# Is it possible that $L$ is not totally ramified over $K$?

Let $$L$$ be a totally ramified extension of the $$p$$-adic field $$\mathbb{Q}_p$$.

Then there is an algebraic number $$\alpha$$ which is a root of an Eisenstein polynomial so that $$L=\mathbb{Q}_p(\alpha).$$

Assume an intermediate field $$K$$ i.e., satisfying $$\mathbb{Q}_p \subset K \subset L$$.

Is it possible that $$L$$ is totally ramified over $$\mathbb{Q}_p$$ but not over $$K$$ ?

If we assume $$K$$ is a totally ramified extension of $$\mathbb{Q}_p$$, does it help ?

In that case $$K$$ is totally ramified at every prime of $$\mathbb{Q}_p$$. On the other hand, $$L$$ is totally ramified at every prime of $$\mathbb{Q}_p$$.

• Here totally ramified means having the same residue field as $\Bbb{Q}_p$ Commented Dec 17, 2021 at 19:45

We have $$f_{L \mid \mathbb{Q}_p} = f_{L \mid K} \cdot f_{K \mid \mathbb{Q}_p},$$ and $$L \mid \mathbb{Q}_p$$ being totally ramified means by definition that $$f_{L \mid \mathbb{Q}_p} = 1$$. Since the $$f$$'s are positive integers, this is possible only if $$f_{L \mid K} =1$$, thus if $$L$$ is totally ramified over $$K$$.