Let $K$ be a *complete* non-archimedean field. A norm on a *finite dimensional* vector space $V$
is a function $| \cdot | : V \to \mathbf{R}$ which satisfies the usual norm properties (with the non-archimedean triangle inequality). If the valuation of $K$ is discrete, one can prove that the norm has a very simple form: there exists a basis $e_i$ of $V$ as a $K$-vector space such that
$$
\left| \sum_i \lambda_i e_i \right| = \max |\lambda_i| |e_i|
$$
(we can freely choose the $|e_i|$ of course). One call these norms splittable.

When $K$ is no longer discrete, there should be counterexamples, but I cannot find anywhere in the literature. I am wondering if anyone has written a counter-example down anywhere. Of course, any example must be at least two-dimensional.

For example if one considers $V^{\leqq 1} \subset V$ the $\mathcal{O}_K$-submodule consisting of elements of norm $\leqq 1$, then this is torsion-free hence flat over the valuation ring $\mathcal{O}_K$. If this is finitely generated it is therefore free and even one sees that the rank must agree with the dimension of the vector space. This then implies that the norm splits (and gives a proof in the discrete valuation case), so one sees that this phenomenon has to do with the non-noetherianness of the ring of integers.