0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ It does not satisfy the property (i) either. $\endgroup$ – Mindlack Jan 17 at 7:58
  • $\begingroup$ @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? $\endgroup$ – M. A. SARKAR Jan 17 at 8:06
  • 1
    $\begingroup$ Unless I am deeply mistaken, $\|v_2\|_{sup}\|v_2\|_{sup}=1 \cdot 1=1$, but $\|v_2v_2\|_{sup}=\|pv_1\|_{sup}=p^{-1} \neq 1$. $\endgroup$ – Mindlack Jan 17 at 9:24
  • $\begingroup$ @Mindlack, very nice $\endgroup$ – M. A. SARKAR Jan 17 at 9:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.