I know that the two norms: $p$ – adic norm and the usual norm ($\left| \cdot \right|$) defined on $\mathbb{Q}$ are not quivalent. This is clearly because the $p$ adic norms staisfies the strong triangle inequality while the usual norm does not. However, it contradicts a theorem that I have studied.
"On a finite dimensional vector space, any two norms are equivalent."
We know that a field over itself is a one – dimensional vector space and therefore, $\mathbb{Q}$ is a vector space. From the above statement $p$ adic norm and usual norm should be equivalent.
I know that there is something hiding behing the "$p$ – adic" norm, rather numbers. But I am not able to figure it out. I would like some help regarding the same.
Best Answer
The theorem you quote carries a bunch of implicit assumptions, more specifically, it only considers special kinds of norm. Specifically, the norms it concerns are norms $\|\cdot\|$ on a finite-dimensional vector space $V$ over real numbers, which further satisfies the homogeneity property $\|av\|=|a|\|v\|$ for all $v\in V,a\in\mathbb R$.
As you see, the example of the standard and of the $p$-adic norm show that this is not true much more generally. In fact, both requirements are necessary: the norm must both be homogeneous (otherwise there are wild counterexamples even on $\mathbb R$) and the vector space must be over $\mathbb R$ (we can find homogeneous examples over $\mathbb Q$. Actually, some other fields will work in place of $\mathbb R$, for instance $p$-adic numbers $\mathbb Q_p$, but it's an exception more than a rule).