# Computing ramification in extension of complete DVRs

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.

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 $$\prod_{i=1}^d h^{\sigma^i}(x-\sigma^i(\zeta_{q^d-1}))$$.

• 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).
– user208649
May 18, 2020 at 21:16
• Sure I should have set $m=\deg(f)$ thanks May 18, 2020 at 21:20