It is widely known that on a finite-dimensional vector space over a complete field, every norm is equivalent. However, I'm looking for a counterexample over a field which is not complete.


I have found no such counterexample neither by myself nor in the internet, but it is frequently stated that such result does not hold for non-complete fields, without a proof.


**Edit:** The stackexchange app notified me of a comment which I believe is now deleted. Still, I will address its concern in case somebody has the same idea.


I already tried to treat $\mathbb{Q}$ as a vector space over itself and find two non-equivalent norms. As for Ostrowski's theorem we know that, for example, the absolute value and the 2-adic norm are not equivalent. However, I noticed that the $p$-adic norm, though being a norm on the field $\mathbb{Q}$, it is not a norm on the vector space $(\mathbb{Q},|\cdot|)$, and thus is not a valid counterexample.



Non-equivalent norms on finite dimensional vector spaces over a non-complete field
Field $\mathbb Q$ with the usual absolute value $|\cdot|$ from the real numbers.


Two norms on $\mathbb Q^2$ ...
$$
\|(x,y)\|\_1 = |x|+|y|
$$
and
$$
\|(x,y)\|\_2 = \left|\,x+\sqrt{2}\;y\,\right|
$$
In both of these, $|\cdot|$ is still the usual absolute value for the real numbers.


