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Let $O$ be a noetherian local domain with maximal ideal $m$. I want to prove: for a suitable choice of generators $x_1,\dots,x_n$ of $m$, the ideal $(x_1)$ in $O'=O[x_2/x_1,\dots,x_n/x_1]$ is not equal to the unit ideal.

This statement originates from Ex.4.11, Chapter 2 of Hartshorne.

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If one is willing to use the results already proved in Hartshorne in the context of the valuative criterion, that is before exercise 4.11, I see the following approach: there exists a valuation ring $O_v$ of the field $K$ (for the moment I ignore the finite extension $L$ that appears in the exercise) that dominates the local ring $O$. In particular we have $v(x_k)>0$ for any set $x_1,\ldots ,x_n$ of generators of the maximal ideal $m$ of $O$. Suppose $v(x_1)$ is minimal among the values $v(x_k)$. Then $O^\prime\subseteq O_v$ and $q:=M_v\cap O^\prime$, $M_v$ the maximal ideal of $O_v$, is a proper prime ideal of $O^\prime$. By definition $x_1\in q$ and thus $x_1O^\prime\neq O^\prime$. The "suitable choice" is just relabelling the elements $x_k$ if necessary.

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I roughly got your idea,thank you! Here are few comments and questions:(1) For the exercise alone, it is enough to just consider $L=K$ as you did.(2)I do not quite follow the claim"$\dots M_{v}$ the maximal ideal of $O_{v}$ is a proper prime ideal of $O'$" , and I cannot see it is relevant to $x_1O' \neq O'$.(3)Which results in Harshorne did you use ? I guess the only thing you use is the existence of $O_{v}$, which can be derived from "the valuation ring is the maximal local ring in the order of domination". – Li Zhan Jul 6 '11 at 12:39
    
(2) I mean that $q:=M_v\cap O^\prime$ is a proper prime ideal of $O^\prime$. Of course then $x_1O^\prime$ is a proper ideal of $O^\prime$ since it is contained in $q$. (3) Yes. – Hagen Knaf Jul 6 '11 at 15:28
    
Great!Thank you! – Li Zhan Jul 6 '11 at 21:23

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