Skip to main content
Clarified that $R$ is a Noetherian local domain.
Source Link
Kevin
  • 103
  • 3

Suppose $R$ is a local ringNoetherian 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?

Suppose $R$ is a local ring 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)?

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?

Source Link
Kevin
  • 103
  • 3

Generic Rank of R^{1/p}

Suppose $R$ is a local ring 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)?