Eisenstein polynomial of totally ramified extension over $p$-adic field

Let $$p\geq 3$$ be a prime number, $$K$$ be a finite extension of $$\mathbb{Q}_p$$ with no non-trivial unramified subextension, $$f(x)$$ be an irreducible monic polynomial in $$\mathcal{O}_K[x]$$, making $$L=K[x]/(f(x))$$ a Galois totally and wildly ramified extension of $$K$$. How can we find an Eisenstein $$E(x)\in\mathcal{O}_K[x]$$ that generates $$L$$, i.e. $$L=K[x]/(E(x))$$? As far as I know, there are two ways to do this:

1. When the MAGMA system factorizing a polynomial $$f$$ over $$p$$-adic field $$K$$, it has an option to generate the field $$K[x]/(g(x))$$ for every irreducible factor $$g$$ of $$f$$ (thanks for Professor T. Dokchitser telling me this). This field is represented as a totally ramified extension over an unramified extensin over $$K$$, where the totally ramified part is represented as an Eisenstein polynomial. The factorization relies on S. Pauli's Round Four algorithm, which provides some additional arithmetic information of $$g$$ to produce the corresponding Eisenstein polynomial. However, this method (for generating Eisenstein polynomial) is not well documented. Professor Pauli claims that any Okutsu-Montes algorithm can be used to construct such an Eisenstein polynomial, but I don't know how to do it in detail.
2. The third-party SageMath package Henselization implemented by Julian Rüth can generate such an Eisenstein polynomial for the totally ramified extension. His code is based on the "limit valuation" theory by MacLane, which differs from the traditional implementation of $$p$$-adic arithmetic.

My question is: whether there exists a relatively self-contained algorithm to obtain this goal, or this question is essentially related to some complicated theories, like the Okutsu-Montes algorithm and MacLane's valuation theory?

I'd like to view this question as an algebraic number theory question rather than a computational one, since the time complexity is not the main concern here.

There is a super naive algorithm, assuming that you know some uniformizer $$\pi_K$$:

Let $$\alpha\in L$$ be a root of $$f$$ and $$N=\deg(f),M=[K:\Bbb{Q}_p]$$. Normalize the valuation such that $$v(\pi_K)=N$$.

Try each $$d\ge 1$$ and each element $$\bar{b}$$ of $$(\Bbb{Z}[\pi_K]/(p^d,\pi_K^{dM}))^N$$, if $$d$$ is large enough then there will be one $$\bar{b}$$ such that $$v(\sum_{n=0}^{N-1} \bar{b}_n \alpha^n)=v(p^e)+1$$ with $$e< d$$.

ie. taking any $$b\in \Bbb{Z}[\pi_K]^N$$ such that $$b=\bar{b}\bmod (p^d,\pi_K^{md})$$ you'll have that $$\det(x-\frac{\sum_{n=0}^{N-1} b_n \alpha^n}{p^e}\in End_K(L))$$ is the monic degree $$N$$ Eisenstein polynomial in $$O_K[x]$$ you are asking for.

Of course $$\pi_K$$ can be found with the same algorithm in the first place.

Note that knowing $$\pi_K$$ is useful mainly for checking if $$h=\det(x-\frac{\sum_{n=0}^{N-1} b_n \alpha^n}{p^e}\in End_K(L))$$ is Eisenstein. From any basis of $$K/\Bbb{Q}_p$$, for $$d>l$$ large enough you can enumerate the elements of $$p^l O_K/p^d O_K$$, and you can check instead if $$\det(x-\frac{\sum_{n=0}^{N-1} b_n \alpha^n}{p^e}\in End_{\Bbb{Q}_p}(L))$$ is Eisenstein, if it is then $$\det(x-\frac{\sum_{n=0}^{N-1} b_n \alpha^n}{p^e}\in End_K(L))$$ will be Eisenstein as well.

• So basically it's a brute-force search? Jan 29 at 6:20