Orders in extension of $Q_p$

Question

Suppose we have field extensions $L/K/\mathbb{Q}_p$. If $R$ is a ring such that $$\mathcal{O}_K \subset R \subset \mathcal{O}_L$$ (where if $M$ is an extension of $\mathbb{Q}_p$ then $\mathcal{O}_M$ is the integral closure of $\mathbb{Z}_p$ in $M$) and $R$ has rank $[L : K]$ as an $\mathcal{O}_K$-module, then can we say that $R = \mathcal{O}_L$? In the number field case, this isn't true, but I'm wondering if it is true in the local case.

Answer 1

The ring of integers of $ \mathbf Q_2(\sqrt{5}) $ is not $ \mathbf Z_2[\sqrt{5}] $, it's $ \mathbf Z_2[(1 + \sqrt{5})/2] $, so that's a counterexample to your claim.

In general, you can turn global counterexamples to this into local ones by recalling that the discriminant captures the ramification information of a number ring, so if you pick an order lying inside it whose discriminant includes a prime factor $ p $ which is unramified in the actual number ring, then you'll get a counterexample after taking completions of the rings at $ p $. In the example I gave, the discriminant of $ \mathbf Z[(1 + \sqrt{5})/2] $ is $ 5 $, but the discriminant of the order $ \mathbf Z[\sqrt 5] $ is $ 20 = 2^2 \cdot 5 $, so you get a counterexample after completing at the prime $ p = 2 $.