My tutor recently mentioned the following result:

"For any complete discretely valued extension $K$ of $\mathbb{Q}_p$ with perfect residue field $k$, the ring of integers $\mathcal{O}_K$ can be written as a quotient of $W(k)[[T]]$ by a principal ideal generated by an Eisenstein polynomial"

I have tried looking at various books on local fields but this has not helped. Many hours of Internet search have not helped either. Could someone please tell me where I could find a proof of this fact?


When $k$ is a finite field (ie. $K/\Bbb{Q}_p$ is algebraic) then $W(k)$ is a fancy way to say $\Bbb{Z}_p[\zeta_{q-1}]$ where $q=|k|$, the ring of integers of the largest unramified extension $\subset K$.

$\zeta_{q-1}\in K$ by Hensel lemma and $$O_K=\Bbb{Z}_p[\zeta_{q-1}][[\pi_K]]=\Bbb{Z}_p[\zeta_{q-1},\pi_K]\cong \Bbb{Z}_p[\zeta_{q-1}][T]/(f(T))$$ where $\pi_K$ is a generator of the maximal ideal of $O_K$ and $f(T)\in \Bbb{Z}_p[\zeta_{q-1}][T]$ is its minimal polynomial, which is Eisenstein, because $\pi_K^e = pu$ for some $e$ and $u\in O_K^\times$,

from which $\deg(f)= [K:\Bbb{Q}_p[\zeta_{q-1}]]=e$, which follows from $$O_K=\{ \sum_{j\ge 0} c_j \pi_K^j, c_j\in 0\cup \{\zeta_{q-1}^l\}\} = \sum_{j=0}^{e-1}\pi_K^j \Bbb{Z}_p[\zeta_{q-1}]$$

When $k$ is a more general residue field then the idea is the same: by Hensel lemma the unramified polynomials have roots in $K$ so there is a largest unramified extension $L$ such that $O_L/(p)=k$ and $O_K=O_L[[\pi_K]]\cong O_L[[T]]/(f(T))$.

  • $\begingroup$ Thank you! I want to include a reference for this. Do you know any textbook where I could find this? $\endgroup$ – marlasca23 2 days ago
  • $\begingroup$ I never opened a book of $p$-adic number theory. $\endgroup$ – reuns 2 days ago
  • $\begingroup$ Also, why do you have $\mathcal{O}_K = \mathbb{Z}_p[\zeta_{q-1},\pi_K]$? $\endgroup$ – marlasca23 2 days ago

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.