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Let $R$ be a regular local ring of dimension $d$ and prime characteristic $p$, with maximal ideal $m = (x_1, \cdots, x_d)$. Let also $(f) \subsetneq R$ be a principal ideal in R, and $e \in \mathbb{N}$ be some integer. We denote $m^{[p^e]}$ for $(x_1^{p^e}, \cdots, x_d^{p^e})$.

I need to show that $\lim\limits_{e \to \infty} \frac{1}{p^{ed}} l(R/(f+m^{[p^e]})) = 0$. In other words, that $l(R/(f+m^{[p^e]}))$ grows "slower" than $p^{ed}$ as $e$ grows. (Here $l(\cdots)$ denotes the length as a module).

I tried many things but I'm stucked... Can anyone help ?

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    using $\dim R/(f) = d-1 < d$. – 2017-01-05
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    Thank you for answering but I don't know how to use it here, because I'm working on length and not dimensions. – 2017-01-05
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    Since $\dim R/(f) = d-1$ so we have the limit $\lim_{e \to infty} \frac{1}{p^{e(d-1)}} \ell (R/(f+ m^{[p^e]}))$ is just the Hilbert-Kunz multiplicity of $R/(f)$ (see this survey https://arxiv.org/pdf/1409.0467.pdf). Thus we have your question. – 2017-01-05
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    I thank you a lot for your answer that seems to end my problem ! However, I didn't work on Hilbert-Kunz multiplicity a lot (I could cite an article though). Do you have an explicit definition for it ? I tried to look at arxiv.org/pdf/1409.0467.pdf but no explicit definition seems to be given. I just need to ensure that my quantity is indeed $HK(R/(f))$. Thank you – 2017-01-05
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    Actually, doing what you said, I'm getting on one hand : $\lim\limits_{e \to \infty} \frac{1}{p^{e(d-1)}} l(R/(f+m^{[p^e]})) = \lim\limits_{e \to \infty} \frac{1}{p^{e(d-1)}} l(\frac{R/(f)}{n^{[p^e]}})$ where n is the maximal ideal of $R/(f)$ of dim $d-1$. Thus if $R/(f)$ is regular (which I don't know) I can use Kunz's Theorem and this is equal to $l(R/(f))$. On the other hand, I have $e_{HK} (R/(f)) = \lim\limits_{e \to \infty} \frac{1}{p^{e(d-1)}} l(R/(f)^{[p^e]})$. I don't see how it is equal to $l(R/(f))$, and why $l(R/(f))$ is regular... – 2017-01-06

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For any $n$, it is easy to see that $\mathfrak{m}^{nd} \subseteq (x_1^n, \ldots, x_d^n) $, where $\mathfrak{m} = (x_1, \ldots, x_d)$. Thus $$\ell(R/(f + \mathfrak{m}^{[p^e]})) \le \ell (R/ f + \mathfrak{m}^{d p^e}) = \ell (R'/ \mathfrak{m}^{d p^e}),$$ where $R' = R/(f)$ is a local ring of dimension $d-1$. Since $\ell (R'/ \mathfrak{m}^{d p^e})$ is the Hilbert polynomial of $R'$ for $e \gg 0$. This polynomial can be press as a polynomial of $p^e$ of degree $d-1$. So we can prove your question easily.

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    This looks really clear, thank you Pham Hung Quy ! Last thing that I need : How do you know that $R/(f)$ is also regular ? Assuming $(f)$ is a random principal ideal of $R$. – 2017-01-06