Let $(R,m)$ be a regular local ring of dimension $d$ and char $p>0.$ Let $F^e:R\longrightarrow R$ defined by $r\longrightarrow r^{p^e}$be the Frobenius map.

How to compute $l(R/m^{[p^e]})?.$

I know the answer is $p^{ed}$ but I do not know how to prove it.

  • 1
    $\begingroup$ Take the completion w.r.t. the maximal ideal and use the Cohen structure theorem, which reduces the task to formal power series rings, for which the result is evident. $\endgroup$ – Wille Liou Oct 23 '17 at 11:16
  • $\begingroup$ @WilleLiou It would be helpful if you kindly explain it more. $\endgroup$ – Cusp Oct 23 '17 at 12:17
  • 2
    $\begingroup$ Let $\hat R$ denote the completion of $R$. Then for all $n\in \mathbf{N}$ we have $R/\mathfrak{m}^n\cong \hat R/\hat{\mathfrak{m}}^n$. We know that $\mathfrak{m}^{dp^e}\subseteq \mathfrak{m}^{[p^e]}$ and $\hat{\mathfrak m}^{dp^e}\subseteq \hat{\mathfrak{m}}^{[p^e]}$. The image of $\mathfrak{m}^{[p^e]}$ in $R/\mathfrak{m}^{dp^e}$ is identified with the image of $\hat{\mathfrak{m}}^{[p^e]}$ in $\hat R/\hat{\mathfrak{m}}^{dp^e}$. It amounts thus to determine the length for $(\hat R, \hat{\mathfrak{m}})$. Finally, the Cohen structure theorem implies $\hat R\cong (R/\mathfrak{m})[[x_1, ..., x_d]]$. $\endgroup$ – Wille Liou Oct 23 '17 at 12:56

This is false. The noetherian local ring $R = \mathbb{F}_3[[X,Y]]/(Y^2 - X^3)$ has dimension one, and if $x,y$ are the images of $X,Y$ in $R$ then the sequence $$ R \supseteq (x,y) \supseteq (x^2,y) \supseteq (y) \supseteq (y^2) = \mathfrak{m}^{[3]} $$ shows that $R / \mathfrak{m}^{[3]}$ does not have $R$-length $3$.

  • $\begingroup$ it should be regular local ring. I have edited. $\endgroup$ – Cusp Oct 23 '17 at 11:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.