In the book of $\text{Neal Koblitz}$ on p-adic numbers, p-adic analysis and zeta-function, the following exercise is given:
Exercise: Let $V=\mathbb{Q}_p(\sqrt p)$, $ \ v_1=1, \ v_2=\sqrt p \ $. Here $V=\mathbb{Q}_p(\sqrt p)$ is a vector space over the p-adic field $\mathbb{Q}_p \ $. Show that $\text{sup-norm}$ is not a field norm.
The purpose of this example is to show that a vector space norm may not be a field norm.
In the hintz of this book it is given that-
$v_2 \cdot v_2=pv_1$, but $||v_2||_{sup} \cdot ||v_2||_{sup}=1$, $||pv_1||_{sup}=|p|_p=\frac{1}{p}$.
That is all given in the hintz.
How does it conclude that $sup-norm$ is not a field norm ?
Now if $K$ is a finite extension of the field $F$, then the properties of the field norm $||.||$ is given below:
$(i) \ ||xy||=||x||||y||, \ x,y \in K$,
$(ii) \ ||ax||=a^n ||x||, \ a \in K, \ \forall x \in L, \ n=[K:F]$.
Here probably property $(ii)$ does not hold. Because,
$a=p \in V=\mathbb{Q}_p(\sqrt p)$ and $v_1=1 \in \mathbb{Q}_p$ but $||pv_1||_{sup}=\frac{1}{p} \neq p^{[\mathbb{Q}_p(\sqrt p): \mathbb{Q}_p]} ||v_1||_{sup}=p^2||v||_{sup}.$
So the $sup-norm$ does not satisfy the field norm property $(ii)$.
But I am not sure.
Can you please help me?