Can a generator of the ring of integers of local fields can be chosen so that it is also a uniformizer at the same time?

Question

Let $L/K$ be an extension of local fields. We can find $\alpha$ such that $\mathcal{O}_L=\mathcal{O}_K[\alpha]$. What do we know about this generating element? I think that this $\alpha$ can be selected in such a way that, in addition to the above property, it is also a uniformizing parameter at the same time. But I can not prove it.

Answer 1

If you look at a proof that $\mathcal O_L = \mathcal O_K[\alpha]$ for some $\alpha$, such as Prop. 3, Chapter III, Section 1 in Lang's "Algebraic Number Theory" then you should be able to see that you can take as $\alpha$ either $\zeta$ or $\zeta + \pi$ where $\zeta$ is an arbitrary generator (modulo the maximal ideal of $\mathcal O_L$) of the residue field extension and $\pi$ is an arbitrary choice of uniformizer. When $L/K$ is unramified we can take as $\pi$ a prime from $K$ and thus $\mathcal O_L = \mathcal O_K[\zeta + \pi] = \mathcal O_K[\zeta]$. When $L/K$ is totally ramified, we can use $\zeta = 1$ (or $\zeta = 0$), so $\mathcal O_L = \mathcal O_K[\pi]$.

Lang proves the result for an extension of DVRs with a separable residue field extension; no assumption of completeness, so in particular no use of Hensel's lemma. But the assumption that the DVR downstairs has an integral closure in the extension of its fraction field that is also a DVR is not at all typical in a "global" situation such as the localization of a Dedekind domain, as there can be more than one prime upstairs lying over the prime downstairs. In the complete setting things are nicer since every valuation ring of a local field is a DVR.

Answer 2

$O_L=O_K[\pi_L]$ iff $f(L/K)=1$ and $O_L=O_K[\zeta_{q_L-1}]$ iff $e(L/K)=1$. In general it is $$O_L=O_K[\zeta_{q_L-1}+\pi_L]$$

Hensel lemma is needed to construct the root unity, then (in the non-trivial case $e\ne 1,f\ne 1$) we need a closed-ness and density argument, investigating $\zeta_{q_L-1}+\pi_L-(\zeta_{q_L-1}+\pi_L)^q$

Answer 3

Hint: If $\mathcal{O}_L = \mathcal{O}_K[\alpha]$ with $\alpha$ a uniformiser of $\mathcal{O}_L$, then the canonical inclusion of their residue fields $\mathcal{O}_K/(\pi_K) \hookrightarrow \mathcal{O}_L/(\alpha)$ is surjective.

Conversely, if the extension is totally ramified (which is what the above says), then ...

reuns' answer says the same and more.