Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.