# Non-equivalent norms on finite dimensional vector spaces over a non-complete field

It is widely known that on a finite-dimensional vector space over a complete field, every norm is equivalent. However, I'm trying (and failing) to find a counterexample over a field which is not complete.

My first try was 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 $(\mathbb{Q},|\cdot |)$.

I have tried to find such norms in $\mathbb{Q}^2$ and other (not complete) fields to no avail. I'm content with any example, but it would be great if it's accompanied by the proof of their non-equivalence.

• Maybe it is more helpful if you regard an easier field like $\mathbb{Z}/p\mathbb{Z}$ for $p \in \mathbb{P}$ then you can choose an adequate topology. – Nathanael Skrepek Mar 13 '17 at 22:29
• Try taking $V=\Bbb Q[x]/(f)$, where $f$ is an irreducible polynomial where one root $\alpha$ of $f$ is in $\Bbb C\setminus\Bbb R$ and one root $\beta$ is in $\Bbb R$. Then $\Bbb Q[x]/(f)\cong\Bbb Q(\alpha)\cong\Bbb Q(\beta)$ as fields and therefore as $\Bbb Q$-vector spaces (even $(\Bbb Q,\left|\cdot\right|)$-vector spaces), but if the topologies are different (e.g. the one on $\Bbb Q(\beta)$ is the order topology, and the one on $\Bbb Q(\alpha)$ cannot be ordered), the norms cannot be equivalent. – Stahl Mar 13 '17 at 22:30
• Do you require that the norms take only values in the same field that the vector space is over? I thought of a $2$-dimensional example with a vector space over the field $\mathbb Q$ but one of my norms takes real values that are not all in $\mathbb Q.$ – DanielWainfleet Mar 14 '17 at 8:19
• I'm defining a norm as a real-valued function so that's not a problem :) – Garoal Mar 14 '17 at 9:05