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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.

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    $\begingroup$ 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 Liu
    Commented Oct 23, 2017 at 11:16
  • $\begingroup$ @WilleLiou It would be helpful if you kindly explain it more. $\endgroup$
    – Cusp
    Commented Oct 23, 2017 at 12:17
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    $\begingroup$ 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 Liu
    Commented Oct 23, 2017 at 12:56

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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 $3$.

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  • $\begingroup$ it should be regular local ring. I have edited. $\endgroup$
    – Cusp
    Commented Oct 23, 2017 at 11:00

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