at 3:00 comment added Martin Brandenburg Thanks a lot for your answer, ayanta. Unfortunately I don't have enough background in algebraic number theory. I didn't imagine that such a simple question requires so much theory. There are many number theorists on MO, perhaps one of them can confirm the proof?
Mar 23, 2013 at 1:16 comment added user30180 Dear Aakumadula: Here is the argument I had in mind. Since "ideal generated by" is inverse to the operation of "intersection" when considering maximal ideals relatively prime to the index (so to speak), we just use a bit of transitivity: if $A'' \subset A' \subset A$ is a containment of orders and $P$ is a maximal ideal of $A$ whose residue characteristic doesn't divide $[A:A'']$ then the prime ideals $P' = A' \cap P$ and $P'' = P' \cap A'' = P \cap A''$ satisfy $P'' A' = P'$ (!) so if $P \cap A''$ is principal then such a generator in $A''$ also generates $P'$ in $A'$.
Mar 20, 2013 at 7:37 comment added Venkataramana I sort of agree, but it is not clear that for the order $A={\mathbb Z[x]/(F)$, the intersection of the principal maximal ideal with $A$ is actually principal even if it is so for the smaller ring ${\mathbb Z}+MO_K$. .
Mar 20, 2013 at 4:34 history answered user30180 CC BY-SA 3.0