Let $K \supset \mathbb{Q}_p$ be the $p$-adic field with ring of integers $\mathcal{O}_K$ and $\pi_K$ be its uniformizer . Let $L$ be an unramified extension of $K$ of dgree $d$ and ring of integers $\mathcal{O}_L$. Let $\pi_L$ be an uniformizer of $\mathcal{O}_L$.

What is the cardinality of $\mathcal{O}_L/\pi_L^n \mathcal{O}_L$ for some $n \in \mathbb{N}$ ?

Since $L$ is an unramified extension of degree $d$, we have $[ \mathcal{O}_L/\pi_L \mathcal{O}_L:\mathcal{O}_K/\pi_K \mathcal{O}_K]=d$, where $q=| \mathcal{O}_K/\pi_K \mathcal{O}_K|$.

Therefore $|\mathcal{O}_L/\pi_L^n \mathcal{O}_L|=(q^d)^n=q^{dn}$.

Am I correct ?


1 Answer 1


Your final result seems correct, but your argument seems incomplete and contains at least one error.

Namely, if we abbreviate the residue fields as $l := \mathcal{O}_L/\pi_L \mathcal{O}_L$ and $k := \mathcal{O}_K/\pi_K \mathcal{O}_K$, then the extension being unramified of degree $d$ means more or less by definition that $[l:k]=\color{red}{d}$ (not $q^d$). Together with $q := card(k)$ this of course gives $card(l) = q^d$.

So that would solve the case $n=1$. To conclude for $n \ge 2$, I suggest induction, and using certain exact sequences. But maybe some other method would work too. Can you fill in these details?

  • $\begingroup$ Yes, it was an error. I corrected it. Thanks $\endgroup$
    – MAS
    Aug 26, 2021 at 16:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .