Hello is there anyone that would know where I can find an example of a noetherian N-1 ring that is not a Nagata ring. (See the Wikipedia article "Nagata ring" for the definitions of N-1 ring and Nagata ring.)
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3$\begingroup$ Did you check Nagata's "Local rings"? $\endgroup$– Hailong DaoCommented Mar 27, 2011 at 22:40
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4$\begingroup$ The title does not make sense... $\endgroup$– Mariano Suárez-ÁlvarezCommented Mar 28, 2011 at 7:56
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$\begingroup$ there is no counter example to the above question in Nagata's Local rings I checked $\endgroup$– user13953Commented Apr 6, 2011 at 1:18
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$\begingroup$ There is such a counter example on pages 206 - 207. $\endgroup$– HagenCommented Apr 6, 2011 at 8:32
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There is a discrete valuation ring $R$ (hence trivially N-1) of characteristic $p>0$ whose completion $\widehat{R}$ contains an element $x\not\in R$ such that $x^p\in R$. Such a ring cannot be N-2.
answered Mar 28, 2011 at 7:56
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$\begingroup$ I dont understand why such a ring cannot be n-2 furthermore why must we look at the completion to find such an x i mean why not take x= y^(1/p) where y $in$ R $\endgroup$ Commented Apr 3, 2011 at 23:19
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$\begingroup$ A discrete valuation ring $R$ has finite integral closure $S$ in the finite extension $L$ of the fraction field $K=\mathrm{Frac}(R)$ if and only if: [ \sum\limits_{w|v}e(w|v)f(w|v)=[L:K] ] where $v$ is the valuation associated to $R$ and the sum is taken over all extensions of $v$ to $L$. For $L=K(x)$ with $x\in\widehat{R}\setminus R$ and $x^p\in R$ the following facts are known: there is only one extension $w$ of $v$ to $L$ ($L/K$ is purely inseparable). The equations $e(w|v)=f(w|v)=1$ holds, because the completion is unramified over $R$ and has the same residue field as $R$. $\endgroup$– HagenCommented Apr 6, 2011 at 8:08