# Generic Rank of R^{1/p}

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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Is $R$ noetherian domain? – BCnrd Sep 9 '10 at 14:20
Yup. R is a Noetherian local domain. – Kevin Sep 9 '10 at 14:46
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. – Kevin Sep 9 '10 at 15:01
@Kevin: did you check Kunz's paper on Noetherian rings of char p? – Hailong Dao Sep 9 '10 at 21:30
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! – Kevin Sep 9 '10 at 22:51