# How does it conclude that sup-norm is not a field norm?

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.

• @Mindlack, How? I think $||pv||_{sup}=||v_2||_{sup} \cdot ||v_2||_{sup}=1$ while $||pv_1||_{sup}=|p|_p=\frac{1}{p}$. But then what? Can you finish it? – M. A. SARKAR Jan 17 at 8:06
• Unless I am deeply mistaken, $\|v_2\|_{sup}\|v_2\|_{sup}=1 \cdot 1=1$, but $\|v_2v_2\|_{sup}=\|pv_1\|_{sup}=p^{-1} \neq 1$. – Mindlack Jan 17 at 9:24