Suppose $R$ is a local Noetherian domain of dimension $d$ in characteristic $p>0$. Suppose $R^{1/p}$ is a finitely generated $R$-module, and suppose $k$ is the residue field of $R$. Is the generic or torsion free rank of $R^{1/p}$ (i.e. the rank of this module after tensoring up to the fraction field) always equal to $[k:k^p] \cdot p^d$ (which is true at least when $R$ is complete)? What if, in addition, the completion of $R$ along its maximal ideal is also known to be a domain?
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$\begingroup$ Is $R$ noetherian domain? $\endgroup$– BCnrdCommented Sep 9, 2010 at 14:20
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$\begingroup$ Yup. R is a Noetherian local domain. $\endgroup$– KevinCommented Sep 9, 2010 at 14:46
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$\begingroup$ For what it is worth, I should remark that the requirement that $R^{1/p}$ is a finitely generated $R$-module automatically implies that $R$ is excellent. $\endgroup$– KevinCommented Sep 9, 2010 at 15:01
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1$\begingroup$ @Kevin: did you check Kunz's paper on Noetherian rings of char p? $\endgroup$– Hailong DaoCommented Sep 9, 2010 at 21:30
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$\begingroup$ I have now ... the answer to the question is yes, and it follows from Proposition 2.3 in Kunz's paper "On Noetherian rings of characteristic p" as suggested. Thanks! $\endgroup$– KevinCommented Sep 9, 2010 at 22:51
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