Assume I am given a finite primitive extension of complete discretely valued fields $L=K(\alpha)/K$, say with monic integral minimal polynomial $f$ for $\alpha$.

How does one systematically compute the ramification and inertia indices $e$ and $f$?

Do I know that the ring of integers $A_L$ is automatically generated over $A_K$ by $\alpha$? If so, then I can reduce $f$ to obtain the extension of residue fields, and compute the inertia that way. I know that such a generator exists, but it isn’t clear the $\alpha$ I started with works.

I am mostly concerned with finite extensions of the $p$-adics. It seems like this local theory should be computationally straightforward, but a systematic procedure for computing the ramification eludes me.

My thoughts:

Ramification is known if you know a uniformizer, or know the valuation on $L$ that extends the valuation on $K$, but both of these require the ramification index to compute.

In the Galois case, the filtration by higher ramification groups gives some restriction on what subgroups of $G$ can be the inertia group, but this data is insufficient in general.

I know that I can express $L$ as adjoining a root of an Eisenstein polynomial if and only if the extension is totally ramified. But this feels like a limited trick, as the substitutions required to get an Eisenstein polynomial (and thus a uniformizer upstairs) are not usually obvious. More generally, the Newton polygon can give you some bounds on ramification, but I don’t see how it can tell you the exact value of ramification in general.


1 Answer 1


Given $K/Q_p$ a finite extension and $f\in O_K[x]$ monic irreducible and $L=K[x]/(f)$.

With the size of the residue fields $q=|O_K/(\pi_K)|,q^d=|O_L/(\pi_L)|$ then $\zeta_{q^d-1}\in L$ and $L/K(\zeta_{q^d-1})$ is totally ramified of degree $e=[L:K]/d$.

Let $m=\deg(f)$.

It means that the first step is to factorize your polynomial in $K(\zeta_{q^m-1})$.

Concretely find $r$ such that $f\in O_K/(\pi_K^r)[x]$ is separable and set $R=(\deg(f)r [K:\Bbb{Q}_p])!$ (I never recall the optimal constant here), then (extended) Hensel lemma holds in $O_K[\zeta_{q^m-1}]/(\pi_K^R)$, following the same gradient descent algorithm.

It means that you can factorize in the finite ring $$f=\prod_{j=1}^D f_j\in O_K[\zeta_{q^m-1}]/(\pi_K^R)[x]_{monic}$$ the factorization will lift uniquely to $O_K[\zeta_{q^m-1}][x]$ and the lifts are irreducible.

Then the ramification index is $e=[L:K]/d=\deg(f_j)$ and $d= [L:K]/\deg(f_j)=D$.

If $p\nmid e$ then $L/K$ is tamely ramified and $L=K(\zeta_{q^d-1},(\zeta_{q^d-1}^l\pi_K)^{1/e})$. Otherwise the extension $L/K$ is wildly ramified and finding the uniformizer $\pi_L$ (its $K(\zeta_{q^d-1})$ minimal polynomial $h$) is more complicated. Finally $O_L=O_K[\zeta_{q^d-1}+\pi_L]$ (show the RHS is complete and has the same uniformizer and residue field as $O_L$) and the $K$ minimal polynomial of $\zeta_{q^d-1}+\pi_L$ is $\prod_{i=1}^d h^{\sigma^i}(x-\sigma^i(\zeta_{q^d-1}))$.

  • $\begingroup$ Are you sure about choosing $\zeta_{q^m-1}$? For instance, I am worried slightly $L/K$ being an unramified extension of degree $p$. I would be inclined to pass to the maximal unramified extension of $K$ (or the unramified extension of degree $\deg f$ if one wants to keep it finite). $\endgroup$
    – user208649
    May 18, 2020 at 21:16
  • $\begingroup$ Sure I should have set $m=\deg(f)$ thanks $\endgroup$
    – reuns
    May 18, 2020 at 21:20

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