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Is it true that every prime ideal of height one is principal ?

Please help

user26857
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    http://math.stackexchange.com/questions/1186419/is-a-height-one-ideal-in-a-ufd-principal – user40276 Jul 28 '15 at 05:33
  • is every principal prime ideal height 1? – JMP Jul 28 '15 at 05:37
  • No, that is something different –  Jul 28 '15 at 05:37
  • @JonMarkPerry $(0)$ is not height 1 in a domain and is principal (and prime). I think what you are looking for is Krull's principal ideal theorem. In a Noetherian ring, any principal proper ideal has rank at most 1. https://en.wikipedia.org/wiki/Krull%27s_principal_ideal_theorem – CPM Jul 28 '15 at 05:52
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    Rank one valuation domains are not necessarily noetherian, so their maximal ideal can be even infinitely generated. – user26857 Jul 28 '15 at 06:33
  • Exactly. Strange stuff happens when rings are not Noetherian, but that makes them fun to study! – CPM Jul 28 '15 at 06:38

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Any Dedekind domain is dimension one, but not all Dedekind domains are PID's. For instance $\mathbb{Z}[\sqrt{-5}]$ is a standard example. The ideal $(2,1+\sqrt{-5})$ is maximal and therefore prime and is not the zero ideal, so it has rank $1$, but is not principal.

CPM
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